Dynamics - ETH Z

Institute for Mechanical SystemsChair in Nonlinear Dynamics

Prof. George Haller

Dynamics

Institute for Mechanical SystemsETH Zürich

Based on the lecture notes of Prof. George HallerEdited by:Lucas Liebenwein, David Öttinger,Shobhit Jain, Sten Ponsioen,Dr. Paolo Tiso

Version from October 5, 2017

Contents

Nomenclature iii

0 Introduction and Newton’s three axioms 3

I Dynamics of a single particle 7I.1 Linear momentum principle . . . . . . . . . . . . . . . . . . . . . . . . . . 8I.2 Angular momentum principle . . . . . . . . . . . . . . . . . . . . . . . . . 9I.3 Work-energy principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

I.3.1 Work-energy principle for conservative systems . . . . . . . . . . . 13

II Examples of single-particle systems 21

III Applications of single-particle dynamics 31

IV Particle motion in nonstandard coordinates 47IV.1 Particle motion in curvilinear coordinates . . . . . . . . . . . . . . . . . . 47IV.2 Particle motion in non-inertial frames . . . . . . . . . . . . . . . . . . . . 51

V Dynamics of a system of particles I: Set-up 61V.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62V.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

V.2.1 Holonomic constraints . . . . . . . . . . . . . . . . . . . . . . . . . 66V.2.2 Forces and constraints . . . . . . . . . . . . . . . . . . . . . . . . . 66

V.3 Degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67V.4 Center of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

VI Dynamics of systems of particles II: Momentum & work-energy prin-ciples 73VI.1 Linear momentum principle . . . . . . . . . . . . . . . . . . . . . . . . . . 73VI.2 Angular momentum principle . . . . . . . . . . . . . . . . . . . . . . . . . 74VI.3 Work-energy principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

VI.3.1 Work-energy principle for rigid body systems . . . . . . . . . . . . 77VI.3.2 Work-energy principle for conservative systems . . . . . . . . . . . 79

VII Examples of particle system dynamics 83

VIIIDynamics of rigid bodies 93VIII.1Rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93VIII.2Velocity transfer formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

VIII.2.1 Relative velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

i

VIII.2.2 Angular velocity of a rigid body . . . . . . . . . . . . . . . . . . . 95VIII.2.3 Velocity transfer formula . . . . . . . . . . . . . . . . . . . . . . . 95

VIII.3Planar rigid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96VIII.3.1 Angular velocity for planar rigid body motion . . . . . . . . . . . 96VIII.3.2 Instantaneous center of rotation . . . . . . . . . . . . . . . . . . . 97

VIII.4Example problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98VIII.5Three-dimensional rotations . . . . . . . . . . . . . . . . . . . . . . . . . . 107

VIII.5.1 Angular velocity of time-dependent rotations . . . . . . . . . . . . 111VIII.6Uniqueness of the angular velocity if a rigid body . . . . . . . . . . . . . . 114

IX Kinetics of rigid bodies 119IX.1 Degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119IX.2 Linear momentum principle . . . . . . . . . . . . . . . . . . . . . . . . . . 124IX.3 Angular momentum principle . . . . . . . . . . . . . . . . . . . . . . . . . 127IX.4 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

IX.4.1 Calculation of the angular momentum when Point B is fixed orcoincident with the center of mass C . . . . . . . . . . . . . . . . . 128

IX.4.2 Calculation of the angular momentum when point B is arbitrary . 129IX.5 Properties of the moment of inertia tensor . . . . . . . . . . . . . . . . . . 131

IX.5.1 Two-dimensional motion . . . . . . . . . . . . . . . . . . . . . . . . 131IX.5.2 Additivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133IX.5.3 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134IX.5.4 Parallel axes theorem (Steiner’s theorem) . . . . . . . . . . . . . . 135IX.5.5 Principal axes of a body B . . . . . . . . . . . . . . . . . . . . . . 137IX.5.6 Tables with moment of inertia tensors . . . . . . . . . . . . . . . . 139

IX.6 Work-energy principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140IX.7 Rotating frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

IX.7.1 Rotation transformation . . . . . . . . . . . . . . . . . . . . . . . . 143IX.7.2 Differentiation of quantities in rotating frames . . . . . . . . . . . 145

X Examples of rigid-body dynamics in 2D 151

XI Examples of rigid body dynamics in 3D 173

XII Gyroscopes 185XII.1 Euler equations for gyroscopes . . . . . . . . . . . . . . . . . . . . . . . . 185

XIIIVibrations 197XIII.1One-DOF vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

XIII.1.1 Free vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200XIII.1.2 Forced vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

XIVMulti-DOF vibrations 213XIV.1Free, undamped vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . 214

XIV.1.1 Case 1: Purely imaginary pair of eigenvalues (λ = ±iωj) . . . . . . 216XIV.1.2 Case 2: A pair of zero eigenvalues (λ = 02) . . . . . . . . . . . . . 216XIV.1.3 General solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

XIV.2Modal decomposition of free, undamped vibrations . . . . . . . . . . . . . 221XIV.3Multi-DOF forced, damped vibrations . . . . . . . . . . . . . . . . . . . . 223

Nomenclature

Acronyms and abbreviations

AM angular momentumAMP angular momentum principleCM center of massdim dimensionsDOF degrees of freedomLM linear momentumLMP linear momentum principleODE ordinary differential equationPDE partial differential equationw.r.t. with respect to

Latin symbols

a acceleration [m/s2]A amplitude of particular solution [m]B rigid body [-]c damping coefficient [kg/s]C damping matrix [kg/s]dm infinitesimal mass element [kg]dA infinitesimal area element [m2]dV infinitesimal volume element [m3]D Lehr’s damping [-][ i, j, k] inertial frame [-][ e1, e2, e3] principal frame [-]E total mechanical energy [J]f

0normalized force vector [m/s2]

F resultant force [N]F ext resultant external force [N]F int resultant internal force [N]Fnp resultant non-potential force [N]F p resultant potential force [N]

iii

F 0 force vector [N]G gyroscopic matrix [kg/s]HB angular momentum w.r.t. B [kg m2/s]IB mass moment of inertia w.r.t. B [kg m2]I B mass moment of inertia tensor w.r.t. B [kg m2]K stiffness coefficient [N/m]K stiffness matrix [N/m]Kij force exerted by particle j on particle i [N]m mass of a particle [kg]M mass of a rigid body [kg]M mass matrix [kg]MB resultant torque w.r.t. B [N m]M ext

B resultant external torque w.r.t. B [N m]M int

B resultant internal torque w.r.t. B [N m]N follower force matrix [N/m]O(3) orthogonal group in 3D [-]P linear momentum [kg m/s]r position vector [m]rAB vector pointing from point A to point B [m]rAB relative velocity of point B w.r.t. point A [m/s]SO(3) special orthogonal group in 3D [-]T kinetic energy of a system [J]u eigenvector [-]U modal matrix [-]v velocity [m/s]vB/A relative velocity of point B w.r.t. point A [m/s]V potential energy of a system [J]V magnification factor [-]W12 work done on a system between state 1 and 2 [J]W ext

12 external work done on a system between state 1 and 2 [J]W int

12 internal work done on a system between state 1 and 2 [J]x deviation from stable equilibrium (dim = 1) [m]x deviation from stable equilibrium (dim≥ 2) [m]y modal coordinate vector [m]

Greek symbols

δ characteristic damping [1/s]η frequency ratio [-]λ eigenvalues [1/s2]ϕ phase shift [-]ρ density of a rigid body [kg/m3] or [kg/m2]%B

vector from reference point B to generic position r [m]ω angular velocity of rigid body [rad/s]ω forcing frequency [rad/s]ω0 natural frequency [rad/s]Ω angular velocity matrix [rad/s]

Indices

B reference pointC center of massh homogeneous solution to a linear ODEp particular solution to a linear ODE


Prof. George Haller

Dynamics

0. Introduction and Newton’s threeaxioms




Chapter 0

Introduction and Newton’s threeaxioms

In previous mechanics courses, you have learned about rigid bodies in equilibrium (Me-chanics I - Statics) and about deforming bodies (Mechanics II - Strength of Materials).In contrast, this course treats the dynamics of mechanical systems, i.e. how a system ofrigid bodies reacts when certain forces/moments are applied to it. Think, for example,about a car engine where the piston in each cylinder is displaced due to the forces arisingfrom combustion. Other common examples include the rotation of a shaft under externaltorques or the motion of a car. In summary, Dynamics investigates how rigid bodies moveunder the forces applied to them.

In the 17th and 18th century, Sir Isaac Newton (1642-1726) proposed a relation betweenthe motion of a particle and the forces exerted on it when he observed apples falling downa tree due to gravity. He summarized his observations in three axioms, commonly calledNewton’s three axioms or Newton’s laws. Based on these three axioms, Leonhard Euler,Joseph-Louis de Lagrange, William Rowan Hamilton and others extended Newton’s the-ory to today’s understanding of dynamics. These extended axioms cover systems of rigidbodies, not just particles. Newton’s original three axioms can be summarized as follows.

Newton’s three axioms

Axiom 1. A force-free particle is at rest or moves uniformly along a straight line.

Axiom 2. The resultant force F (t) acting on a particle is proportional to its acceler-ation a(t). The factor of proportionality is the mass m:

F (t) = ma(t) .

Axiom 3. Action and reaction forces are equal and act in opposite directions.

As Newton’s laws are based on a vectorial description of the system, a reference frame isneeded to evaluate them. Newton’s laws however, are not valid in an arbitrary referenceframe: we must select an inertial frame, as defined below. Consider e.g. , a candle attached

3

4

to a rotating platform, described w.r.t. to a frame rotating with the platform. In this frameof the platform, the flame of the candle will exhibit motions that cannot be explained byall known active forces into Newton’s 2nd axiom. The frame of the rotating platform is,therefore ,not an inertial reference frame.

Definition. An inertial frame is a reference frame in which Axioms 1-3 hold true.

In this course, we will start out by extending Newton’s laws to describe the motion ofparticles. Namely, we will derive the linear momentum principle, the angular momentumprinciple and the work-energy principle. A crucial step in a dynamics problem is to find theequations of motion that govern the system’s motion, which will generally form a nonlinearmulti-dimensional 2nd -order system of ordinary differential equations (ODEs), with timeacting as independent variable. First, single particles are treated in Chapter I, then thederived equations are generalized to an arbitrary number of particles in Chapter V.In the second part of the course, we will study the dynamics of a rigid body. A rigid bodyis characterized by the property that any two points on the body remain at a constantdistance from each other. This implies a lack of deformation and hence great simplificationto the underlying equations of motion. The kinematics of a rigid body are treated inChapter VIII and the kinetics of a rigid body in Chapter IX.In the last part of the course (see Chapter XIII), we introduce the concept of vibrations, i.e.small oscillations around a stable equilibrium. Since only small motions are considered,the equation of motion can be linearized, allowing us to use tools from linear algebra toexamine the characteristics of oscillations near the equilibrium.


Prof. George Haller

Dynamics

I. Dynamics of a single particleContents

I.1 Linear momentum principle . . . . . . . . . . . . . . . . . . . . . 8I.2 Angular momentum principle . . . . . . . . . . . . . . . . . . . . 9I.3 Work-energy principle . . . . . . . . . . . . . . . . . . . . . . . . 12

I.3.1 Work-energy principle for conservative systems . . . . . . . 13




Chapter I

Dynamics of a single particle

In this chapter, we begin by describing the motion of a single particle. We apply Newton’sthree axioms and find a relation between the forces applied to a particle and its motion.As a direct consequence of Newton’s 2nd law, we state the principle of linear momentumin Section I.1. Afterwards, we also state the implications Newton’s 2nd law for the angularmomentum (see Section I.2) and for the total mechanical energy (see Section I.3). InSection II, these principles are applied to various examples. In Section III, real-worldproblems are discussed that can be approached using single-particle dynamics. Finally,we will see how Newton’s principles can be applied in curvilinear coordinates and in non-inertial frames (see Section IV).

Figure1.1

z

y

x

0

r(t0)m

r(t)

F (t)

Figure I.1: A particle of mass m is shown moving along a trajectory r(t), subjected to theforce F (t). We describe the system w.r.t. the inertial frame of (x, y, z) coordinates alongthe unit vectors

[i, j, k

]. The vector r(t0) refers to the initial position of the particle.

Consider a single particle of mass m, as shown in Figure I.1, moving in time t along atrajectory r(t) ∈ R3, with velocity

v(t) =d

dtr(t) = r ,

and acceleration

a(t) =d2

dt2r(t) = r .

7

8 I.1. Linear momentum principle

All quantities are expressed w.r.t. an inertial frame[i, j, k

], i.e.

r(t) = x(t) i+ y(t) j + z(t) k ,

v(t) = x(t) i+ y(t) j + z(t) k ,

a(t) = x(t) i+ y(t) j + z(t) k .

Further, the particle is subjected to n forces, F i ∈ R3, i = 1, . . . , n. We refer to the sumof all forces as the resultant force F ,

F =n∑i=1

F i .

I.1 Linear momentum principle

To express Newton’s 2nd law in a more general form, we use Euler’s formulation, for whichwe first define the linear momentum.

Definition. The linear momentum of a particle is defined as

P := mv . (I.1)

When differentiating the linear momentum w.r.t. time, we obtain the linear momentumprinciple (LMP).

The principle of linear momentum (LMP) for a single particle:

P = F (I.2)

P . . . linear momentum of the particleF . . . resultant force acting on the particle

Note: Conservation of linear momentum: When F = 0, we obtain P = const. fromI.2. Therefore, linear momentum is conserved under vanishing resultant force.

Example I.1: Block sliding without frictionConsider a block of mass m that slides without friction, as shown in Figure I.2.The mass is subject to gravity and to a normal reaction force N from the ground.This reaction force must equal the force of gravity, because there is no verticaldisplacement. Thus, we find that the resultant force is equal to 0, i.e.

F = mg +N = 0 .

Hence, linear momentum is conserved and the velocity of the block remains constant.

Chapter I. Dynamics of a single particle 9

Figure1.2

m

N

mg

v

frictionless

Figure I.2: A block of mass m sliding without friction on a horizontal plane.

I.2 Angular momentum principle

We still consider a particle of mass m with trajectory r(t), as shown in Figure I.3. Wenow introduce an arbitrary, possibly moving point B with trajectory rB(t). The vector %

Bfrom the point B to the particle is

%B

(t) = r(t)− rB(t) . (I.3)

Furthermore, we define the angular momentum HB of the particle w.r.t. B and the resul-tant torque MB of the resultant force F w.r.t. B as follows:

Figure1.3

z

y

x

Fp = mv

r(t)

%B

(t)

B

m

rB(t)

O

Figure I.3: A particle of mass m is moving along a trajectory r(t). The reference point Bis arbitrary and may move along a path rB(t).

Definition. The angular momentum HB(t) of the particlem w.r.t a point B is definedas the cross product of the vector %

Bbetween the point B and the particle with the

linear momentum P ,

HB(t) := %B

(t)× P (t). (I.4)

Here, B can be an arbitrary point in space, either fixed or moving.

10 I.2. Angular momentum principle

Definition. A force F i acting on a particle m at position r(t) produces a torque ormoment M iB w.r.t. a point B of the form

M iB = %B× F i , (I.5)

where × denotes the cross product. The resultant torque MB w.r.t. B is then the sumof all torques w.r.t. B:

MB :=

n∑i=1

M iB =

n∑i=1

%B× F i = %

B×

n∑i=1

F i ,

which is equal to the torque of the resultant force F w.r.t. point B:

MB := %B× F . (I.6)

Similarly to the LMP, we can now relate the time derivative of the angular momentumHB to the resultant torque MB. We first differentiate the angular momentum HB in timeusing the product rule,

HB =d

dtHB =

d

dt

(%B× P

)= %

B× P + %

B× P .

Then we apply the linear momentum principle (I.2) to obtain

HB = %B× P + %

B× F .

Next, we replace %B× F by (I.6) and % by (I.3) to obtain

HB = %B× P + %

B× F

= %B× P +MB

= (r − rB)× P +MB

= (v − vB)× P +MB .

Furthermore, since the vector product of two parallel vectors vanishes,

v × P = v × (mv) = 0 ,

which simplifies the above equation to

HB = v × P − vB × P +MB ,

= −vB × P +MB .

This equation represents the angular momentum principle (AMP), derived originally byEuler.


The principle of angular momentum (AMP) for a single particle:

HB + vB × P = MB (I.7)

HB . . . angular momentum of the particle w.r.t. the point BvB . . . . velocity of the point BP . . . . . linear momentum of the particleMB . . . resultant torque w.r.t. the point B

Specifically, HB = MB if and only if either of the following two conditions is met:

• vB = 0 (B is fixed)

• vB ‖ vNote: Conservation of angular momentum: If MB = 0 and (vB = 0 or vB ‖ v), thenHB = 0 and hence HB = const.. Therefore, the angular momentum is conserved.

Example I.2: Mass in circular motionConsider a mass m, as shown in Figure I.4, that is constrained to move on thehorizontal (x, y)-plane. It is attached to a string of constant length |r| = const.,resulting in a circular motion. The other end of the string is fixed at B. Theposition of the mass is described by r and its velocity by r = v. We now show thatthe mass must move with constant speed |v| = const.

Figure1.4

Bstring

rm

vy

x

Figure I.4: A mass m, attached to a string, performs circular motion in the hori-zontal x, y-plane.

To show this, we apply the angular momentum principle w.r.t. the point B, i.e.

HB + vB × P = MB . (I.8)

Since B does not move (vB = 0) and no forces are exerted on the body (F = 0),there is no torque produced about B (MB = r × F = 0), the angular momentumHB = const. is conserved. This can also be seen from (I.8), when we plug inMB = 0 and vB = 0 to obtain HB = 0. Next, we compute the angular momentumHB using (I.4) as

HB = %B× P = r × (mv) = m |r| |v| k .

12 I.3. Work-energy principle

In this last step, we used the fact that both r and v lie on the (x, y)-plane, and,hence, their cross product must point in the k-direction perpendicular to the plane.Since HB = const., the speed |v| of the mass must also be constant. This, however,does not imply a constant velocity v 6= const. Rather, the velocity vector v(t)changes direction in time, always pointing in a direction tangent to the circularpath of the particle.

I.3 Work-energy principle

Suppose that a particle of mass m experiences a force F (t) as it moves along its path r(t)as shown in Figure I.5. Between the positions r1 = r(t1) and r2 = r(t2), the work W12

done by F (t) on the particle is

W12 =

∫ r2

r1

F · dr , (I.9)

where a · b indicates the scalar product between the vectors a and b.

Figure1.5

r1

r2

m

r(t)

F (t)

Figure I.5: A particle of mass m subjected to the force F (t) moving from position r1 tor2.

Substituting

dr = v dt

into (I.9), we obtain

W12 =

∫ t2

t1

F · v dt .

Using the linear momentum principle (I.2), we can also express the force F in terms of thevelocity v to obtain

W12 =

∫ t2

t1

d

dt(mv) · v dt ,

then evaluate the integral using the chain rule as

W12 =

∫ t2

t1

d

dt(1

2m v · v) dt

=1

2m |v2|2 −

1

2m |v1|2 . (I.10)

The term 12m |v|

2 is just the kinetic energy T of a particle, a measure of the amount ofwork one has to perform to accelerate the particle to the velocity v.


Definition. The kinetic energy T of a particle of mass m and velocity v is defined as

T :=1

2m |v|2 . (I.11)

From (I.10), we obtain that the change in kinetic energy (T2− T1) of a particle is equal tothe work W12 done by all forces on the particle (work-energy principle).

Work-energy principle for a single particle

W12 = T2 − T1 (I.12)

W12 . . . work done on the particle between r1 and r2

Ti . . . . kinetic energy of particle associated with velocity vi

I.3.1 Work-energy principle for conservative systems

To further simplify the work-energy principle (I.12), we now introduce the concept of apotential force.

Definition. A potential force (or conservative force) F i is a force that can be ex-pressed as the negative gradient of a scalar function Vi(r), called potential or (potentialfunction),

Fi = −∂Vi∂r

= −∇Vi(r) . (I.13)

Here ∇Vi(r) denotes the gradient of the function Vi(r), defined as

∇Vi(r) =(∂Vi∂x ,

∂Vi∂y ,

∂Vi∂z

)Tin the Cartesian coordinate system [x, y, z].

Note: For a given potential force F i, the potential Vi(r) is determined only up to a constantC since

Fi = −∇Vi(r) = −∇(Vi(r) + C

).

The constant C can be chosen arbitrarily, thus we can select the potential to be 0 at anydesired position r0.

Example I.3: Which force is conservative?We review here typical forces encountered in models of mechanical systems anddiscuss whether they are conservative, or not. Remember that potential functions


are only unique up to an arbitrary constant C.

(a ) Gravity

When a particle is subjected to gravity (see Figure I.6), a force

F = mg = −mg j

is exerted on the particle. This force is conservative with the corresponding poten-tial

V (r) = mgy .

Indeed, we have

−∇V (r) = 0 · i−mg · j = F .

Figure1.6

i

jg F = mg = mgj

m

Figure I.6: A particle of mass m subjected to gravity.

(b ) Linear spring force

A block of mass m attached to a (linear) spring (see Figure I.7), is subjectedto the spring force

F = −kx i

caused by the spring elongation.

Figure1.7

mk

x

F = kxi

mF

F F

Figure I.7: A block attached to a spring with stiffness k, and the correspondingfree-body diagrams

This force is also conservative with the corresponding potential function

V (r) =1

2kx2 .

Indeed, we can write

−∇V (r) = −kx i− 0 j = F .


(c ) Viscous damping force

A (linear) damper exerts the force

F = −cx i

on a particle (see Figure I.8). This force is not potential since we cannot find apotential V (r) such that

F (r) = −∇V (r) .

F = cxi

Figure1.7

m

x

mc F

FF

Figure I.8: A block attached to a damper with damping coefficient c. Also shownare the corresponding free-body diagrams

Indeed, the gradient of an r-dependent function V (r) cannot depend on r.


We now introduce the definition of conservative systems.

Definition. A conservative system is a system where all forces are either potential ordo no work.

To capture the effect of all potential forces acting on a system, we introduce the potentialenergy of a system.

Definition. The potential energy V (r) of a mechanical system is defined as the sumof the potentials Vi(r) of all potential forces F i acting on the system:

V (r) :=

n∑i=1

Vi(r) , i = 1, . . . , n . (I.14)

The resultant force

F p :=n∑i=1

F i

of all potential forces can then be expressed as

F p = − d

drV (r) = −∇V (r) . (I.15)

We now recall formula (I.10), which gives the work done by the resultant force F along acertain path:

W12 =

∫ r2

r1

F · dr .

We would like to evaluate this expression for a conservative system. The resultant force Fof any system can be split into a potential part F p and a non-potential part Fnp as

F = F p + Fnp ,

which can be inserted into the above expression for W12 to yield:

W12 =

∫ r2

r1

(F p + Fnp) · dr ,

=

∫ r2

r1

F p · dr +

∫ r2

r1

Fnp · dr .

By definition, in a conservative mechanical system, only potential forces do work. There-fore,

Fnp · dr = 0 ,


and hence

W12 =

∫ r2

r1

F p · dr .

Using the fact that the potential forces F p can be expressed with the potential energy V (r)of the system as

F p = − d

drV (r) ,

we can evaluate the work integral as

W12 =

∫ r2

r1

F dr =

∫ r2

r1

− d

drV dr

=

∫ r2

r1

d

dr(−V ) dr

= V (r1)− V (r2) = V1 − V2 . (I.16)

Combining the result (I.16) with (I.12) yields the work-energy principle for conservativesystems in the form

V1 − V2 = T2 − T1 .

Work-energy principle for a conservative systemThe total mechanical energy

E := T + V

of a conservative systems is conserved along all motions,

T1 + V1 = T2 + V2 (I.17)

or equivalently,

d

dtE(t) = 0 .

E . . . total mechanical energyTi . . . kinetic energy of the system in state iVi . . . potential energy of the system in state i

Note: (I.16) implies that, for a conservative system, the work done to move a particlefrom position r1 to r2 does not depend on the path taken. This is not true for non-conservative systems, for which the work done does depend on the path taken. Consider,e.g., a non-conservative system subjected to friction where any displacement results in aloss of energy.



Prof. George Haller

Dynamics

II. Examples of single-particlesystems




Chapter II

Examples of single-particle systems

Example II.1: String pulled slowly from belowA mass m moves on a plane while attached to a taut string, as shown in Figure II.1. Thestring runs through a hole B in the plane while the other end of the string is subjectedto a force F (t). The initial length l(0) of the planar section of the string and the initialvelocity v(0) of the mass m are given. Indicate with l(t) the length of the string at anarbitrary time t while the string is being pulled. What is the velocity |v(t)| of the mass mat that time?

Figure1.9

k

l(t)B

F (t)

r(t)

m

v(t)

F (t)

frictionless plane

Figure II.1: The mass m shown here is constrained to move on a frictionless plane whileattached to a string pulled downwards by a force F . Since the string is massless, we have|F | = |F |.

Solution:

First, we pick the reference point B ≡ 0 with velocity vB ≡ 0, then compute the mo-ment of the resultant force w.r.t. B as

MB(t) = r(t)× F (t) = 0 ,

because

r ‖ F .

Here r(t) denotes the vector between the hole B and the mass m, and hence |r(t)| = l(t).The torque of the gravitational force and the torque of the reaction force w.r.t. B cancel

21

22

each other out and hence do not contribute to the resultant torque MB. By the AMP, wetherefore obtain that HB = const. and hence conservation of angular momentum holds forall times. In the following, we denote the angle between the vectors r and v by α. We canwrite:

HB(0) = HB(t) ,

r(0)×mv0 = r(t)×mv(t) ,

l(0) |v(0)| sinα(0)︸︷︷︸=1 since r(0)⊥v0

k = l(t) |v(t)| sinα(t)︸︷︷︸=1 since r(t)⊥vt∀t

k ,

|v(t)| = l(0) |v(0)|l(t)

Thus, we can determine the velocity |v(t)| at time t using conservation of angular momen-tum.

Example II.2: Forced and damped pendulumIn Figure II.2 (a) a mass m is shown that is attached to a massless rod of length l, which

Figure1.10

B

l

mc

ex

ey

ez

ByB(t)

'

0

c'ex FB

K

mg

(a) (b)

Figure II.2: (a) A forced and damped pendulum is shown, (b) The free-body diagram.

is, in turn, hinged at B. We use the inertial reference frame[ex, ey, ez

], as indicated in

the figure, to describe the problem. The point B is forced to move along the path yB(t)in time along ey and a linear, angular damper with damping coefficient c acts on the rod.We call this mechanism a forced, damped pendulum. In the following, we are interested infinding the equation of motion governing the system’s motion.

(i) Free-body diagram (FBD)The free-body diagram of the system is shown in II.2 (b). The damper creates a freetorque T = −cϕ i. Further, the joint at B causes a reaction force FB, and, similarly,the connection between the rod and the mass creates a constraint force K = −FB. Thesystem is also subjected to gravity.

Chapter II. Examples of single-particle systems 23

(ii) Angular momentum principleSince we are interested in finding the equation of motion, but not the unknown forces FBand K, it is preferable to apply the momentum principles, in a way that does not engagethese forces. A handy choice is to apply the angular momentum principle w.r.t. the hingepoint B. In this case, neither K, nor FB produce a torque about B. The overall torqueMB about B is

MB = (−mgl sinϕ− cϕ) ex .

The angular momentum principle w.r.t. B states

HB + vB × P = MB . (II.1)

We now need to find all remaining quantities for the AMP. The linear momentum P ofthe system is

P = mv = mr = m(

(yB + l cosϕϕ)ey + l sinϕϕez

),

with the position vector

r = (yB + l sinϕ)ey − l cosϕez

found from the geometry. Next, we compute the angular momentum HB:

HB = %B× P = m

(l2ϕ+ l cosϕyB

)ex ,

and, similarly the time derivative HB (not detailed here). The last term to be computedis

vB × P = yBey ×m(

(yB + l cosϕϕ)ey + l sinϕϕez

)= ml sinϕyBϕex ,

where the velocity of B is vB = yBey. Finally, we plug everything into (II.1) and obtain

ml2ϕ+ cϕ+mlyB cosϕ+mgl sinϕ = 0 . (II.2)

As expected, this result is independent of the unknown force FB = −K. The aboveequation represents the equation of motion of the damped-forced pendulum, a 2nd order,non-autonomous ODE. It can be solved together with the initial conditions ϕ(t0) = ϕ0

and ϕ(t0) = ϕ0.

(iii) Remark: Why is FB = −K for a massless rod?In this example, we used the fact that K = −FB and that these forces are collinear withrAB. To show that this is true for any massless rod, we apply the LMP and AMP to sucha massless rod. We denote the end points by A, respectively B and the forces at each endby FA and FB respectively, as seen in Figure II.3.

24

Figure1.11

B

A

m = 0(massless)

FB

FA

Figure II.3: The free-body diagram of a massless rod.

When applying the LMP, i.e.

P =∑

F ,

we find that∑F = 0 , (II.3)

since for a massless rod (M = 0), P = Mv = 0. Next, we apply the AMP w.r.t. B, i.e.

HB + vB × P =∑

MB .

Note that HB = 0 since the rod is massless (M = 0). Hence, we have that∑MB = 0 . (II.4)

However, (II.3) and (II.4) are only true when FA = −FB and both forces are collinearwith rAB. We stress that this result is only true for a massless rod.

Example II.3: Mass accretion problemConsider a particle of mass m(t), as shown in Figure II.4, with velocity v(t). The particleaccretes (i.e. , gradually attains) over the time ∆t an additional mass of ∆m that has initialvelocity u(t). Find the equation of motion.

Figure1.12

mu(t)

v(t)

m(t)

v(t + t)

m(t + t)

Figure II.4: A particles of massm(t) accretes the additional mass ∆m over the time interval∆t.


Solution:

To find the equation of motion, we apply the linear momemtum principle

P = F .

For the case of a particle with constant mass m, the time derivative P is P = ma. Thisis, however, for a particle of varying mass m(t). Instead, we recall the definition of thederivative

P (t) = lim∆t→0

P (t+ ∆t)− P (t)

∆t, (II.5)

which relies on the linear momentum at the two time instances t and t+ ∆t for the wholesystem. These momenta are

P (t) = m(t)v(t) + ∆mu(t) ,

and

P (t+ ∆t) = m(t+ ∆t)v(t+ ∆t)

=(m(t) + m(t)∆t+O(∆t2)

)(v(t) + v(t)∆t+O(∆t2)

)= mv + ∆t(mv +mv) +O(∆t2) .

Note that we Taylor-expanded the term P (t + ∆t) around ∆t = 0 as this is the limit ofinterest in (II.5). Next, we insert both momenta into (II.5) to obtain:

P = lim∆t→0

mv + ∆t(mv +mv) +O(∆t2)−mv −∆mu(t)

∆t= mv +mv − mu .

Inserting the above quantities into the LMP yields the equation of motion for a massaccretion problem:

mv +mv − mu = F . (II.6)

Alternatively, the equation can be stated in terms of the relative velocity vrel = u − v ofthe additional mass w.r.t. the particle:

mv = F + mvrel . (II.7)

Example II.4: Raindrop falling through floating vaporA raindrop of mass m(t) is falling towards the earth due to gravity, as depicted in Fig-ure II.5. Due to the floating vapor, the drop accumulates mass at a rate m = kmv = kmy,where k = const., [k] = 1/m is a constant and v = y is the downfall velocity of the rain-drop. What is the asymptotic behavior of the falling raindrop? (Hint: The results fromExample II.3 may be used to find the governing equation of motion. Then, consider theasymptotic behavior of the equation. The accumulated mass has zero velocity u = 0 priorto attaching to the drop.)

26

Figure1.13

gm

v

v

2arrowsinthescript??

+

–y

(a) (b)

Figure II.5: (a) A raindrop of massm(t) falling through floating vapor. (b) The accelerationv is plotted w.r.t. the velocity v.

Solution:

Using equation (II.6) from Example II.3 with zero initial velocity u = 0 of the accumulatedmass, we obtain:

mv +mv = mg .

Taking the y-component of above equation and inserting the given relation for m yields

my +my = mg ,

respectively,

kmy2 +my = mg .

The mass m cancels out in above equation, leading to the equation of motion

y + ky2 = g ,

which is a 2nd -order nonlinear ODE. The asymptotic velocity v0 =√

gk is an equilibrium

of the equivalent 1st order ODE

v = g − kv2

for the velocity v(t). In fact, if v > v0, then v < 0 and therefore the particle tends to v0.Likewise, if v > v0, then v > v0, as can be seen from Figure II.5 (b). Thus the v ≡ v0

equilibrium of the ODE for v(t) is asymptotically stable.

Example II.5: Rocket motionConsider a rocket of mass m(t) that burns fuel to accelerate. The dynamics shows thissystem may be described using the general mass accretion equation. Figure II.6 shows thesituation, with the rocket subject to forces F (t). The relative velocity of the exhaust gas


w.r.t. the rocket is denoted by c , and the vertical position of the rocket by y. What is thenecessary condition for lift-off when drag is neglected?

Figure1.14

m(t)

y

F

c

Figure II.6: A rocket is loosing mass by burning fuel. The velocity c of the exhaust gas isdefined relative to the motion of the rocket.

Solution:

The system falls into the general framework of the mass accretion problem with nega-tive accretion m < 0. Using (II.7), the relevant equation of motion stated in terms of therelative velocity vrel = c, we obtain the following equation for the rocket:

mv = F + mc .

The term mc denotes the thrust produced by the rocket, perceived as an additional forceon the rocket. When drag is ignored, the only force exerted on the rocket is gravity, thus,F = mg. Taking the y-component of the above equation, therefore, yields

my + mc+mg = 0 .

For lift-off, we need y = v > 0, which is only possible when

−mg − mc > 0 ,

or, equivalently,

|m| c > mg .

If this lift-off condition is not fulfilled, the rocket will sit on the launch pad and burn fueluntil the lift-off condition is satisfied.

28


Prof. George Haller

Dynamics

III. Applications of single-particledynamics




Chapter III

Applications of single-particledynamics

Example III.1: Car stunt in a circular ramp with foot off the pedalConsider a car of mass m that approaches a vertically looping ramp and attempts to runthrough it without falling off, as shown in Figure III.1. Suppose that the car rolls withoutfriction along the track, modeled as a point mass. The ramp is massless and has radius R.Furthermore, the car is subject to gravity and the driver does not accelerate the car onceit enters the loop (foot is off the pedal). For all computations, the inertial frame

[i, j, k

]will be used.

(a ) What is the minimum entry velocity so that the car does not fall off?

(b ) What is the maximum g-force acting on the driver?

Figure1.15

g2

1

R

m

Figure III.1: A car of mass m is approaching a vertical, looping track. Point 1 denotes theentry point and point 2 is the turn-over point.

(a ) Minimum entry velocity

31

32

(i) Free-body diagram (FBD)The only two forces acting on the car are the constraint force N and the gravity mg (apotential force), as seen in the free-body diagram in Figure III.2.

Figure1.16

iN

mg

Figure III.2: The free-body diagram of the system is shown. The blue line is tangent tothe track (black-dashed line).

(ii) Conservation of energyThe constraint force N does no work because it always remains normal to the track, andtherefore

N · dr = 0 ,

implying

WN =

∫N · dr = 0 .

Therefore the car motion is conservative and the work-energy principle for conservativesystems (I.17) can be applied. The potential energy V (r) of the car is the potential ofthe gravitational force,

V (r) = V (y) = mgy .

The critical point where the car is likely to fall off is the turning point of the ramp, point2. This prompts us to apply the conservation of energy between the critical point 2 andthe entry point 1:

T1 = T2 + V2 ,

=⇒ 1

2m |v1|2 =

1

2m |v2|2 +mg 2R , (III.1)

where V1 = 0 since y = 0 at position 1. From (III.1), the relation between v1 and v2 is

v2 =√v2

1 − 4gR . (III.2)

This relation is only valid if the car is still on the track. At any time, the track pushesthe car with a force N . As long as N > 0 the car remains on the track. The limiting caseis, therefore, N = 0.

(iii) Linear momentum principle

Chapter III. Applications of single-particle dynamics 33

Figure1.17

i

j

r

eN

'

2

N

v

mg

Figure III.3: The car at an arbitrary position on the looping ramp, marked by the angleϕ.

To relate the velocities v1 and v2 to the normal force N and therefore obtain a conditionfor the minimal entry velocity, we apply the linear momentum principle

P = F = N +mg . (III.3)

We use the unit normal vector

eN = cosϕ i+ sinϕ j

of the ramp to express the trajectory r(t) of the car as

r(t) = R eN (t) .

The angle ϕ(t) in eN (t) is defined in Figure III.3. The speed of the car is

|v| = v = Rϕ .

To evaluate (III.3), we must first calculate P and P :

P = mv = md

dtr (III.4)

= md

dt(R cosϕ i+R sinϕ j)

= mR ϕ(− sinϕ i+ cosϕ j) ,

P = −mR (ϕ sinϕ+ ϕ2 cosϕ) i

+mR (ϕ cosϕ− ϕ2 sinϕ) j . (III.5)

Now, we project (III.3) onto eN by taking the dot product of III.3 with eN :

P = N +mg ,

=⇒ P · eN = (N +mg) · eN .

We insert in (III.5) into the last equation to obtain

−mR ϕ2 = −N −mg sinϕ ,

=⇒ −mv2

R= −N −mg sinϕ .

34

Substituting at ϕ = π2 and solving for N2 = N(ϕ = π

2 ), we obtain

N2 = m

(v2

2

R− g). (III.6)

For the car to be still on the track (barely) at point 2, we must have N2 ≥ 0. Thisinequality, combined with III.6 gives

v2 ≥√g R .

Using (III.2), v2 can be expressed in terms of the initial speed v1:√v2

1 − 4 g R ≥√g R ,

=⇒ v1 ≥√

5 g R (III.7)

Using realistic values for a 2R = 40 ft = 12.192 m-high ramp, and g = 9.81 m/s2 yields

v1 ≥√

5 · 9.81 m/s2 · 1

2· 12.192 m =

√299 m2/s2 ,

≥ 17.29 m/s = 38.67 mph .

The movie (https://youtu.be/wiZoVAZGgsw) shown in the lecture suggests the car mustbe faster than 36 mph (v1 ≥ 36 mph), which corresponds to a height of 35 ft.

(b ) Maximal g-force

For this part, we seek the maximal g-force a driver of mass md experiences during theride. The normal force N(ϕ) exerted on the passenger at an arbitrary angle ϕ can beexpressed using (III.6) as

N(ϕ) = mdv2(ϕ)

R−md g sinϕ . (III.8)

In order to obtain v(ϕ), we apply again the conservation of energy but this time for anarbitrary angle ϕ:

Tϕ + Vϕ = T1 + V1 ,

=⇒ Tϕ = T1 + V1︸︷︷︸V (y=0)=0

−Vϕ ,

=⇒ 1

2mdv

2(ϕ) =1

2mdv

21 −mdgR(1 + sinϕ) ,

=⇒ v2(ϕ) = v21 − 2gR(1 + sinϕ) . (III.9)

Plugging (III.9) into (III.8) gives

N(ϕ) =md

R

(v2

1 − 2gR(1 + sinϕ))−mdg sinϕ

=md

Rv2

1 − 2mdg − 3mdg sinϕ .


The the maximum of N(ϕ) over the desired interval ϕ ∈[−π

2 ,3π2

]is obtained as

maxϕ∈[−π

2, 3π

2]N(ϕ) = N

(−π

2

)= N

(3π

2

)= md

(v2

1

R+ g

)

arising at the entry of the loop. When plugging in the minimal necessary entry speedfrom (III.7) into this last expression, we obtain

maxϕ∈[−π

2, 3π

2]N(ϕ) = md(5g + g) = 6mdg

for the maximal force the driver feels. The passenger, therefore, experiences "6g"; a typi-cal person can handle about 5g without passing out. This explains why the stunt driverpracticed in a looping airplane before the stunt, as shown in the movie.

How is a roller coaster loop is designed in practice?The curvature of a roller-coaster loop increases linearly to ease the transition into the loop(Euler spiral). The loop then has a similar curvature as the loop shown in Figure III.4.Highway exits are designed in a similar fashion.

Figure III.4: A roller-coaster loop has linearly increasing curvature to ease the transition.

Example III.2: Motion of a rocking chairThe rocking chair shown in Figure III.5 is modeled as a single point mass m with aweightless frame that models the bottom part of the chair. The mass of the chair, as wellas of the person sitting on it, is concentrated in C. The chair rocks without slipping, i.e.exhibits pure rolling. The bottom part has radius R, and the concentrated mass m is at adistance a away from the center of the circle. The origin of the coordinate system

[i, j, k

]is placed at this center. The system is subjected to gravity.

36

(a)

Figure1.19

1.19a->image,nochanges1.19b->seebelow

g

a

R

m

O

(b)

Figure III.5: (a) An actual rocking chair. (b) The corresponding mechanical model. Thechair is modeled as a point mass supported by a weightless frame attached to a weightlesscircular profile of radius R.

(a ) What is the equation of motion for the point mass m?(b ) Calculate the contact forces between the chair and the ground.(c ) Determine a comfortable design for the rocking chair, i.e. find an R for the chair to

rock at an adequate rate of 30 rocking periods per minute.

(a ) Equation of motion

(i) Free-body diagramConsider the free-body diagram shown in Figure III.6. Point B hereby denotes the moving(!) point of contact between the rocking chair and the ground, N is the normal reactionforce. (Note: N 6= −mg; common mistake from statics); S is the tangential reaction force(friction), which is necessary to prevent the chair from sliding on the ground (S 6= 0).

Figure1.20

i

j

g

mgN

a

'C

B

xB

sinitialposition

O

Figure III.6: The free-body diagram for the mechanical model of the chair at an arbitraryangle ϕ. Orange arrow indicates the position of the contact point B at ϕ = 0.

(ii) Angular momentum principleSince we are interested in finding the equation of motion but not the forces it is preferable


to choose a reference point where the resultant torque does not depend on unknown forces.This is why we choose B as reference point. The only force producing a torque about Bis known (gravitational force), whereas S and N produce no torque w.r.t. B. The AMPw.r.t. point B is

HB + vB × P = MB . (III.10)

To evaluate (III.10), all quantities involved must be determined first. The velocity of thecontact point B is

vB = Rϕ i ,

because the arclengths of the two marked curve sections are equal. As a result, thevelocity vB is the time derivative of the arclength Rϕ covered by B. (More informationon rolling without slipping is provided in Example VIII.2.) The remaining quantities canbe computed as follows:

P = mvC = md

dtrC = m

d

dt

((Rϕ− a sinϕ) i+ (R− a cosϕ) j

)= mϕ

((R− a cosϕ) i+ a sinϕ j

),

HB = rBC × P = rBC × (mvC)

= m(− a sinϕ i+ (R− a cosϕ) j

)× ϕ

((R− a cosϕ) i+ a sinϕ j

)= mϕ

(− a2 −R2 + 2aR cosϕ

)k ,

HB =(m(2aR cosϕ− a2 −R2)ϕ− 2maR sinϕ ϕ2

)k ,

vB × P = Rϕ i× (mvC) = aRmϕ2 sinϕ k ,

MB = mga sinϕ k .

Inserting all these terms into (III.10) and simplifying the resulting expression yields theequation of motion

m(2aR cosϕ− a2 −R2)ϕ−maRϕ2 sinϕ−mga sinϕ = 0

=⇒ (R2 + a2 − 2aR cosϕ)ϕ+ aR sinϕ ϕ2 + ag sinϕ = 0. (III.11)

This equation of motion is a 2nd-order nonlinear ordinary differential equation (ODE).Since the ODE is nonlinear it is hard or even impossible to find an analytical solution.Instead, the ODE can be solved numerically, e.g. using MATLAB’s ODE45 solver. Forthat purpose, the equation needs to be first transformed into a system of 1st-order ODEs(phase space representation), see below.

The phase space variables x1 = ϕ and x2 = ϕ are introduced via

x =

(x1

x2

).

38

The phase space representation is then

x = f(x)

=⇒ x =

(x1

x2

)=

x2

− a sinx1(Rx22 + g)

R2 + a2 − 2aR cosx1

.

The initial conditions for the system are

x(0) =

(ϕ0

ϕ0

).


(b ) Contact forces

To find the contact forces, we use the linear momentum principle

P = F . (III.12)

(Note: We could also apply the AMP again w.r.t. a point where the forces N and S pro-duce a torque but this would unnecessarily increase the complexity of the derivation.)

(i) LMP in x-directionEvaluating (III.12) in the x-direction gives us the tangential reaction force S:

mxC = S ,

xC = Rϕ− a sinϕ ,

=⇒ S = md

dt

((R− a cosϕ)ϕ

)= m

(a sinϕϕ2 + (R− a cosϕ)ϕ

). (III.13)

Formula (III.11) is used to eliminate ϕ from (III.13):

S = m

(a sinϕ ϕ2 − (R− a cosϕ)(aR sinϕ ϕ2 + ag sinϕ)

R2 + a2 − 2aR cosϕ

).

(ii) LMP in y-directionSimilarly, evaluating (III.12) in the y-direction yields an equation for N:

myC = N −mg ,yC = R− a cosϕ ,

=⇒ N = m(yC + g) = m(a cosϕ ϕ2 + a sinϕ ϕ2 + g

).

(c ) Comfortable chair

A rocking chair is considered comfortable when it rocks with about 30 rocks (periods)per minute. To obtain a relation between the geometric parameters R and a that ensurethis comfortable rocking, we linearize the equation of motion (III.11) at its ϕ = 0 equi-librium. Afterwards the eigenfrequency of the linearized equation is set to the desiredrocking frequency.The equilibrium is indeed ϕ = 0 (plugging ϕ ≡ 0 into (III.11) yields ϕ = 0 and ϕ = 0). Aslong as only small oscillations around the equilibrium are considered, it is a valid assump-tion to work with linearized equations. For that, (III.11) is Taylor-expanded and only thelinear terms are kept. Since the equation of motion is

(R2 + a2 − 2aR cosϕ)ϕ+ aR sinϕ ϕ2 + ag sinϕ = 0,

its Taylor-expansion is given by(R2 + a2 − 2aR+O(ϕ2)

)ϕ+ aR

(ϕ+O(ϕ3)

)ϕ2 + ag

(ϕ+O(ϕ3)

)= 0 .

40

Dropping all terms of 2nd or higher order, we obtain

(R− a)2ϕ+ agϕ = 0 (III.14)

for the linearized equation of motion around the equilibrium position ϕ = 0. Now, wewill bring this equation to a form that reveals directly the eigenfrequency of the linearizedmotion:

(R− a)2ϕ+ ag ϕ = 0 ,

=⇒ ϕ+ag

(R− a)2ϕ = 0 ,

=⇒ ϕ+ ω20 ϕ = 0 .

Therefore, the eigenfrequency of the linearized equation is

ω0 =

√ag

R− a , (III.15)

which we equate with the desired rocking frequency. Specifically, 30 rocks per minutecorrespond to

ω0 =2π

60/30s−1 = πs−1. (III.16)

We use the results obtained from (III.15) and (III.16) to relate the radius R and thedistance a of a comfortable rocking chair. This relation can be used as a design principle:

R = a+

√ag

π.

The units of the quantities are [R] = m, [a] = m, [g] = m/s2 and [π] = s−1.

Example III.3: Design of a curb-tipped rollover test for carsA car of mass m is skidding sideways with skidding velocity v0 into a curb, then rolls-overwhen hitting the curb, as shown in Figure III.7. To assess the damage on the car, a crashtest must be properly designed. The question is the minimal skidding velocity v0 leadingto a roll over. For this purpose, the car is modeled as a point mass m at the center ofmass C located in the geometric center of a weightless frame with dimensions a and b.Furthermore, any motion tangent to the curb is neglected. The contact point B where thecar hits the curb is smooth and the height between the curb and the street is negligible aswell. The collision is assumed to be perfectly inelastic (no rebound) with gravity actingon the car.


(a)

Figure1.21

a

g

b

C

1.21a->image,nochanges1.21b->seebelow

v0

m

Bnegligibleheight (b)

Figure III.7: (a) A car hitting a curb and starting to roll-over. (b) The car is modeledas point mass with a weightless frame. The surface is assumed smooth and the heightbetween the street and the curb (marked red) is negligible. The curb point is denoted byB.

We divide the roll-over into two phases: a short phase including the collision, and a phasestarting after the collision, describing the roll-over.

(a ) Phase 1: Collision

Collisions are often hard to describe, which leads to various misconceptions about them.In general, energy is not conserved during a collision. Energy is not transformed into an-gular momentum either. Similarly, linear energy is not transformed into rotational energyand linear momentum is not transformed into angular momentum.

(i) Free-body diagramLet us consider the forces acting on the particle in the moment t = 0 of the collision asshown in Figure III.8. The car is subject to gravity mg as well as the normal force N .From the moment the car hits the curb, the force FB(t) is exerted as well. Neither thedirection nor the magnitude of FB(t) is known at this point.

Figure1.22

Ci

j

a b

a

b

B

B

a/2

N

mg

v0

v(0+)

(a) (b)

Figure III.8: The free-body diagram and the velocity after the impact is shown. (a) Thefree-body diagram depicts when the car hits the curb. (b) After the collision, the carrotates around the contact point B and the velocity is denoted by v(0+). As the carrotates around B, the velocity vector v(0+) is perpendicular to the vector rBC .

42

(ii) The Linear momentum principleThe LMP for the sketched situation is

P = F = mg +N + FB .

However, FB(t) is unknown and complicated to express. Therefore, we will not apply theLMP.

(iii) The Angular momentum principleSince force FB does not produce a torque about point B, we can apply the AMP w.r.t.B without the knowledge of FB(t). Since vB = 0 we obtain

HB + vB × P = MB ,

=⇒ HB = MB (III.17)

for the AMP w.r.t. B. Next, we integrate (III.17) over t ∈ [0−, 0+], where t = 0− denotesthe instance just before the collision and t = 0+ just after it:∫ 0+

0−HB dt =

∫ 0+

0−MB dt

=⇒ HB(0+)−HB(0−) =

∫ 0+

0−rBC × (N +mg) dt . (III.18)

Next, we evaluate the right-hand side of (III.18) and let the integration interval go tozero. Since N and mg are bounded in time (non-impulsive), we obtain

lim0+→0−

∫ 0+

0−rBC × (N +mg) dt = 0 .

Therefore, the angular momentum w.r.t. B is conserved during the collision:

HB(0+) = HB(0−) . (III.19)

The angular momenta before and after the collision can be expanded to

HB(0−) = rBC × P (0−) = rBC × (mv0) = −a2mv0 k

and

HB(0+) = rBC × P (0+) = rBC × (mv+) = −mv+

√a2 + b2

2k ,

respectively, with the velocities v0 = v(0−) just before and v+ = v(0+) just after thecollision (see Figure III.8). The initial velocity v0 is given. As stated before in Figure III.8,we have rBC ⊥ v+ . Therefore, the vector product

rBC × (mv+)

simply yields

rBC × (mv+) = −m |rBC |∣∣v+

∣∣ k = −mv+

√a2 + b2

2k .


Inserting all those terms into (III.19) gives

−a2mv0 k = −mv+

√a2 + b2

2k ,

which can be solved for the velocity v+ after the collision:

v+ =a√

a2 + b2v0. (III.20)

(b ) Phase 2: Roll-over

After the collision the point mass rotates around B in a circular motion, starting from po-sition 1 with initial velocity v+. For the car to roll-over, the point mass needs to pass overthe curb point B (see position 2 in Figure III.9). This can only happen for a sufficientlylarge velocity v+, otherwise the car rolls back. The goal is to find the necessary minimalvelocity v+ by applying the appropriate principle.

(i) Free-body diagramAfter the collision, the only forces acting on the car are gravity and the force K at thecontact point (see Figure III.9). During this phase, angular momentum w.r.t. B is notconserved anymore sincemg will produce a torque about B, and the time interval regardedis not infinitesimally small (which was the reason why conservation of angular momentumabout B could be applied before). Therefore we cannot relate position 1 and 2 to eachother without solving the equation of motion, which is rather cumbersome.

Figure1.23

^i

j

B

1

2

m

mg

a/2

K

pa2 + b2

2

Figure III.9: Free-body diagram for the situation after the collision. The zero potentiallevel is on the ground. Recall that the height difference between the ground and the curbis neglected.

(ii) Conservation of energyInstead, we exploit the fact that the system is conservative since gravity is a potentialforce and the constraint force K does no work by

K · dr = 0

during the roll-over. The total mechanical energy of the system is conserved betweenposition 1 and 2 (work-energy principle for conservative systems). The conservation ofenergy states

T1 + V1 = T2 + V2 , (III.21)

44

where

V (y) = mgy

is the potential energy of the system (zero-level assumed on the ground). We are interestedin the limiting case, in which v+ is just large enough to cause the car to roll-over. In thatcase,

T2 = 0

because

v2 = 0 .

Evaluating (III.21) for this case yields

1

2mv2

+ +mga

2= 0 +mg

√a2 + b2

2,

which can be solved for v+ to yield

v+ =

√g(√a2 + b2 − a) . (III.22)

Combining (III.22) with the relation (III.20) for the velocities before and after the collisiongives the minimum necessary skidding speed that results in roll-over:

vcrit0 =

√g(√

a2 + b2 − a)(

1 +b2

a2

). (III.23)

We see from this result that flat cars (b a) are more stable, since vcrit0 is larger forthem. This agrees with our intuition and experience from driving, as flat cars with alower center of mass are more stable.


Prof. George Haller

Dynamics

IV. Particle motion in nonstandardcoordinates

ContentsIV.1 Particle motion in curvilinear coordinates . . . . . . . . . . . . . 47IV.2 Particle motion in non-inertial frames . . . . . . . . . . . . . . . . 51




Chapter IV

Particle motion in nonstandardcoordinates

In this section, we restate the momentum principles in different, nonstandard coordinateframes. First, we show how the momentum principles can be written in curvilinear coor-dinates (see Section IV.1) and then we derive the momemtum principles for non-inertialframes (see Section IV.2).

IV.1 Particle motion in curvilinear coordinates

In some situations, it is preferable to write the momentum principles w.r.t. a differentcoordinate system, such as a curvilinear one. This does not require any new principle: itsimply provides an alternative way to approach a problem. For many problems involvingrotational motion, curvilinear coordinates can reduce the complexity of the expressions forthe forces and the equations of motion. In the following, we discuss a general approach tohow the momentum principles can be transformed to curvilinear coordinates.

Consider the position vector r in Cartesian coordinates,

r =

xyz

.

We could also express r as a function of some curvilinear coordinates c ∈ R3 as

r = f(c) .

The linear momentum principle states that

md2

dt2r = F (r, r, t)

or, expressed in curvilinear coordinates,

md2

dt2f(c) = F

(f,

d

dtf , t

)= F (c, c, t) . (IV.1)

47

48 IV.1. Particle motion in curvilinear coordinates

Note: Since f = f(c(t)) has no explicit dependence on the time t, the time derivative off(c) is simply given by

d

dtf(c) = ∇cf c =

(∂f

∂c1

∂f

∂c2

∂f

∂c3

)c =

∂f1

∂c1· · · ∂f1

∂c3...

. . ....

∂f3

∂c1· · · ∂f3

∂c3

c1

c2

c3

.

Here ∇cf denotes the Jacobian of f w.r.t. the curvilinear coordinates c, and c =d

dtc

denotes the time derivative of the coordinates c.Using this time derivative in (IV.1), we obtain

md

dt

(∇cf c

)= F .

Carrying out the remaining differentiation in the same way, we obtain the linear momentumprinciple in curvilinear coordinates as

m

((∇2cf c

)c+∇cf c

)= F , (IV.2)

where the second derivative ∇2cf is a three-tensor. To write IV.2 out in component form,

we will use the notation

αiβi =∑i

αiβi .

which implies summation over repeated indices. Furthermore, we denote the partial deriva-tive of fi w.r.t. cj by

fi,j =∂fi∂cj

.

With all this notation, (IV.2) becomes

m(fi,jk cj ck + fi,j cj

)= Fi, i = 1, 2, 3 , (IV.3)

where i denotes the component index. Summarizing:

The principle of linear momentum in curvilinear coordinates

m

((∇2cf c

)c+∇cf c

)= F

(f,

d

dtf , t

)= F (c, c, t) , (IV.4)

or, in index notation,

m

(fi,jk cj ck + fi,j cj

)= Fi

(f,

d

dtf , t

)= Fi (c, c, t) , i = 1, 2, 3 . (IV.5)

Chapter IV. Particle motion in nonstandard coordinates 49

c . . . . . . curvilinear coordinates on R3

f(c) . . . position of particle expressed in the curvilinear coordinates c∇cf . . . Jacobian of f in cm . . . . . mass of the particleF . . . . . resultant force acting on the particleF . . . . . resultant force acting on the particle expressed in curvilinear coordinates

Example IV.1: Cylindrical coordinatesA common example for curvilinear coordinates are cylindrical coordinates (see Fig-ure IV.1), collected in the vector.

c =

%θz

.

Figure1.24

^

z

y

x

m

r

z

Figure IV.1: The cylindrical coordinates are indicated w.r.t. an inertial frame.

We can express the position vector r in terms of the cylindrical coordinates c as

r =

xyz

=

% cos θ% sin θz

,

and use (IV.4) or (IV.5) to apply the LMP in cylindrical coordinates. With thisapproach, we can efficiently solve problems with cylindrical symmetry, such as Ex-ample III.2.

Example IV.2: Flexible pendulumConsider a pendulum bob of mass m attached to the hinge point B via a masslessrod, as shown in Figure IV.2. The bob can slide along the rod without friction.There is a linear spring of unstretched length l0 connecting the hinge point and the

50 IV.1. Particle motion in curvilinear coordinates

bob. We describe this two dimensional system using polar coordinates (r, ϕ) withcorresponding (moving) unit vectors er and eϕ, as illustrated in the figure. Theinertial reference frame is denoted by

[i, j]. What is the equation of motion in the

(r, ϕ) coordinates?

Figure1.25

^

g y

x

er

e'm

B

masslessrod

k : sti↵nessof spring'

r

Figure IV.2: A flexible pendulum that not only swings, but also oscillates along themassless rod.

Solution:

(i) Free-body diagram

The various forces acting on the bob are indicated in the free-body diagram depictedin Figure IV.3 (a), including the spring force K = −k(r − l0) er. Note that thereis no friction force, since we assume that the mass slides without friction.

Figure1.26

^C

C

B

N

mg'

K = k(r l0)er

N

(a) (b)

Figure IV.3: (a) The free-body diagram of the pendulum bob. (b) The normal forceN = 0 vanishes for a massless rod.

(ii) Angular momentum principle for the massless rodWhen we cut free the massless rod, as depicted in Figure IV.3 (b), we see that theonly force acting on the rod with a non-zero torque w.r.t. B is the normal force


N (the reaction forces at the joint B does not produce a torque about B). Theangular momentum principle w.r.t.the non-moving joint B states that

HB = MB .

Because the rod is massless, HB = 0 holds, thus,

MB = rBC ×N = 0 .

But rBC ∦ N , thus this last equation can only hold true if

N = 0 .

With this result at hand, we drop N from any subsequent equations.

(iii) Linear momentum principleTo find the equation of motion of the system, we now apply the linear momentumprinciple to the mass m, i.e.

mx = K +mg ,

with x = r sinϕ i − r cosϕ j denoting the position of the mass, described usingthe polar coordinates in the inertial basis. We now project the linear momentumprinciple onto er = sinϕ i − cosϕ j and eϕ = cosϕ i + sinϕ j, to obtain,respectively

(mx) · er =(K +mg

)· er ,

(mx) · eϕ =(K +mg

)· eϕ .

Evaluating these expressions gives the equations of motion in terms of the cylin-drical coordinates:

m(r − rϕ2) = −k(r − l0) +mg cosϕ ,

m(rϕ+ 2rϕ) = −mg sinϕ .

This is a relatively simple 2nd-order coupled system of nonlinear ODEs in (r, ϕ).The equations of motion would be substantially more complicated in the Euclideancoordinates (x, y) coordinates.

IV.2 Particle motion in non-inertial frames

Sometimes it is convenient to describe the motion of a system w.r.t. a moving frame ofreference. Consider two reference frames, an inertial frame [x1, x2, x3] with origin O and amoving frame [y1, y2, y3] with origin Ot, as shown in Figure IV.4. Since the frame [y1, y2, y3]is not an inertial frame, we cannot apply the linear and angular momentum principlesdirectly. However, we can express these principles in the inertial frame [x1, x2, x3], thenexpress the resulting equations using the coordinates of the non-inertial frame. The ideais, therefore, similar to our handling of curvilinear coordinates in Section IV.1.The general form of transformations from the y-frame to the x-frame is

52 IV.2. Particle motion in non-inertial frames

Figure1.27

x3

x2

x1

x

y

F

m

b(t)

y1

y2y3

0t

O

Figure IV.4: The motion of a particle can be described by the vector x w.r.t. an inertialreference frame [x1, x2, x3] or by the vector y w.r.t. a moving, non-inertial frame [y1, y2, y3]with origin 0t

x = Q (t)y + b(t) . (IV.6)

where b(t) ∈ R3 is the translational vector between the two origins of the coordinatesystem; y is the position of the particle in the y-frame; x is the position of the particle inthe inertial frame x; and Q (t) ∈ R3×3 is the rotation matrix that aligns the y axes withthe x axes. As a rotation matrix, Q (t) satisfies

Q T = Q−1 , det(Q)

= 1 .

The group of matrices satisfying these two properties is called the special orthogonal groupSO(3). They describe rotations in 3D-space, an operation that is length, angle and orien-tation preserving, also see Section VIII.5.Note: Suppose that the x and y-frame share the same origin. Then b(t) ≡ 0, and hencethe coordinate transformation simplifies to:

x = Q (t)y.

This coordinate change rotates a vector represented in the y-frame into its representationin the x-frame. The inverse transformation

y = Q−1(t)x = Q T (t)x

does the opposite, i.e. , rotates the vector from the x-frame to the y-frame. In case the tworeference frames are also shifted relative to reach other, we need the additional translationvector b(t) to describe the transformation, as in (IV.6). The vector b is represented in theinertial frame x.Next, we insert the transformation (IV.6) into the LMP

mx = F (x, x, t) ,

which gives us

md2

dt2(Qy + b

)= F

(Qy + b,

d

dt(Qy + b), t

).


Carrying out the differentiation in this equation, we obtain

md2

dt2(Qy + b

)= m

d

dt(Q y +Q y + b) = m(Q y + 2Q y +Q y + b) = F .

Multiplying from the left by Q T and rearranging some terms, we find the LMP expressedin the y-frame as

my = Q TF −mQ T Q y − 2mQ T Q y −mQ T b .

Although only the term Q TF describes forces that are actively exerted on the particle,additional three terms (inertial terms) have appeared due to the motion of the referenceframe. It is important to note that the inertial terms are not actual active forces. Theyarise solely due to the the more complicated nature of time derivatives in a non-inertialframe.

The principle of linear momentum in a non-inertial frame

my = Q TF −mQ T Q y − 2mQ T Q y −mQ T b . (IV.7)

The transformation of the vector y into its inertial representation x is

x = Q (t)y + b(t) . (IV.8)

m . . . . . . mass of the particley . . . . . . . position of the particle expressed in the non-inertial frame [y1, y2, y3]

x . . . . . . . position of the particle expressed in the inertial frame [x1, x2, x3]

Q . . . . . . rotation matrix from y to x

b . . . . . . . linear translation of the frames expressed in the inertial frameQ TF . . . active forces expressed in the moving frame

The angular momentum principle can similarly be expressed in non-inertial coordinate,but that will be omitted here.


Example IV.3: Planar motion in a rotating frameConsider a particle of mass m that moves in the (x, y) - plane. The particleis subject to forces with resultant F (t), as shown in Figure IV.5. The inertialframe is denoted by [x1, x2]. An additional frame y rotates with angular velocityΩ = Ωx3 = φ x3 about the origin enclosing an angle φ(t) with the inertial frame.

Figure1.28

x2

x1

y2F

yx y1

(t) = tF cor y

(a) (b)

O O

Figure IV.5: (a) The motion of a particle in the x-frame can also be described in arotating y frame. (b) The Coriolis force F cor always acts perpendicular to the axisof rotation and to the relative velocity y.

The particle position in the inertial frame x is represented by

x =

(x1

x2

),

and in the moving frame y by

y =

(y1

y2

).

We can relate x and y by the transformation

x = Qy

where

Q =

(cosφ − sinφsinφ cosφ

).

(Note: We have b ≡ 0 here since both coordinate frames share the same origin)In the following, we are interested in computing the inertial terms(i.e. , perceivedinertial forces) arising in the equations of motions written in the y frame (cf. IV.7).The active forces are

F = Q TF .

To evaluate the inertial terms, we note that

Q = φ

(− sinφ − cosφcosφ − sinφ

)︸︷︷︸

P (φ)

= φP (φ) ,


and

Q = φP − φ2Q = ΩP − Ω2Q .

Note that Q , P ∈ SO(2), i.e. , both matrices belong to the special orthogonalitygroup in 2D, describing two-dimensional rotations. Now we can evaluate the firstinertial term in IV.7 as

−mQ T Q y = −m(

ΩQ TP y − Ω2Q TQy).

The matrix

Q TP =

(0 −11 0

)represents a counter-clockwise rotation by π

2 . Instead of using this matrix we canalso use the vector product

e3 × y =

(0 −11 0

)y

to express the same rotation, where e3 denotes the unit vector perpendicular to the(x1, x2)-plane. Also, using Q−1 = Q T , or, equivalently,

Q TQ = I ,

we obtain

−mQ T Q y = −mΩ× y +mΩ2y .

The first term

F eul = −mΩ× y

in the above equation is called Euler "force", which arises only when the non-inertialy frame rotates with varying angular velocity Ω 6= const. The Euler "force" is alwaysperpendicular to y. The second term

F cf = mΩ2y (IV.9)

describes is the Centrifugal "force" that arises in any rotating frame. It acts parallelto y, drawing the particle away form the center O. This effect is utilized by manymachines (e.g. ,centrifuges). The last term that arises in (IV.7) is called Coriolisforce:

F cor = −2mQ T Q y

= −2mΩQ TP y

= −2mΩ

(0 −11 0

)y . (IV.10)


Again, we may express this matrix-vector multiplication as a cross product, whichgives

F cor = −2mΩ× y . (IV.11)

Notice that the Coriolis force F cor is orthogonal to the axis of rotation as well as tothe velocity (see Figure IV.5). This term arises only when the particle moves relativeto the non-inertial reference frame. The last inertial term in (IV.7) −mQ T b givesno contribution in our present setting since b ≡ 0 in this case. We can, therefore,restate the linear momentum principle with the above introduced terms:

my = F + F cf + F cor + F eul , (IV.12)

where F denotes the active forces expressed in the rotating frame; F cf ‖ y is thecentrifugal "force"; F cor ⊥ y is the Coriolis "force", and F eul ⊥ y is the Euler"force".

Example IV.4: How non-inertial is the earth as a rotating frame?In most practical cases, we assume the earth to be an inertial frame. However, this isnot exactly true since the earth rotates around its own axis while also encircling thesun (see Figure IV.6). We now want to estimate the error we make when consideringthe earth as an inertial frame. For that we can use the results obtained from theprevious Example IV.3 to estimate the Coriolis force Fcor and the centrifugal forceF cf due to the rotation of the earth. There is no Euler force, as the earth is assumedto rotate with constant angular velocity Ω in our analysis.Figure1.29

y1

y2

y3

Figure IV.6: The earth itself is a rotating frame with angular velocity Ω.

The angular velocity of the earth is

Ω =2π

24 h= 7.3 · 10−5 rad/s ≈ const. .


Inserting this value into (IV.11), we obtain an estimate for the Coriolis force,

|F cor| ≤ 2m |Ω| |v|= 1.46 · 10−4m |v| ,

where v denotes the velocity of a mass m relative to the earth. Using (IV.9), andnoting that the radius of the earth is R = 6.4 · 106 [m], we can also estimate thecentrifugal force acting on a mass m moving on the surface of the earth as∣∣F cf ∣∣ =

∣∣mΩ2y∣∣

≤ m |Ω|2R= (7.3)2 · 10−10 · 6.4 · 106m = 3.4 · 10−2m.

The centrifugal force, which draws the mass away from the center of the earth, istherefore much smaller than the gravitational force drawing the mass towards thecenter of earth,∣∣F cf ∣∣ = 3.4 · 10−2m 9.81m = gm .

Therefore, for ordinary velocities, both the Coriolis and the centrifugal "forces" canbe neglected. However, for large masses such as those in the study of ocean or windflows, these effects cannot be neglected. Similarly, the Coriolis "force" cannot beneglected for ordinary masses that travel at very large velocities (e.g. , the bulletfired from a rifle).



Prof. George Haller

Dynamics

V. Dynamics of system of particlesI: Set-up

ContentsV.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62V.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

V.2.1 Holonomic constraints . . . . . . . . . . . . . . . . . . . . . 66V.2.2 Forces and constraints . . . . . . . . . . . . . . . . . . . . . 66

V.3 Degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . 67V.4 Center of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69




Chapter V

Dynamics of a system of particles I:Set-up

In this chapter, we generalize the results from Chapter I to systems made of an arbitrarynumber of particles. For that, we first introduce in Section V.1 the concept of internaland external forces. Then we discuss how particles can be constrained to each other (Sec-tion V.2) and what this implies for the degrees of freedom of the system (Section V.3).The concept of the center of mass as a distinct geometric point for each particle system isexplained in Section V.4. Finally, we generalize the momentum and work-energy princi-ples for a system of particles (Section VI.1 - Section VI.3) and show various examples inSection VII.

Figure2.1

z

y

x

F i

mi

rB(t)B

m1

mn

i(t)

O

m2

m3ri(t)

Figure V.1: The system shown consists of n particles, with each particle of mass mi

subjected to a resultant force F i(t). The reference point B at rB(t) is arbitrary andpossibly moving.

Consider n particles described w.r.t. an inertial frame[ex, ey, ez

]as shown in Figure V.1.

The ith particle of the system has mass mi and is located at position ri. The resultantforce acting on mi is denoted by F i, for i = 1, . . . , n. As in the single-particle case, wechoose an arbitrary and possibly moving reference point B at position rB(t) and denotethe distance from B to mi as

%i(t) = ri(t)− rB(t) .

61

62 V.1. Forces

V.1 Forces

In this section we examine the resultant force F i on the mass mi more closely. This willlater help us simplify the momentum principles for a system of particles. We first splitthe resultant force into two parts, the external force F exti and the internal force F inti , byletting

F i = F exti + F inti . (V.1)

Whether a force is classified as external or internal depends on whether the reaction forceit generates acts on another particle of the system or not.

Definition. An external force F exti is a force whose reaction force does not act on anyparticle j of the system, i, j = 1, . . . , n.

Definition. An internal force F inti is a force whose reaction force acts on some particlej of the system, i, j = 1, . . . , n.

We can split each internal force F inti into force contributions from each of the other parti-cles, i.e., we have

F inti =n∑j=1j 6=i

Rij , (V.2)

where Rij is the force exerted by mj on mi. By Newton’s 3rd law, action and reactionforces are parallel but act in opposite directions, therefore

Rij = −Rji . (V.3)

When we sum over the internal forces of all particles in the system, we therefore obtain

n∑i=1

F inti =n∑i=1

n∑j=1j 6=i

Rij . (V.4)

Using (V.3), the right-hand side of (V.4) equals

n∑i=1

n∑j=1j 6=i

Rij = 0 . (V.5)

Therefore, the sum of all internal forces on a system of particles is zero by (V.4) and (V.5).

Chapter V. Dynamics of a system of particles I: Set-up 63

Resultant force on a system of particlesThe total resultant force F acting on a system of n particles,

F =

n∑i=1

F exti +

n∑i=1

F inti ,

is equal to the total resultant force F ext of all external forces,

F = F ext =

n∑i=1

F exti , (V.6)

becausen∑i=1

F inti = 0 .

F exti . . . Resultant external force acting on particle iF inti . . . Resultant internal forces acting on particle i

A similar result can be established for the torque w.r.t. an arbitrary point B resulting fromall internal forces Rij , i, j = 1, . . . , n. We now show this for two masses mi and mj , n = 2,but the result can be generalized to any number of particles.Consider the torque M int

Bi w.r.t. the point B, generated by the force Rij that particle jexerts on particle i, i.e.

M intBi = %

i×Rij ,

where %iis defined as

%i

= ri − rB .

Similarly, the torque Rji w.r.t. point B is,

M intBj = %

j×Rji .

The total resultant torque M intB of all internal forces w.r.t. point B is

M intB =

n∑i=1

M intBi =

n∑i=1

n∑j=1j 6=i

%i×Rij

since we have assumed n = 2, we specifically have

M intB = %

i×Rij + %

j×Rji .

By eq. (V.3), we obtain

M intB = %

i×Rij + %

j× (−Rij) = (%

i− %

j)×Rij = 0 , (V.7)

64 V.2. Constraints

where we also used the fact that

(%i− %

j) ‖ Rij ,

and hence the cross product of these two vector is zero (see Figure V.2).

Figure2.2

B

mi

mj

i

j

Rij

Rji = Rij

Figure V.2: Geometry of the internal forces for a system of two particles. Observe that(%i− %

j) ‖ Rij .

Resultant torque about B for a system of particlesThe resultant torque MB about an arbitrary point B for a system of particles, writtenas

MB = M intB +M ext

B =n∑i=1

M intBi +

n∑i=1

M extBi ,

is equal to the resultant torque of all external forces, i.e. ,

MB = M extB =

n∑i=1

M extBi =

n∑i=1

%i× F exti (V.8)

because

M intB =

n∑i=1

M intBi =

n∑i=1

n∑j=1j 6=i

%i×Rij = 0 .

M intBi . . . The torque of the internal force F inti w.r.t. point B

M extBi . . . The torque of the external force F exti w.r.t. point B

V.2 Constraints

The motion of a particle is often limited by effects external to the particle. For example,the particle may only allowed to move in a certain direction or must stay on the ground(2D-motion). Such a limitation on the motion is called a constraint. We have brieflytouched upon various types of constraints for a single particle. This section treats theconcept more formally and gives a few examples of common constraints. The constraintsdiscussed in this section will also be relevant for the motion of rigid bodies.


Definition. A constraint is a scalar relation that limits the possible displacements ofparticles, i.e. , prohibits the particle from moving arbitrarily in space.

We may have constraints acting on the whole system, constraints acting on single particlesand constraints acting among particles. All these types of constraints can be described byscalar relations, as we see in the following examples.

Example V.1: Different types of constraintsWe now discuss examples of common constraints.

Figure2.3

r mimi

r(t)

y

x

R

mi

mi

zi = 0

x

y

z

(a) (b)

(c) (d)

e(t)

mj

Figure V.3: Typical constraints. (a) A particle is constrained to be at a fixeddistance r away from the link. (b) A particle is constrained to move within arotating tube whose orientation e(t) is prescribed. (c) A particle is constrained tomove on the plane z = 0. (d) A particle is constrained to stay at the center of awheel that is only able to roll along the x axis

(a ) Rigid link

A rigid link, as shown in Figure V.3 (a) sets a fixed distance r between two particlesi and j, i.e. , satisfies the scalar relation

r2 = (xi − xj)2 + (yi − yj)2 + (zi − zj)2 . (V.9)

Therefore a rigid link constraint can be fully characterized by one scalar equation.

(b ) Rotating tube

Consider now a mass mi in a tube that rotates about the origin O in 3D, asshown in Figure V.3 (b). The current orientation of the tube, described by the unit

66 V.2. Constraints

vector e(t) in the direction of the tube, is prescribed. The distance of the mass mi

from the origin is denoted by ri(t). We can express the position (xi, yi, zi)T of the

mass in terms of the constrained orientation of the tube asxiyizi

= ri(t)e(t) .

This constraint can, therefore, be characterized by two independent scalar equa-tions.

(c ) Planar motion

A particle of mass mi is constrained to move on the plane z = 0, as shown inFigure V.3 (c). This is characterized by one constraint,

zi = 0 ,

where zi denotes the position of the particle in z.

(d ) Planar rolling

A cylinder with mass mi concentrated in C, rolls on the plane z = 0 in x-direction,as shown in Figure V.3 (d). The y-position yi of C is constrained to be

yi = R ,

since the cylinder is rigid and does not penetrate into the ground. This constraintis therefore characterized by one scalar equation.

V.2.1 Holonomic constraints

We distinguish between two basic types of constraints: holonomic and non-holonomic. Aholonomic constraint only depends on the current position of the particles subjected to theconstraint, and possibly on the current time. All constraints presented in Example V.1 are,therefore, holonomic. A non-holonomic constraint may also dependent on other quantitiessuch as the velocity of the particle.

Definition. A holonomic constraint is a constraint that only depends on the currentposition of the particles involved and the current time. A non-holonomic constraintalso depends on time-derivatives of the motion.

V.2.2 Forces and constraints

So far we have only treated constraints from a purely kinematic view, i.e. we have focusedon their geometric impact on the motion of the particle. However, in order to keep themotion of the particle constrained the constraint must exert a force on the particle, whichis a kinetic aspect of the constraint. This constraint force is to be distinguished from activeforces that do not arise from constraints.


Definition. A constraint force is a force that arises to keep the motion constrainedaccording to the imposed constraints. An active force, on the other hand, is activelyexerted on the body independently of the constraints.

Example V.2: Particle moving in a planeA particle of mass m is constrained to move in a horizontal plane and is alsosubjected to gravity, as shown in Figure V.4. In this case, the normal force Nexerted by the ground on the particle is a constraint force. It arises because theparticle is restricted to move on the plane. The gravitational force mg, however, isan active force since it acts on the particle independent of the constraints.

Figure2.4

N : constraintforce

G = mg : activeforce

Figure V.4: A particle constrained to move in the horizontal plane. The activegravity force G and the constraint force N are shown.

For a constrained system, both the constraint forces and the active forces must be includedin the principles of linear and angular momentum. Constraint forces generally depend onthe motion, thus these forces must be eliminated if one’s interest is to obtain the equationof motion. We will illustrate this procedure through several examples.

V.3 Degrees of freedom

To describe positions for a system of particles we need a set of coordinates. Typically,a particle is described by a coordinate vector (x, y, z)T in 3D or by (x, y)T in 2D. Thesevectors represent three/two variables for each particle that are used to fully determine itsposition. Similarly, one might choose cylindrical coordinates (see Section IV.1) to describethe position of the particle. We call any set of variables that fully describe the motion,generalized coordinates.

Definition. A set of variables used to uniquely describe the configuration of a systemare called generalized coordinates. This can comprise, e.g., distances or angles.

Assume now that the particle is subject to a holonomic constraint. In this case, thegeneralized coordinates are not independent anymore since they are related to each otherby the constraint equation. As a result of each scalar constraint equation, one of thegeneralized coordinates becomes expressible as a function of the rest of the coordinates.

68 V.3. Degrees of freedom

Any additional holonomic constraint reduces further the number of independent generalizedcoordinates. The final number of independent generalized coordinates needed to describethe motion of the system uniquely is called the degree of freedom of the system.

Definition. The degree of freedom (DOF) of a system is defined as the number ofindependent generalized coordinates necessary to describe the motion of the systemcompletely and uniquely.

The DOF of a particle system with holonomic constraints is

DOF = dim · n− k , (V.10)

where dim denotes the dimension of the space (2 or 3); n is the number of particles inthe system; and k is the number of independent holonomic constraints applied to thesystem.

Note: For an unconstrained system, DOF= dim · n, i.e. for each particle we need dimgeneralized coordinates to describe its motion.Note: The relation (V.10) is specifically for systems with holonomic constraints. It gener-ally fails for non-holonomic constraints.

Example V.3: Planar double pendulumConsider a double pendulum in 2D with two particles of mass m1 and m2 as shownin Figure V.5. The two masses are linked to each other by a rigid, massless barof length l2, and the mass m1 is linked to a smooth joint by a rigid, massless barof length l1. Determine the DOF of the system, and identify a set of independentgeneralized coordinates that fully describe the motion.

Figure2.5

g

m1

m2

l2

l1

'1

'2

Figure V.5: The considered double pendulum is shown, consisting of two massesm1 and m2 linked to each other.

(a ) Degrees of freedom

We first need to determine the number k of independent holonomic constraints.Mass m1 is attached to the joint via a rigid bar. Similarly, mass m2 is attached


to mass m1 via a rigid bar. Therefore, by Example V.1 (a), k = 2. Then for-mula (V.10) gives the DOF as

DOF = dim · n− k = 2 · 2− 2 = 2 .

Therefore, this system has two degrees of freedom.

(b ) Generalized coordinates

Since we have two degrees of freedom, we need two independent generalized co-ordinates to describe the motion of the double pendulum. For example, we canpick ϕ1 and ϕ2 that describe the orientation of the bars holding masses m1 andm2 relative to the vertical as generalized coordinates. We could alternatively selectthe vertical positions y1, y2 of the masses instead.

V.4 Center of mass

The center of mass of a mechanical system is uniquely defined as the geometric pointw.r.t. which the total mass moment of the system is zero. This point plays a central rolein formulating the momentum principles for a system of particles.

Definition. The center of mass of a system of particles is defined as the geometricpoint C w.r.t. which the total moment of all masses m1, . . . ,mn of the system is zero.

Note: The center of mass is, therefore, the point about which the resultant torque of thegravitational forces vanishes, provided that the gravitational acceleration g is constantalong all masses of the system.

Figure2.6

mi

ri

O

C

rC

ri rC

Figure V.6: The center of mass C, located at the endpoint of the vector rC in the figure,is shown for a system of particles. Only one generic particle mi is shown.

According to the above definition, the center of mass is located at the endpoint of a vectorrC that satisfies

n∑i=1

mi(ri − rC) = 0 , (V.11)

where ri denotes the position of particle i with mass mi (see Figure V.6). Using the totalmass

M =n∑i=1

mi . (V.12)

70 V.4. Center of mass

of the system of particles, we can rewrite (V.11) as:

n∑i=1

mi(ri − rC) = 0 ,

⇒n∑i=1

miri −n∑i=1

mirC = 0 ,

⇒n∑i=1

miri −MrC = 0 ,

⇒ rC =1

M

n∑i=1

miri .

Location of the center of mass for a system of particles

rC =1

M

n∑i=1

miri (V.13)

rC . . . location of the center of massM . . . total mass of the system of particlemi . . . mass of particle iri . . . position of particle i


Prof. George Haller

Dynamics

VI. Dynamics of systems ofparticles II: Momentum &

work-energy principlesContents

VI.1 Linear momentum principle . . . . . . . . . . . . . . . . . . . . . 73VI.2 Angular momentum principle . . . . . . . . . . . . . . . . . . . . 74VI.3 Work-energy principle . . . . . . . . . . . . . . . . . . . . . . . . 76

VI.3.1 Work-energy principle for rigid body systems . . . . . . . . 77VI.3.2 Work-energy principle for conservative systems . . . . . . . 79




Chapter VI

Dynamics of systems of particles II:Momentum & work-energy principles

VI.1 Linear momentum principle

Next, we deduce the linear momentum principle for a system of n particles. First, we definethe linear momentum P of a particle system as the sum of the individual linear momentaP i, i.e.,

P :=n∑i=1

P i =n∑i=1

mivi . (VI.1)

From (V.13) we obtainn∑i=1

miri = MrC ,

which we differentiate in time, to obtain

d

dt

n∑i=1

miri =d

dtMrC =

n∑i=1

mivi = MvC , (VI.2)

where vC denotes the velocity of the center of mass. Using (VI.2), we can rewrite (VI.1)as

P =

n∑i=1

mivi = MvC .

Definition. The linear momentum P of a system of n particles, total mass M andcenter of mass rC equals the sum of the individual linear momenta P i of each particle,

P :=

n∑i=1

P i =

n∑i=1

mivi , (VI.3)

which is equal to

P = MvC . (VI.4)

73

74 VI.2. Angular momentum principle

Similarly, the linear momentum principle for a particle system is the linear momentumprinciple (I.2) for a single particle, i.e.,

P i = F i = F exti + F inti , (VI.5)

summed over all particles, leading to

P =

n∑i=1

P i =

n∑i=1

F i . (VI.6)

We can now use (V.6) to simplify (VI.6) to

P =n∑i=1

F i =n∑i=1

(F exti + F inti ) =n∑i=1

F exti = F ext .

Therefore, using (VI.4), we obtain the following:

The principle of linear momentum for a system of particles

P = MaC =n∑i=1

F exti = F ext (VI.7)

P . . . . . linear momentum of the system of particlesM . . . . total mass of the systemaC . . . . acceleration of the center of mass CF exti . . . external forces acting on particle iF ext . . . resultant external force acting on the system

Note: If F ext = 0 then the system linear momentum is conserved, i.e. P = const. and vC =const. Even then, however, the individual linear momenta are generally not conserved, i.e.,P i 6= const.

VI.2 Angular momentum principle

With reference to Figure VI.1, we define the angular momentum of a system of particlesas follows:

Definition. The angular momentum of a system of n particles w.r.t. an arbitrary,possibly moving point B is defined as the sum of all individual angular momenta HBi,

HB :=

n∑i=1

HBi =

n∑i=1

%i× P i , (VI.8)

where %i(t) = ri(t)− rB(t) is a vector pointing from the point B to particle i.

Chapter VI. Dynamics of systems of particles II: Momentum & work-energy principles75

Figure2.1

z

y

x

F i

mi

rB(t)B

m1

mn

i(t)

O

m2

m3ri(t)

Figure VI.1: The necessary quantities to define the angular momentum w.r.t. an arbitrarypoint B at rB(t).

Differentiating (VI.8) in time yields

d

dtHB =

n∑i=1

%i× P i +

n∑i=1

%i× P i .

Next, we replace %iin the first term of the above equation with

%i

= ri − rB = vi − vB,

to obtain

HB =

n∑i=1

(vi − vB)× P i +

n∑i=1

%i× P i

=

n∑i=1

vi × P i − vB ×n∑i=1

P i +

n∑i=1

%i× P i

= −vB × P +

n∑i=1

%i× P i . (VI.9)

Here we used the definition of linear momentum (VI.1) and the relation vi ‖ P i = mvi,which implies vi × P i = 0. In the second term of (VI.9), we can replace P i by (VI.5) toobtain

HB = −vB × P +

n∑i=1

%i× P i = −vB × P +

n∑i=1

%i× (F exti + F inti ) . (VI.10)

At the same time, (V.8) can be written as

MB = M extB =

n∑i=1

%i× F exti =

n∑i=1

%i× (F exti + F inti ) , (VI.11)

detailing the resultant torque for a system of particles. The principle of angular momentumfor systems of particles then arises from a comparison of (VI.10) and (VI.11):

76 VI.3. Work-energy principle

The principle of angular momentum for a system of particles

HB + vB × P = M extB (VI.12)

HB . . . . angular momentum of a particle system w.r.t. a point BvB . . . . . velocity of point BP . . . . . . linear momentum of the particle systemM ext

B . . . resultant external torque w.r.t. point B

Note: If M extB = 0 and one of the three following conditions is fulfilled

• vB = 0 or

• B = C or

• vB ‖ vC ,then the angular momentum of the particle system is conserved: HB ≡ const.

VI.3 Work-energy principle

Assume that particle i in a system of n particles is subject to a resultant force F i(t). Thework Wi,12 done by F i(t) between positions ri,1 = ri(t1) and ri,2 = ri(t2) is given by

Wi,12 =

∫ ri,2

ri,1

F i · dri .

Summing over all particles, we obtain the total work W12 done on all particles by all forcesbetween states 1 and 2:

W12 =n∑i=1

∫ ri,2

ri,1

F i · dri . (VI.13)

It is useful at this point to introduce the kinetic energy T of a system of particles.

Definition. The kinetic energy T of a system of n particles is defined as the sum ofthe individual kinetic energies of all particles, i.e. ,

T :=

n∑i=1

1

2mi |vi|2 . (VI.14)

Using this definition, (VI.13) becomes

W12 = T2 − T1 , (VI.15)

where Tj = T (tj) . Formula (VI.15) offers a simple way to compute the total work doneon a system of particles without evaluating the integrals appearing in (VI.13).


Work-energy principle for a system of particles

W12 = T2 − T1 (VI.16)

W12 . . . work done on the particles by all forces between states 1 and 2Tj . . . . kinetic energy of the particle system associated with state j, for j = 1, 2

VI.3.1 Work-energy principle for rigid body systems

As has been shown before, the force F i acting on particle i may be split into an internaland an external contribution. Similarly, we can split the work W12 as

W12 =n∑i=1

∫ ri,2

ri,1

F i · dri =n∑i=1

∫ ri,2

ri,1

F inti · dri +n∑i=1

∫ ri,2

ri,1

F exti · dri (VI.17)

= W int12 +W ext

12 .

This is done to see whether we can further simplify the work-energy principle to facilitateits evaluation in practice. First, we express the vector element dri as dri = vi dt, to obtain

W int12 =

n∑i=1

∫ t2

t1

F inti · vi dt .

From (V.2) we know that

F inti =n∑j=1j 6=i

Rij ,

where Rij denotes the force exerted on particle i by particle j. Combining the last twoequations gives

W int12 =

∫ t2

t1

n∑i=1

F inti · vi dt

=

∫ t2

t1

n∑i=1

n∑j=1j 6=i

Rij · vi dt

=

∫ t2

t1

n∑i=1

n∑j<i

(Rij · vi +Rji · vj) dt . (VI.18)

From Newton’s 3rd law, we know that

Rij = −Rji ,

so we can rewrite (VI.18) as

W int12 =

∫ t2

t1

n∑i=1

n∑j<i

Rij · (vi − vj) dt . (VI.19)


This relationship can only be simplified further if we assume that any two particles remainat a constant distance from each other, i.e., the system is a rigid-body system of particles.

Definition. A rigid-body system of particles is a particle system in which any particleis at a constant distance from all the other particles, i.e.,

d

dt

∣∣ri − rj∣∣2 = 0 ∀i, j = 1, . . . , n . (VI.20)

This is equivalent to the assumption that each particle is connected to all others viarigid links.

Note: For any particle system that is not a rigid body system one must go back to (VI.15)to compute the work done. From here on, we discuss the case of rigid body systems.Condition (VI.20) can be evaluated as,

d

dt

((ri − rj) · (ri − rj)

)= 2 (ri − rj) · (vi − vj) = 0 . (VI.21)

Implying that (ri − rj) ⊥ (vi − vj). Also, recalling that

(ri − rj) ‖ Rij

holds for a rigid body system (cf. Figure V.2), (VI.21) can be rewritten as

Rij · (vi − vj) = 0 . (VI.22)

Consequently, by (VI.22) the right hand side of (VI.19) vanishes, giving

W int12 = 0 .

Consequently, for rigid-body systems of particles, eq. (VI.17) simplifies to

W12 = W ext12 .

In other words, the total work performed by internal forces on a rigid-body system ofparticles is zero

Work-energy principle for a rigid body system of particles

W12 = W ext12 = T2 − T1 (VI.23)

W12 .. . . work done on the particles between states 1 and 2W ext

12 . . . work done by the external forces between states 1 and 2Tj . . . . . kinetic energy of the particle system associated with state j


VI.3.2 Work-energy principle for conservative systems

Similarly to Section I.3.1, we define a potential energy from which all potential forces actingon the particle system can be derived. These forces can be both internal and external.

Definition. The potential energy V (r1, . . . , rn) of a system of n particles is defined asthe sum of the potentials Vk of all (internal and external) potential forces F k actingon the system for k = 1, . . . ,K:

V (r1, . . . , rn) :=

K∑k=1

Vk . (VI.24)

Note: For a rigid-body system of particles, we only need to consider the potentials ofexternal forces, since the potential of the internal force Rij and the potential of Rji canceleach other out. In case Rij and Rji are not potential forces, they do not contribute to thepotential energy anyway. Therefore, for rigid-body systems of particles, we obtain

V =

K∑k=1

V extk . (VI.25)

As in Section I.3.1, we can deduce that the total mechanical energy E of a conservativeparticle system, for which all forces are either potential or do no work, is conserved.

Work-energy principle for a conservative system of particlesThe total mechanical energy

E := T + V

of a conservative system of particles is conserved along all motions, i.e.,

T1 + V1 = T2 + V2 (VI.26)

or, equivalently

d

dtE(t) = 0 .

E . . . total mechanical energyTj . . . kinetic energy of the system associated with state jVj . . . potential energy of the system at state j

Note: For a rigid-body system of particles to be conservative, it is sufficient that all externalforces F exti are either potential or do no work. This is because we have already shown thatthe internal work vanishes for such a system and therefore does not contribute to the totalenergy balance.


For more discussion on potential forces and potentials, please refer to Section I.3.1.


Prof. George Haller

Dynamics

VII. Examples of particle systemdynamics




Chapter VII

Examples of particle systemdynamics

Example VII.1: Elastic collision of a system of particles with a wall

Figure2.8

m1

m2

m3

rigid barsv

Figure VII.1: A rigid body system of three particles approaching a rigid wall with a velocityv perpendicular to the wall.

Consider a rigid body system of three particles uniformly approaching a wall with velocityv, as shown in Figure VII.1. The particles are connected to each other via rigid bars andhence hit the wall at the same time. The collision with the wall is assumed to be perfectlyelastic, which means the collision forces derive from elastic potentials, and therefore thetotal mechanical energy is conserved over the collision. Furthermore, the particles are notsubject to any external forces beyond those associated with the impact.(a ) Apply the conservation of energy to the system and see what conclusions can be

drawn from it.(b ) Discuss qualitatively the case when the particles are not connected via rigid bars but

via springs and identify the consequences for the conservation of energy.

(a ) Conservation of energy

The system is not subject to any external force before and after the collision. Also,the collision forces do no work as the displacement of their point of attack is zero duringthe collision. Therefore, the potential energy of the system before V− and after V+ thecollision is 0,

V− = V+ = 0 .

83

84

The conservation of mechanical energy implies

T− + V− = T+ + V+ ,

and, therefore,

T− = T+

since the potential energy is zero in both states. Inserting the definition of kinetic energygives

1

2

3∑i=1

mi

∣∣vi−∣∣2 =1

2

3∑i=1

mi

∣∣vi+∣∣2and since the velocities are individually constant before and after the collision (uniformapproach and departure) we can conclude that

1

2M∣∣v−∣∣2 =

1

2M∣∣v+

∣∣2 ⇒ 1

2M∣∣vC−∣∣2 =

1

2M∣∣vC+

∣∣2 . (VII.1)

In the last step, we used the fact that vC = vi, for i = 1, 2, 3, because all particles movewith the same velocity and thus the center of mass must also move with the same velocity.From (VII.1) it follows that∣∣vC−∣∣ =

∣∣vC+

∣∣ . (VII.2)

Therefore, the absolute value of the velocity of the center of mass remains the same overthe collision.

(b ) Non-rigid-body system of particles

If we replace the rigid links with springs, as depicted in Figure VII.2, we can still ap-ply the conservation of energy because the spring forces are potential (and we can simplyinclude them in the potential energy of the system). However, this means |vC | needs notto be constant anymore, given that T + V = const but not T = const., as before.

Figure2.9

m1

m2

m3

Figure VII.2: If we replace the rigid links with springs, the potential of the springs mustbe taken into account and therefore vC is not necessarily constant anymore.

Chapter VII. Examples of particle system dynamics 85

Example VII.2: Two-dimensional collision

Figure2.10

m1m2

v1 v2

Figure VII.3: Two particles of mass m1 and m2 are approaching each other and willeventually collide.

Consider two spherical particles of mass m1 and m2, initially moving with velocities v1

and v2, as illustrated in Figure VII.3. At time t−, the two particles collide. At the impact,the collision forces are impulsive and act in the direction normal to the colliding surfaces(there is no tangential collision due to the smoothness of the surfaces). All external forcesunrelated to the collision are non-impulsive, and hence remain bounded. Determine thevelocities u1 and u2 at time t+ after the impact.

(a ) Linear momentum principle

We first evaluate the linear momentum principle for the whole system, i.e. particles 1and 2 combined:

P =

2∑i=1

F exti .

Integrating this equation between t− and t+ gives∫ t+

t−P dt =

∫ t+

t−

2∑i=1

F exti dt ⇒ P (t+)− P (t−) =

∫ t+

t−

2∑i=1

F exti dt .

Since all external forces are bounded and the time interval [t−, t+] is infinitesimally small,the right-hand side of above equation vanishes, as t+ → t−:

limt+→t−

∫ t+

t−

2∑i=1

F exti dt = 0 . (VII.3)

Note: The impulsive, possibly unbounded collision forces are internal and therefore do notneed to be included in the above equation!

We can deduce from (VII.3) that the linear momentum of the particle system is conservedduring collision:

P (t+) = P (t−) (VII.4)

Next, we divide our equations into a component normal to the collision surface (superscriptn) and a component tangential to it (superscript t), then evaluate the linear momentumprinciple in both directions. The respective velocities in normal and tangential directionare shown in Figure VII.4 .

86

Figure2.11

(a) (b)

vt1

vn1 vn

2

vt2 ut

2

un2un

1

ut1

Figure VII.4: Velocities just before and just after the collision of the two spherical particles.(a) The two masses just before the collision. Note that vn1 > vn2 , otherwise there is nocollision. (b) After the collision, we must have un2 ≥ un1 otherwise mass m1 penetrates intomass m2.

(i) Tangential directionFor the tangential direction, we assumed no collision forces due to the lack of friction.Therefore, there are no impulsive internal forces acting in tangential direction and thelinear momentum is not only conserved for the overall system, but also for each particleindividually, i.e.

P t1 = const. , P t2 = const.

From the above equations it follows that the velocity in tangential direction does notchange during the collision:

ut1 = vt1

ut2 = vt2

(VII.5)(VII.6)

For more detail on this argument, see the derivation leading to (VII.3).

(ii) Normal directionIn the normal direction, the linear momentum of the particle system is conserved, becausethe collision force acting in this direction is an internal force. Evaluating (VII.4) in thenormal direction gives

Pn(t−) = Pn(t+) ⇒ m1vn1 +m2v

n2 = m1u

n1 +m2u

n2 . (VII.7)

(b ) Coefficient of restitution

Formula (VII.7) provides one equation for the two unknowns un1 and un2 . We now in-troduce the parameter e, the coefficient of restitution, to describe the imperfection of thecollision. The parameter e is defined as the ratio between the relative normal velocitiesbefore and after the collision, i.e.,

e = − relative normal velocity afterrelative normal velocity before

= −un2 − un1vn2 − vn1

,

⇒ e =un2 − un1vn1 − vn2

. (VII.8)


The coefficient of restitution e ∈ [0, 1] depends on various physical parameter, such assurface quality, material, shape, etc. For e = 0, (perfectly inelastic collision), both particlesmove with the same normal velocity after the impact, while e = 1 corresponds to theperfectly elastic collision case. This is the only case in which energy is conserved, otherwiseenergy is not conserved! This is the reason why we generally cannot assume conservationof energy in an impact.We now combine use (VII.7) and (VII.8) to obtain an equation for un1 and un2 as a functionof e:

un1 =m1 − em2

m1 +m2vn1 +

(1 + e)m2

m1 +m2vn2 ,

un2 =(1 + e)m1

m1 +m2vn1 +

m2 − em1

m1 +m2vn2 .

(VII.9)

(VII.10)

Note: This result is generally valid for any 2D collision problem that satisfies the assump-tions stated at the beginning of the example, (two particles, non-deformable, impulsivecollision force, non-impulsive external forces, smooth surfaces).

Example VII.3: Dynamics of a double pendulumConsider a double pendulum shown in Figure VII.5, swinging in the

[ex, ey

]- plane under

the effect of gravity. The masses m1 and m2 are linked to each other by a rigid, masslessbar of length l2 and m1 is also linked to a smooth joint via a rigid, massless bar of lengthl1.(a ) Draw a free-body diagram and decide how many generalized coordinates are needed

to describe the problem. Hint: Show that a massless bar can only support tangentialforces.

(b ) Find the equation of motion by applying the appropriate momentum principles.Figure'2.12'

g

m1

m2

l2

l1

'1

'2

r1r2m1

m2

(a) (b)

m2g

m1g

R

OO ex

ey

R21

R12

Figure VII.5: The geometry of a planar double pendulum. (a) The double pendulum withthe generalized coordinates ϕ1 and ϕ2. (b) The free-body diagram of the system.

88

(a ) Free-body diagram and generalized coordinates

(i) Free-body diagramWhen separating the particles involved, we find the forces shown in Figure VII.5. Massm1 is subject to gravity m1g, to the constraint force R exerted by the joint and to theinternal force R12 between the two particles. Similarly, mass m2 experiences the forcesm2g and R21.

(ii) Generalized coordinatesAs already discussed before in Example V.3, a double pendulum in 2D possesses twodegrees of freedom. We will choose ϕ1 and ϕ2 as generalized coordinates to describe theorientation of the two bars relative to the vertical.

(iii) Massless barsA massless bar can only support forces that are parallel to it. Furthermore, those forcesmust be collinear with the bar, as indicated in Figure VII.5 for the forces R12 and R21.To prove this we apply the linear and angular momentum principles to a massless beamwith m→ 0, with the center of mass C in the middle of the bar.Assume that the reaction forces at the ends of the bar have nonzero normal componentsN12 and N21 at the respective ends r1 and r2. When we now apply the linear momentumprinciple and take into account that the bar is massless, we obtain:

P = mrC = N12 +N21, ⇒ 0 = N12 +N21, ⇒ N12 = −N21 , (VII.11)

therefore the two assumed normal force components are negatives of each other. Further-more, the angular momentum principle w.r.t. point C yields

HC + vC × P = r1 ×N12 + r2 ×N21 , ⇒ 0 = r1 ×N12 + r2 ×N21 . (VII.12)

where the left-hand side is 0 because the bar is massless. Combining (VII.11) and (VII.12)gives

(r1 − r2)×N12 = 0 ,

which can only hold if

N12 = N21 = 0 , (VII.13)

given that r1 − r2 6= 0 and N12 ⊥ (r1 − r2). Therefore, the reaction forces at the endsof a massless bar can indeed admit components parallel to the bar. These two parallelcomponents must also cancel each other out by the linear moment principle.

(b ) Equation of motion

To find the equation of motion we apply the angular momentum principle w.r.t. pointO and the work-energy principle. We could also apply the linear momentum principlebut for that choice the constraint force R would appear in the equations. In other words,applying the LMP would give us an additional equation but also an additional unknown, R.


(i) Angular momentum principle w.r.t. OAs stated above, we apply the angular momentum principle about B ≡ O since theconstraint force R does not produce a torque about O. Also, vB ≡ 0 makes the evaluationof the AMP easier. The AMP states

HB + vB × P = M extB , ⇒ HB = M ext

B . (VII.14)

First, we find HB to be

HB = r1 × P 1 + r2 × P 2, (VII.15)

with

r1 = l1(sinϕ1ex − cosϕ1ey) ,

r2 = (l1 sinϕ1 − l2 sinϕ2)ex − (l1 cosϕ1 + l2 cosϕ2)ey ,

which we have obtained from the geometry. Differentiating (VII.15) w.r.t. time yields

HB = r1 × P 1 + r2 × P 2 + r1 × P 1 + r2 × P 2 ,

which can be simplified to

HB = r1 × P 1 + r2 × P 2

= r1 × (m1r1) + r2 × (m2r2), (VII.16)

since

ri ‖ P i = mri, r = 1, 2.

The only forces that contribute to the resultant torque M extB are the gravitational forces:

M extB = r1 × (−m1gey) + r2 × (−m2gey) . (VII.17)

Plugging (VII.16) and (VII.17) into (VII.14) gives the equation of motion

m1r1 × r1 +m2r2 × r2 = −m1g(r1 × ey)−m2g(r2 × ey) (VII.18)

a 2nd -order nonlinear, ordinary differential equation (ODE) in ϕ1 and ϕ2.Note: (VII.18) only gives a single scalar equation since it only possesses a non-vanishingcomponent in the ez - direction (the other two directions yield the trivial result 0 = 0).It remains to compute the expressions r1 and r2, as follows:

r1 = l1ϕ1(cosϕ1ex + sinϕ1ey) ,

r1 = l1ϕ1(cosϕ1ex + sinϕ1ey)− l1ϕ21(sinϕ1ex − cosϕ1ey) ,

r2 = (l1ϕ1 cosϕ1 − l2ϕ2 cosϕ2)ex + (l1ϕ1 sinϕ1 + l2ϕ2 sinϕ2)ey ,

r2 = l1(ϕ1 cosϕ1 − ϕ21 sinϕ1)ex − l2(ϕ2 cosϕ2 − ϕ2

2 sinϕ2)ex

+ l1(ϕ1 sinϕ1 + ϕ21 cosϕ1)ey + l2(ϕ2 sinϕ2 + ϕ2

2 cosϕ2)ey .

90

(ii) Work-energy principleSince there is a rigid link between particles (or, alternatively, there is no internal workdone between particles), we can apply the work-energy principle for a rigid-body system.The work-energy principle for rigid-body systems of particles states that

W ext12 = T2 − T1 .

Since all external forces are potential (gravity) or do no work (constraint force R is normalto velocity v1), the system is also conservative. Therefore, the total mechanical energyE = T + V must be conserved, i.e.,

T + V = const.

or, equivalently,

d

dt(T + V ) = 0 . (VII.19)

The kinetic energy of the system is

T =1

2m1 |r1|2 +

1

2m2 |r2|2 ,

and the potential energy is

V = −m1gl1 cosϕ1 −m2g(l1 cosϕ1 + l2 cosϕ2),

with the zero-level of the potential energy set at O. All necessary quantities have alreadybeen computed and just need to be inserted into (VII.19) to obtain the 2nd order nonlinearODE, describing the motion.

(iii) Equations of motionSubstitution of the relevant quantities into equations (VII.18) and (VII.19) gives thefull set of ODEs for the equations of motion, i.e. a two-dimensional 2nd order system ofnonlinear ODEs in ϕ1 and ϕ2:(

(m1 +m2)l1 −m2l1 cos2(ϕ2 − ϕ1))ϕ1 =

1

2l1m2ϕ

21 sin

(2(ϕ2 − ϕ1)

)+m2g sinϕ2 cos(ϕ2 − ϕ1)

+m2l2ϕ22 sin(ϕ− 2− ϕ1)

− (m1 +m2)g sinϕ1,

((m1 +m2)l2 −m2l2 cos2(ϕ2 − ϕ1)

)ϕ2 = −1

2l2m2ϕ

22 sin

(2(ϕ2 − ϕ1)

)+ (m1 +m2)

(g sinϕ1 cos(ϕ2 − ϕ1)

− l1ϕ21 sin(ϕ2 − ϕ1)− g sinϕ2

).

One can now solve this system numerically after writing it as a four-dimensional systemof 1st order ODEs for the state vector (ϕ1, ϕ1, ϕ2, ϕ2)T .Note: The above procedure produces two equations and is therefore suitable for 2-DOFproblems. For examples involving more than two particles, one has to write out themomentum-equations for individual particles to gain further equations of motions.


Prof. George Haller

Dynamics

VIII. Dynamics of rigid bodiesContents

VIII.1 Rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93VIII.2 Velocity transfer formula . . . . . . . . . . . . . . . . . . . . . . . 94

VIII.2.1 Relative velocities . . . . . . . . . . . . . . . . . . . . . . . 94VIII.2.2 Angular velocity of a rigid body . . . . . . . . . . . . . . . 95VIII.2.3 Velocity transfer formula . . . . . . . . . . . . . . . . . . . 95

VIII.3 Planar rigid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . 96VIII.3.1 Angular velocity for planar rigid body motion . . . . . . . 96VIII.3.2 Instantaneous center of rotation . . . . . . . . . . . . . . . 97

VIII.4 Example problems . . . . . . . . . . . . . . . . . . . . . . . . . . 98VIII.5 Three-dimensional rotations . . . . . . . . . . . . . . . . . . . . . 107

VIII.5.1 Angular velocity of time-dependent rotations . . . . . . . . 111VIII.6 Uniqueness of the angular velocity if a rigid body . . . . . . . . . 114




Chapter VIII

Dynamics of rigid bodies

In the previous chapters, we have only considered systems consisting of particles, i.e. pointmasses with no physical dimensions. This is clearly an idealized setting, which neverthelessturns out to be surprisingly effective in a number of physical situations, as we have seenin several examples. There also exist, however, a number of mechanical problems in whichthe point-mass simplification is no longer an acceptable approximation, and finite sizeobjects need to be studied. Here we still assume that these objects are rigid bodies, i.e.,objects that do not deform, only move. We will introduce the concept of a rigid bodyin Section VIII.1. Section VIII.2 tackles the kinematics1 of a rigid body. The importantspecial case of planar rigid bodies is introduced in Section VIII.3, then multiple examplesrelated to the kinematics are discussed in Section VIII.4.

Figure3.1

z

y

x

B

B

A

rA

rB

rAB

O

Figure VIII.1: A rigid body B is shown. Two arbitrary points A,B ∈ B and their corre-sponding position vectors rA and rB are shown as well.

VIII.1 Rigid body

Consider a body B and two arbitrary points A, B on the body at positions rA, rB describedw.r.t. the inertial frame

[ex, ey, ez

], as illustrated in Figure VIII.1. Let us denote the vector

from A to B as

rAB = rB − rA .1Kinematics focuses solely on features of the motion, such as velocity and acceleration. In contrast,

kinetics is concerned with the relationship between forces and the motion they induce.

93

94 VIII.2. Velocity transfer formula

Extending the notion of a rigid-body particle system, we require that on a rigid body, thedistance between two arbitrary points A and B must remain constant in time.

Definition. A rigid body B is a body for which the distance |rAB| between two arbi-trary points A, B ∈ B remains constant for all times, i.e. ,

|rAB| = const. (VIII.1)

or

d

dt|rAB| ≡ 0 . (VIII.2)

VIII.2 Velocity transfer formula

On rigid bodies, the velocities of two arbitrary points turn out to be related by a formula.This velocity transfer formula will prove to be very useful in deriving equations of motion.In the following, we derive the velocity transfer formula in detail.

VIII.2.1 Relative velocities

Let us write rB as

rB = rA + rAB .

Differentiation in time gives

d

dtrB = rB = rA + rAB ,

or, equivalently,

vB = vA + rAB . (VIII.3)

When we rewrite the rigid-body condition (VIII.1) as

|rAB|2 = rAB · rAB = const.

and differentiate it in time, we obtain

2 rAB · rAB = 0 .

Therefore, rAB remains orthogonal to rAB for all times:

rAB ⊥ rAB . (VIII.4)

The relative velocity rAB of B w.r.t. A can be interpreted such that the point B constantlyrotates around A, generally with varying rotation axis and speed, at a constant distance|rAB| from A (see Figure VIII.2 (b)).

Chapter VIII. Dynamics of rigid bodies 95

Figure3.2

B

A

rA

rB

rAB

0

rAB

vA

Figure VIII.2: One may partition the velocity of B into a rotation with relative velocityrAB around A plus the velocity vA of A.

VIII.2.2 Angular velocity of a rigid body

Equation (VIII.3) gives a preliminary form of the velocity transfer formula we seek. Wewill ultimately need a more readily computable form of this transfer formula, because rABis generally unknown when A and B are arbitrarily selected.Since rAB is always orthogonal to ωAB (cf. (VIII.4)) there exists a vector ωAB (dependentupon the points A and B) such that

rAB = ωAB × rAB .

We refer to the vector ωAB as the angular velocity at A. Thus, by (VIII.3), we obtain

vB = vA + ωAB × rAB . (VIII.5)

It can be shown (see the Appendix) that the angular velocity vector ωAB is in fact inde-pendent of A, i.e., ωAB = ω gives a uniquely defined angular velocity for the rigid body.

VIII.2.3 Velocity transfer formula

Summarizing the above considerations, we obtain the following velocity transfer formulafor a rigid body:

The velocity transfer formula for a rigid body B

vB = vA + ω × rAB . (VIII.6)

vB . . . . velocity of a point B ∈ BvA . . . . velocity of another point A ∈ Bω . . . . . unique angular velocity of BrAB . . . vector pointing from A to B

Note: Unlike rotations, angular velocities are additive. We can therefore find the angularvelocity ω by first identifying partial angular velocities ωi w.r.t. distinguished axes andthen (vectorially) adding them up, i.e.

ω =∑i

ωi ,

96 VIII.3. Planar rigid bodies

as will be done in Example VIII.4. It is important, however, that all ωi must be expressedw.r.t. the same basis, i.e. , coordinate frame. For more information on how vectors can betransformed from one basis to another, see Section IX.7.

VIII.3 Planar rigid bodies

A common special case is planar rigid-body motion (see Figure VIII.3 (a)) in which allpoints in the body have velocities parallel to the same fixed plane P. The projection ofany point of the rigid body onto this plane can then be tracked by a vector

r(t) = x(t)ex + y(t)ey ,

assuming that the plane of motion is the[ex, ey

]-plane. This yields major simplifications

compared to general three-dimensional motion, mainly due to a simplification in the an-gular velocity ω.

Figure3.3

z

y

B

B

A

rA

rB

rAB

(a)

B

ArAB

rAB

!AB

(b)

ex

ey

ez

P

Figure VIII.3: (a) A planar rigid body B. (b) The angular velocity is perpendicular to theplane of motion P.

VIII.3.1 Angular velocity for planar rigid body motion

By the definition of planar rigid-body motion, for any two points A,B ∈ B, we must havevA, vB ∈ P, and hence vB − vA ∈ P. Since we also have rAB ∈ P, the velocity-transferformula (VIII.41) implies

vB − vA = ω × rAB =⇒ ω ⊥ P .

Indeed, if ω 6⊥ P were the case, then rAB could not lie in P. We, therefore, obtain thefollowing:

Angular velocity ω of a planar rigid body B.For a planar rigid-body motion along the

[ex, ey

]plane, the associated angular velocity

is

ω = ωez,

i.e. , ω is always perpendicular to the[ex, ey

]plane.


How to identify the unique angular velocity ω of a planar rigid B?To find the angular velocity of B we can apply the following steps. First, we put a fixedmark (straight line) on the body and one in the inertial frame, see Figure VIII.4 (a), withθ denoting the angle between the two marks. As the body moves the angle θ between themarks changes, and the time-derivative of θ gives us the single nontrivial angular-velocitycomponent,

ω = θ .

In vectorial form, we therefore have

ω = θez . (VIII.7)

The above result turns out to be independent of the choice of the marks. Indeed, considertwo different marks, where the new mark on B encloses an angle α with the original oneand the new mark in the inertial frame encloses an angle β compared with the original one(see Figure VIII.4 (b)). The evolving angle θ′ measured between the new marks and theangle θ measured between the two original marks are related via

θ′ = θ + α− β .

When we take the time derivative of this relation, we obtain

θ′ = θ ,

and, therefore, ω = θ′ ≡ θ is indeed independent of the choice of markers.

Figure3.4

B

B

↵

0

(a) (b)

Figure VIII.4: Uniqueness of the angular velocity ω. (a) Choose two marks, one body-fixedand one inertial. (b) Select another mark on B, and another mark on the inertial frame.The angles θ′ and θ only differ by a constant angle and hence θ′ ≡ θ.

VIII.3.2 Instantaneous center of rotation

For a planar rigid-body motion, the instantaneous center of rotation O is a point aboutwhich all points of the body rotate instantaneously without any additional translation.Such a point is purely geometric, i.e. , does not have to be an actual material point that ispart of the body. By definition, O must have zero velocity when viewed on the rigid bodyB extended to the whole plane.

98 VIII.4. Example problems

Definition. The instantaneous center of rotation O of a planar rigid body B is definedas the point O ∈ B′ for which

vO = 0 .

Hence the extended rigid body B′ includes the original body B and the instantaneouscenter of rotation O.

Since B′ moves with the same angular velocity as B, any point A ∈ B satisfies, by thevelocity transfer formula,

vA = ω × rOA ,

given that vO = 0. This observation will help us locate O in specific problems.

VIII.4 Example problems

Example VIII.1: Slider-crank mechanismConsider a system of three rigid bodies, consisting of two cranks and one slider as shown inFigure VIII.5. Such mechanism is used to transform a linear, periodic motion of the sliderinto a rotary motion. The same mechanism is also used in the cylinders of car engines,where the piston is connected to the crank shaft via a rod to transform the linear motionof the piston into a rotary motion of the crank shaft. We want to determine the velocityvB for a given instantaneous velocity v0 of the slider in the configuration shown.

Figure3.5

C

A

B

l

2l

a

1

2

3

v0

↵

Slider (prismatic joint)

System of 3 rigid bodies

ex

ey

Figure VIII.5: A slider-crank mechanism consisting of three rigid bodies, two cranks andone slider.

Solution:


Points A and B are part of the crank AB of length l and angular velocity ω1. We canapply the velocity transfer formula between A and B to obtain

vB = vA + ω1 × rAB . (VIII.8)

Point A does not move, therefore

vA = 0 .

We denote the angle between the horizontal and the current position of crank AB by α.We can then write ω1 and rAB as

ω1 = −αez ,rAB = −l cosαex + l sinαey .

When inserting the above into (VIII.8) we obtain

vB = (−αez)× (−l cosαex + l sinαey) . (VIII.9)

Since B is also part of the longer crank BC of length 2l, we can also apply the velocitytransfer formula between B and C to obtain

vB = vC + ω2 × rCB . (VIII.10)

(Note that ω1 6= ω2 as they describe motions of two different bodies.) Using the angle βbetween the horizontal and the current position of crank BC, we can express ω2 and rCBas

ω2 = βez ,

rCB = 2l cosβex + 2l sinβey .

The velocity vC is given as

vC = v0 = −v0ex .

Substitution into (VIII.10) then gives

vB = −v0ex + (βez)× (2l cosβex + 2l sinβey) . (VIII.11)

We now equate (VIII.9) and (VIII.11) to find that

(−αez)× (−l cosαex + l sinαey) = −v0ex + (βez)× (2l cosβex + 2l sinβey) ,

which we rearrange such that all terms parallel to ex are on left side and all terms parallelto ey are on the right side:

ex(lα sinα+ v0 + 2lβ sinβ) = ey(−αl cosα+ 2lβ cosβ) .

Since ex ⊥ ey both sides of the above equation must be zero, yielding two equations for α,α, β and β:

lα sinα+ v0 + 2lβ sinβ = 0 , (VIII.12)

−αl cosα+ 2lβ cosβ = 0 . (VIII.13)


From (VIII.13) we obtain

β =α cosα

2 cosβ,

and from (VIII.12), combined with the above result for β, we obtain

α = − v0

l(sinα+ cosα tanβ). (VIII.14)

Using the geometry of the system, we can relate α and β to each other as

2l sinβ − a = l sinα .

We can then express sinβ and cosβ from this last equation as

sinβ =a+ l sinα

2l, cosβ =

√1− (a+ l sinα)2

4l2.

We use the above relations in (VIII.14) to obtain α as a function of α:

α = f(α) = −v0

l

1

sinα+ cosαa+ l sinα√

4l2 − (a+ l sinα)2

. (VIII.15)

Finally, we substitute (VIII.15) into (VIII.9) to obtain:

vB = −v0

√4l2 + (a+ l sinα)2(sinαex + cosαey)

sinα√

4l2 − (a+ l sinα)2 + (a+ l sinα) cosα. (VIII.16)

The final result (VIII.16) is linear in v0 but highly nonlinear in the geometry.

Example VIII.2: Two-dimensional rollingConsider a disk that rolls without slipping along ex, as shown in Figure VIII.6. We want

Figure3.6

'

CC

R

A0 A0

v0

sA0

O

OA A

Figure VIII.6: The disk depicted in the figure rolls without slipping on the ground. Themoving point A′ denotes the current point of contact on the ground. This point is shownin both the initial and in the final position of the disk. The two arc-lengths indicated inred are equal under pure rolling.

to find the velocity of the center C and of the point A, where A denotes a material pointon the disk that is currently in contact with the point A′ of the ground. On the otherhand, A′ denotes the point on the ground at the location of instantaneous contact between


the disk and the ground.

Solution:

The rolling without slipping conditions states that the covered arc-length Rϕ of the diskmust be equal to the distance sA′ covered by the point A′, i.e. ,

sA′ = Rϕ .

Taking the time derivative of this equation, we obtain the velocity of A′ as

vA′ = Rϕex .

(Note: When applying the velocity transfer formula to the disk, we cannot include A′ sinceA′ is not part of the disk.) Since the center C of the disk is always vertically aligned withA′, we have

vC = vA′ = Rϕex . (VIII.17)

To obtain the velocity vA of the point A on the disk, we can apply the velocity transportformula between C and A to obtain

vA = vC + ω × rCA= Rϕex − (ϕez)× (−Rey) ,= Rϕex −Rϕex ,= 0 .

Therefore, we have found that in case of pure rolling, the contact point on the disk hasinstantaneously zero velocity:

vA = 0 .

This result also implies that any force arising from the rolling friction does no work onthe disk. Also note that A is just the instantaneous center of rotation defined in Sec-tion VIII.3.2. With the help of A, therefore, we can simply compute the velocity of anypoint B on the disk as

vB = vA + ω × rAB = ω × rAB .

This relationship enables us to construct the velocity vB graphically (see Figure VIII.7).We simply exploit the fact that vB ⊥ rAB and compute the length of vB as

|vB| = ω |rAB| =v0

R|rAB| .


Figure3.7

A

rAB

vB

B

Figure VIII.7: Since the velocity vB of B must be perpendicular to the vector rAB, withA denoting the instantaneous center of rotation, we can construct vB graphically.

Example VIII.3: Ladder sliding down a wallConsider a ladder sliding down a wall, as shown in Figure VIII.8 (a). We seek the instan-taneous center of rotation for the configuration shown.

Solution:

By definition, for O to be an instantaneous center of rotation, we must have

vO = 0 .

We now apply the velocity transfer formula between B and O as well as A and O:

vB = vO + ω × rOB = ω × rOB , (VIII.18)vA = vO + ω × rOA = ω × rOA . (VIII.19)

From the above equation, we obtain that

vA ⊥ r0A and vB ⊥ r0B .

These two orthogonality relations allow us to locate the point O graphically, as shown inFigure VIII.8 (a). The point O located in this fashion is viewed as part of an extendedrigid body co-moving with the ladder. This extended body is indicated schematically witha dotted line in Figure VIII.8. As in the previous example, we again construct the velocityvP of any point P graphically, as shown in Figure VIII.8 (b), and find the magnitude ofvP to be

|vP | = ω |rOP | =|vA||rOA|

|rOP | ,

where ω was computed using some known velocity vA and the velocity transport formulabetween O and A .

Chapter VIII. Dynamics of rigid bodies 103Figure3.8

A

B

vA

vB

0 A

B

vA

vB

0

PvP

r0P

a

(a) (b)

Figure VIII.8: A ladder sliding down a wall. (a) Graphical construction of the instanta-neous center of rotation. (b) Graphical construction of the velocity of an arbitrary pointP .

Example VIII.4: Rolling coneConsider a rigid cone that rolls without slipping on the (x, y) - plane with its tip hinged

Figure3.9

r0A

|rAC |

↵0

A

C

00

x

y

z

R

R

Figure VIII.9: A cone rolls without slipping in the shown configuration.

in O = (0, 0, R), as sketched in Figure VIII.9. The cone has radius R at its basis, alongwhich we select a point C, as shown. The angular velocity component of the cone aboutthe z-axis is given as φ. Compute the instantaneous velocity vC .

(a ) Angular velocity

We want to use the velocity transfer formula to infer the the velocity vC from the ve-locity vO = 0. To this end, we first need to identify the angular velocity ω of the cone. We


denote the point of contact with the ground by A. Since the cone rolls without slipping,we must have vA = 0. Similarly, the tip O of the cone has zero velocity: vO = 0. Whenwe apply the velocity transfer formula between O and A, we obtain

vA = vO + ω × rOA = ω × rOA = 0 ,

which in turn implies

ω ‖ rOA .

Therefore, rOA defines the instantaneous axis of rotation for the cone. To find ω, we nowexamine the geometry of the problem more closely (see Figure VIII.10). The angle αdenotes half the angle of the cone; φ is the angle between the x-axis and the projection ofthe axis rAO onto the (x, y)-plane. We denote the unit vector pointing from the origin ofthe coordinate system towards A by

er = cosφex + sinφey ,

and the component of ω in the radial direction er by ωr. We can now express the angularvelocity ω as

ω = ωr er + ωzez ,

where ωz = φ is given. Since ω ‖ rOA, we find that

tanα = −ωzωr

.

Therefore, the angular velocity of the cone can be written as

ω = − φ

tanα

(cosφex + sinφey

)+ φez . (VIII.20)Figure3.10

↵0

A

C 00

x

y

z

R

R

R

!

!r

!z = ez

2

e' er

Figure VIII.10: The detailed geometry of the rolling cone. Note the decomposition of theangular velocity ω into the sum of ωr and ωz.


(b ) Instantaneous velocity of point C

The velocity transfer formula between A and C gives

vC = vA + ω × rAC = ω × rAC , (VIII.21)

where we have recalled that vA = 0. The geometry in Figure VIII.10 gives

rAC = R eφ +Rez ,

where eφ = sinφex − cosφey. Thus, substituting into formula (VIII.21), we obtain thevelocity vC in the form

vC =

(− φ

tanα

(cosφex + sinφey

)+ φez

)×(R sinφex −R cosφey +Rez

)

⇒ vC = Rφ

((cosφ− sinφ

tanα

)ex +

(sinφ+

cosφ

tanα

)ey +

1

tanαez

).

Example VIII.5: Cardan driveThe Cardan drive is a broadly used mechanism that enables the transmission of rotationsbetween two non-parallel axes. It has numerous applications in mechanical engineering(see Figure VIII.11 for examples). We sketch its geometry in Figure VIII.12.

Figure VIII.11: Examples of Cardan drives in engineering applications.

In the figure, we have assumed that the driven shaft lies on the yz plane and forms an angleα with the driving shaft, which is aligned with the z axis. We want to determine the ratio|ω2||ω1|

, given α and ω1. The details of the derivation are left to a homework problem. Here,we only note that the angular velocity of the cross part (Cardan cross) can be expressedin two different ways as

ωcross = ω1 + φeφ = ω2 + γeγ . (VIII.22)

This ultimately leads to the expression

|ω2||ω1|

=cosα

cos2 α+ sin2 α cosψ(VIII.23)


for the ratio between the input and output angular velocities (see homework problem fordetails). Clearly, this ratio is not constant even if ω1 is constant, but depends on therotation angle ψ of the driving shaft. In practice, this unevenness in ψ results in drivelinevibrations that need to be compensated for.

Figure3.12

y

x

ze

e'

↵

!1 = e3

!2

'

drivingshaft

drivenshaft

cardan cross

Figure VIII.12: Cardan drive kinematics.

Appendix

VIII.5 Three-dimensional rotations

In this section, we will review a mathematical approach to three-dimensional rotations.Let us first define what such a rotation is:

Definition. A three-dimensional rotation is a linear transformation that preserveslength and orientation.

Based on this definition, we may represent a rotation as a transformation

r → Rr ,

with r ∈ R3 and R : R3 → R3 (see Figure VIII.13 for an example). Since the transforma-

FigureA.1

e1

e2

e3

R r

r

Figure VIII.13: A vector r is shown before and after it is rotated by the transformationR .

tion R is length-preserving, we must have∣∣Rr∣∣ = |r| , ∀r ∈ R3 ,

or, equivalently,

|r|2 =∣∣Rr∣∣2 .

This last equation can also be expressed in terms of the Euclidean inner product (dotproduct) as

〈r, r〉 =⟨Rr, R r

⟩.

107

108 VIII.5. Three-dimensional rotations

A property of the inner product is that⟨Rr, R r

⟩=⟨r, R TRr

⟩,

and therefore

〈r, r〉 =⟨r, R TRr

⟩.

For the last equation to be true for all r ∈ R3, we must have

R TR = I ,

leading to the following two properties a three-dimensional rotation R :

R T = R−1 , (VIII.24)

det(R TR

)= det

(R T)· det

(R)

= det(R)2

= 1 . (VIII.25)

Further note what happens to two orthogonal vectors r ⊥ s ∈ R3 \ 0 after they aretransformed,⟨

Rr, R s⟩

=⟨r, R TRs

⟩= 〈r, s〉 = 0 , (VIII.26)

where we have used (VIII.24) and the orthogonality of r, s. Thus, by equation (VIII.26),we conclude that two initially orthogonal vectors remain orthogonal under any rotation:

Rr ⊥ Rs .

Such a map (i.e. linear and length-preserving) is called an orthogonal transformation R andis referenced as R ∈ O(3), where O(3) refers to the group of orthogonal transformationsin three-dimensions. Any transformation R ∈ O(3) maps a cube ( e1, e2, e3) based at theorigin into another such cube (see Figure VIII.14 for an illustration). Note, however, thatthe orientation ("handidness") of ( e1, e2, e3) is not necessarily preserved by R ∈ O(3).Indeed O(3) also contains transformations that are reflections, not rotations.

FigureA.2

R

e1

e2

e3

x

y

z

x

y

z

e1

e2e3

Figure VIII.14: A transformation R ∈ O(3) maps a cube at the origin to a rotated cubethat is also located at the origin. Note, however, that the orientation of the cube is notnecessarily preserved.


Example VIII.6: Reflection w.r.t. (y, z) - planeA reflection, R ∈ O(3), is also length-preserving but not orientation-preserving (detR =−1). A reflection w.r.t. (y, z) - plane maps the three unit vectors [e1, e2, e3] as follows:

Re1 = −e1 , R e2 = e2 , R e3 = e3 .

We see that the transformation does not preserve orientation when we consider the deter-minant of [e1, e2, e3] before and after the rotation:

det [e1, e2, e3] = 1 , det[Re1, R e2, R e3

]= −1 .

In general, for three linearly independent vectors a, b, c ∈ R3, we find their determinantafter the transformation to be

det[Ra,R b,R c

]= det[R [a, b, c]] = det

(R)

det [a, b, c] .

For the orientation of [a, b, c] to be preserved, we need

sign (det [a, b, c]) = sign(det[Ra,R b,R c

]),

which holds precisely when

det(R)> 0 .

For any transformation R ∈ O(3), Formula (VIII.25) implies

det(R)

= ±1 .

Therefore, a transformation R ∈ O(3) also preserves orientation beyond length if

det(R)

= 1 .

We call this subset of transformations within the orthogonal group O(3), the special or-thogonal group SO(3). More specifically, we have the definition

SO(3) =R ∈ R3×3 : R TR = RR T = I ; det

(R)

= 1. (VIII.27)

Any transformation R ∈ SO(3) describes a rotation in 3D. We state a few propertiesof R ,S ∈ SO(3) in the following:

• RS , S R ∈ SO(3).

• R−1 = R T ∈ SO(3).

• RR−1 = RR T = R TR = I .

However, 3D rotations do not commute in general, i.e. , for two general rotations R ,S ∈ SO(3), we have

RS 6= S R .

Note: Rotations in two dimensions (i.e. , members of the group SO(2)) do commute.


Rotations in three-dimensional spaceAny rotation of a vector r ∈ R3 about a point can be represented as a linear operation

r → Rr , (VIII.28)

where R ∈ SO(3). This matrix group SO(3) is the special orthogonal group defined as

SO(3) =R ∈ R3×3 : R TR = RR T = I ; det

(R)

= 1. (VIII.29)

Example VIII.7: Rectangle rotated twiceConsider a rectangle A in the (y, z) - plane, as shown in Figure VIII.15. We now rotate Afirst around the x-axis by −90 and then by −90 around the y-axis. This does not yieldthe same result as when we first rotate −90 around the y-axis and then −90 around thex-axis. Specifically,

R−90y R−90

x 6= R−90x R−90

y .

FigureA.3

x

y

z

A

R90

x

R90

x

R90

y

R90

y

Figure VIII.15: The rectangle A in the (z, y)-plane is first rotated around the x-axis andthen around the y-axis, and then the other way around. The resulting rotated rectangleshave different orientations, showing that rotations in 3D are not commutative.

Matrix representation of a 3D rotation

To find the matrix representation R of a given rotation in a given basis [e1, e2, e3], we applythe rotation to every basis vector individually, and place these rotated vectors individuallynext to each other as column vectors of a matrix. The resulting matrix gives the matrixrepresentation R w.r.t. the basis [e1, e2, e3]:[

R][e1,e2,e3]

=[Re1 Re2 Re3

].

In Figure VIII.16, the procedure described above is applied to a rotation around the basisvector e2. The resulting transformation matrix R ϕ

2 can be found in (VIII.31).


FigureA.4

e1

e3

'

'

R'

ye3

R'

ye1

e2 = R'

ye2

Figure VIII.16: To find the matrix representation of R of a rotation by ϕ around the e2

axis, R is applied to all basis vectors.

Rotation around the basis vectors

To rotate a vector by an angle φ around one of the three coordinate axes [e1, e2, e3] in thepositive direction, we use the following matrices representing the corresponding rotationin the frame [e1, e2, e3]:

R ϕx =

1 0 00 cosϕ − sinϕ0 sinϕ cosϕ

, (VIII.30)

R ϕy =

cosϕ 0 sinϕ0 1 0

− sinϕ 0 cosϕ

, (VIII.31)

R ϕz =

cosϕ − sinϕ 0sinϕ cosϕ 0

0 0 1

. (VIII.32)

Note: We can decompose any rotation R ∈ SO(3) into a sequence of three such rotations(with generally different angles) around each basis vector of a given basis [e1, e2, e3].

Representation of rotated vectors

A vector r ∈ R3 represented in the [e1, e2, e3] frame as

r =

3∑i=1

ri ei .

Transformed under R ∈ SO(3), the rotated vector Rr can be represented as

Rr = R

3∑i=1

ri ei =

3∑i=1

riR ei =

3∑i,j=1

Rijrj ei ,

where Rij is the (i, j) element of the matrix representation of R in the [e1, e2, e3] basis.

VIII.5.1 Angular velocity of time-dependent rotations

Consider a vector r(t) rotating about a point 0 in three-dimensional space, as illustratedin Figure VIII.17. We represent r(t) in the coordinate frame [e1, e2, e3]. With the initial


FigureA.5

x

y

z

r(0)

r(t) = R(t)r(0)

r(s) = R(s)r(0)

Figure VIII.17: A rotating vector can be described by a time-dependent rotation familyR (t). The trajectory between 0 and the current time t can then be written using theparameter s ∈ [0, t].

position r(0) of the vector, we can represent the rotated vector r(t) as

r(t) = R(t)r(0) ,

where R (t) ∈ SO(3) rotates the vector r(0) to its current position r(t) for any time t ∈ R.We are interested in computing the velocity r(t) in terms of a the matrix R (t). First,

we take the time derivative of r(t), i.e. ,

d

dtr(t) = r(t) = R (t)r(0) .

We can express r(0) as

r(0) = R−1r(t) = R T r(t)

to obtain

r(t) = R (t)R T(t)r(t) .

Since R−1 = R T , we have

RR T = I .

Differentiating this last equation in time, we obtain

R R T +R R T = 0 ,

⇒ R R T = −R R T . (VIII.33)

Since for any two matrices A ,B , we have

(AB)T = BTAT ,

we find for R R T that

R R T = (R R T )T .

Rormula (VIII.33) can therefore be rewritten as

R R T = −(R R T )T ,


which shows that R R T is a skew-symmetric matrix. We define the angular velocity matrixas

Ω (t) := R R T ∈ R3×3 , (VIII.34)

which is a skew-symmetric matrix (Ω = −Ω T ) defined for any rotation R (t) ∈ SO(3). Wehave therefore obtained an expression for the velocity of r(t) in terms of Ω (t):

Velocity of a rotating vectorA vector r(t) rotating around a point as

r(t) = R(t)r(0) ,

with R(t) ∈ SO(3), has velocity

r(t) = Ω(t)r(t) , (VIII.35)

where Ω (t) := R R T ∈ R3×3, Ω = −Ω T .

In general, any skew-symmetric matrix Ω ∈ R3×3 is of the form

Ω =

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

.

We observe that

Ω r =

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

r1

r2

r3

=

−ω3r2 + ω2r3

ω3r1 − ω1r3

−ω2r1 + ω1r2

=

∣∣∣∣∣∣e1 e2 e3

ω1 ω2 ω3

r1 r2 r3

∣∣∣∣∣∣ .Therefore, with the angular velocity vector ω = ω1 e1 + ω2 e2 + ω3 e3, we can write

Ω r = ω × r .Note: This result holds true for any skew-symmetric matrix.

With the help of the last equation we can rewrite (VIII.35) as

r(t) = Ω r = ω × r .

Angular velocity vector of a rotationThe velocity of a vector r(t) = R(t)r(0) rotating about a fixed point can be describedas

r(t) = ω(t)× r(t) , (VIII.36)

where ω denotes the corresponding angular velocity vector of the rotation familyR (t) ∈ SO(3). The vector ω is uniquely determined by the relationship

Ω r = ω × r , ∀r ∈ R3 ,

for any skew-symmetric matrix Ω ∈ R3×3.

114 VIII.6. Uniqueness of the angular velocity if a rigid body

VIII.6 Uniqueness of the angular velocity if a rigid body

Let us consider three points A,B,C ∈ B of a rigid body B that are not collinear as shownin Figure VIII.18.

FigureA.6

BA

B

CrAC

rAB

rBC

Figure VIII.18: Three non-collinear points A,B,C ∈ B of a rigid body B.

Using the velocity transfer formula (VIII.5), we express the velocity of point C as

vC = vA + ωA × rAC

or, alternatively, as

vC = vB + ωB × rBC .

Subtracting the above two equations from each other gives

vB − vA = ωA × rAC − ωB × rBC . (VIII.37)

We also know from (VIII.36) that

vB − vA = ωA × rAB . (VIII.38)

(Note: When we consider the rotation of any two points B,C ∈ B around the point A ∈ Bthe rotation of these points is described by the same rigid-body rotation with the samelocal angular velocity ωA. Equating (VIII.37) and (VIII.37), we obtain

ωA × rAB = ωA × rAC − ωB × rBC ,

or, equivalently,

ωA × (rAB − rAC) = −ωB × rBC . (VIII.39)

Since

rAB − rAC = −rBC

we can also rewrite (VIII.39) as

(ωA − ωB)× rBC = 0 ,

which implies

(ωA − ωB)× eBC = 0 , (VIII.40)


where eBC =rBC|rBC |

denotes the unit vector pointing from B to C. We observe that

in (VIII.40), the term (ωA−ωB) does not depend on our choice of C. Therefore, by varyingC, we may make eBC point in any direction without influencing the term (ωA−ωB). Thisimplies that (VIII.40) must be true for any unit vector eBC = e ∈ R3, i.e. ,

(ωA − ωB)× e = 0 , ∀e ∈ R3 .

This may only hold true if

ωA = ωB .

Since our choice of the points A,B ∈ B was arbitrary, we conclude that

ω ≡ ωA ≡ ωB ,

i.e. there exists a unique angular velocity ω for the whole body B. We can now rewrite (VIII.5)with this unique angular velocity ω to obtain the velocity transfer formula

vB = vA + ω × rAB (VIII.41)

116 VIII.6. Uniqueness of the angular velocity if a rigid body


Prof. George Haller

Dynamics

IX. Kinetics of rigid bodiesContents

IX.1 Degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . 119IX.2 Linear momentum principle . . . . . . . . . . . . . . . . . . . . . 124IX.3 Angular momentum principle . . . . . . . . . . . . . . . . . . . . 127IX.4 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

IX.4.1 Calculation of the angular momentum when Point B isfixed or coincident with the center of mass C . . . . . . . . 128

IX.4.2 Calculation of the angular momentum when point B isarbitrary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

IX.5 Properties of the moment of inertia tensor . . . . . . . . . . . . . 131IX.5.1 Two-dimensional motion . . . . . . . . . . . . . . . . . . . . 131IX.5.2 Additivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133IX.5.3 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . 134IX.5.4 Parallel axes theorem (Steiner’s theorem) . . . . . . . . . . 135IX.5.5 Principal axes of a body B . . . . . . . . . . . . . . . . . . 137IX.5.6 Tables with moment of inertia tensors . . . . . . . . . . . . 139

IX.6 Work-energy principle . . . . . . . . . . . . . . . . . . . . . . . . 140IX.7 Rotating frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

IX.7.1 Rotation transformation . . . . . . . . . . . . . . . . . . . . 143IX.7.2 Differentiation of quantities in rotating frames . . . . . . . 145




Chapter IX

Kinetics of rigid bodies

Based on the kinematic concepts discussed in the previous chapter, we will now derive themomentum and energy principles for a rigid body. First, we discuss the degrees of freedomof rigid bodies in Section IX.1, then we derive the linear momentum principle for rigidbodies in Section IX.2 and the angular momentum principle in Section IX.3. We will alsotake a more closer look at computing the angular momentum in Section IX.4, since thegeneral formulation is computationally quite involved. Additional practical tools to com-pute the angular momentum will be presented in Section IX.5. The work-energy principlefor general and for conservative systems, respectively, will be discussed in Section IX.6.Subsequently, in Section IX.7, we introduce the concept of rotating frames and illustratetheir use in applying the momentum principles. Afterwards, several example problemsfor planar rigid bodies and three-dimensional rigid bodies are presented in Section X andin Section XI. As an application, in Section XII.1, we discuss the Euler equations for agyroscope, i.e., a three-dimensional rigid body hinged at a point.

IX.1 Degrees of freedom

In Section V.3 we already discussed the degrees of freedom of a system of particles. Recallthat the degrees of freedom for such a system are defined as the number of independentgeneralized coordinates that are necessary to describe completely the motion of the sys-tem. For one particle, in particular, we need three coordinates to describe its unconstrainedmotion in 3D. When we add holonomic constraints (c.f. V.2), each such independent con-straint reduces the degrees of freedom by one.

To derive the degrees of freedom for a rigid body B, we seek a minimal rigid-body systemof particles embedded in the body, such that the motion of the system of particles uniquelydetermines the motion of the body. The DOF of the body is then taken to be the DOFof this particle system. Observe that the rigid triangle formed by the three non-collinearparticles A,B,C ∈ B, constrained by three rigid bars, already determines the position ofa 3D rigid body B uniquely, as shown in Figure IX.1 (a). Using (V.10), we find the DOFof the ABC particle system, to be

DOFB = 3 · 3− 3 = 6

for the unconstrained rigid body B. This result is consistent with our expectations, becausethe motion of a rigid body can be characterized by the position (xC , yC , zC) of its center

119

120 IX.1. Degrees of freedom

of mass C and three orientation angles.Note: We could also choose six independent position coordinates to describe the motionof B by tracking the motion of two points on B.

Figure4.1

BA

B

CB

AB

(a) (b)

y

x

z

y

x

Figure IX.1: Determination of the degrees of freedom of a rigid body. The procedureis based on finding an equivalent rigid-body system of particles fixed on the body thatfully determines the motion of the body. (a) 3D: A rigid-body system of three particlesdescribes the motion of a 3D rigid body uniquely. (b) 2D: For a planar rigid body tworigidly connected particles are sufficient to fully describe its motion.

Similarly, for an unconstrained planar rigid body B we observe that a rigid bodysystem of two points embedded in B suffices to describe the motion of B uniquely, asshown in Figure IX.1 (b). Again, using (V.10), we obtain

DOFB = 2 · 2− 1 = 3

for the DOF of a planar rigid body B. Now, for L unconstrained rigid bodies, respectivelyL unconstrained planar rigid bodies, the degrees of freedom are

3D: DOFB = 6L ,

2D: DOFB = 3L

and for L rigid bodies subject to K independent holonomic constraints

fj(r1(t), . . . , rL(t), t) = 0, j = 1, . . . ,K,

we obtain the degrees of freedom

3D: DOF = 6L−K ,

2D: DOF = 3L−K

The degrees of freedom of a system of rigid bodies

3D: DOF = 6L−K (IX.1)2D: DOF = 3L−K (IX.2)

DOF . . . degrees of freedom of the rigid-body systemL . . . . . . number of rigid bodies in the systemK . . . . . number of independent holonomic constraints on the system

Chapter IX. Kinetics of rigid bodies 121

Note: For a system of DOF degrees of freedom, we need DOF independent 2nd order ODEsto fully describe the motion of the system. This set of differential equations is called theequations of motion of the system.

Example IX.1: Possible holonomic constraints for a planar rigid bodyWe have already seen holonomic constraints before in our discussion of particle sys-tems (see Example V.1). Recall that a holonomic constraint is a scalar equationthat only depends on the current position of the system and time. We review heresome typical holonomic constraints for rigid bodies.

Figure4.2

(a) (b)

Cyc

xc

'

CR

Aex

ey

Figure IX.2: Typical holonomic constraints. (a) A horizontal slider whose verticalmotion and rotations are constrained. (b) Rolling without slipping constraint.

(a ) Slider

Consider a planar rigid body that can slide between two bars, as shown in Fig-ure IX.2 (a). The body cannot move in the y-direction and cannot rotate. Theserestrictions can be formalized as two independent holonomic constraints

yC − const. = 0 ,

and

ϕ = 0 ,

where C denotes the center of mass. Therefore, the degrees of freedom of thistwo-dimensional rigid body reduce to

DOF = 3 · 1− 2 = 1 .

This result is also intuitively deducible as the slider can only move in the x-direction.

(b ) Rolling without slipping in 2D

A disk of radius R rolls without slipping along ex as shown in Figure IX.2 (b).We find the motion to be constrained by two scalar equations. First, the center ofmass C cannot move in ey-direction,

yC = 0 ,

122 IX.1. Degrees of freedom

or, equivalently,

yC(t)−R = 0 ,

where yC denotes the position of the center of mass in the ey-direction. The secondconstraint states that the contact point A is the instantaneous center of rotationand therefore vA = 0. Via the velocity-transfer formula

vC = ω × rAC ,

we find the velocity of the center of mass C in the ex-direction to be

xC = −ϕR ,

or, equivalently,

xC + ϕR = 0 .

This relation appear to be a non-holonomic constraint, but we can actually inte-grate it to a holonomic constraint of the form

xC + ϕR = const.

We call such constraints integrable non-holonomic constraints since they can beintegrated to a form that only depends on the current position and time, as for atrue holonomic constraint. We then have two holonomic constraints in this example,and hence the degrees of freedom are given by

DOF = 3 · 1− 2 = 1.

Example IX.2: Non-holonomic constraints for planar rigid bodiesWe will now state three examples of genuine non-holonomic constraints that cannotbe integrated into holonomic constraints.

Figure4.3

(a)

C

y

x

y

x

y

x

A

'

'

e vA

yA

xA

yC

yT

xC

'vT

v0 = const

(b) (c)

ex

ey

vC = xCex + yCey

vT t

Figure IX.3: Examples of non-holonomic rigid-body constraints. (a) The constraintdescribing the steering of a car. (b) An airplane pursuing another airplane. (c) Ahorizontal slider with imposed velocity.


(a ) Steering/propulsion constraint

We consider a distinguished boundary point A in the line of symmetry of a body,as seen in Figure IX.3 (a). This constraint requires that the velocity of A mustalso fall on this line of symmetry, and hence the body does not skid sideways (thepoint A is the middle point of the rear axis). The velocity

vA = xAex + yAey

of A must therefore point along the axis of symmetry. We let e denote a unit vectoralong the rear axis,

e = − sinϕex + cosϕey ,

where ϕ denotes the angle between ex and the axis of symmetry of the body. Thesteering constraint requires that

vA · e = 0 ,

as these two vectors must be orthogonal to each other. From this relation, weobtain

(xAex + yAey) · (− sinϕex + cosϕey) = 0 ,

or, equivalently,

xA −yA(t)

tanϕ(t)= 0 .

This constraint is non-holonomic and not integrable. As a consequence, it does notdirectly restrict admissible positions, but restricts the admissible velocities.

(b ) Pursuit constraint

An airplane moves with velocity vT in the positive ex-direction. A second airplanewith velocity vC = xCex + yCey chases and constantly targets the first airplane, asshown in Figure IX.3 (b). The position of the first airplane is

rT = vttex + ytey

and the position of the chasing airplane is

rC = xCex + yCey .

We denote the angle between the airplanes by ϕ, and use the following equation todescribe the constraint on the motion of the chasing airplane:

tanϕ = − yCxC

=yC − yTvT t− xC

,

or, equivalently[vT t− xC(t)

]yC(t) +

[yC(t)− yT

]xC(t) = 0

Again this is a truly non-holonomic constraint.

124 IX.2. Linear momentum principle

(c ) Horizontal slider with imposed motion

We look at the same slider as in the previous example, but this time the mo-tion xC in x-direction is imposed to be xC = v0, as shown in Figure IX.3 (c). Wehave already seen that the classic horizontal slider is subject to two constraints,

yC − const. = 0 and ϕC = 0 .

We have now imposed a third constraint

xC = v0 .

When we integrate this relation, we obtain

xC(t) = xC(t0) + v0(t− t0) ,

⇒ xC(t)− v0t− xC(t0) + v0t0 = 0.

Note that the integrated form of this constraint depends on the initial position xC(t)of the slider, and therefore, strictly speaking, this constraint is not holonomic. Ifwe still count it as holonomic constraint, we would obtain

DOF = 3− 3 = 0 .

This would contradict the fact that one can still freely choose the initial positionof the slider. A possibility is to count this constraint as half a constraint:

DOF = 3− 2− 1

2=

1

2.

Note: This correctly suggests that half of a full degree of freedom is still available forthe slider. From above examples we can see that counting DOF in the presence of non-holonomic constraint can be very subtle. For this reason we will not pursue this topic herefurther.

IX.2 Linear momentum principle

To derive the linear momentum principle for a rigid body B of mass M in an inertialframe,

[ex, ey, ez

], we will first divide the body into N parts of mass ∆mi at position ri, as

shown in Figure IX.4. For sufficiently small ∆mi, we can approximate the mass elementsas particles. We can then write the linear momentum P of B as the linear momentum ofa system of N particles, and take the limit ∆mi → 0, N →∞:

P = limN→∞

∆mi→0

N∑i=1

vi∆mi =

∫B

v dm, (IX.3)

where v = v(x, y, z) is the velocity for each point. Further, dm denotes the infinitesimalmass element that can also be expressed in terms of the volume density ρ = ρ(x, y, z) andthe infinitesimal volume element dV as

dm = ρ dV = ρ dx dy dz .

Chapter IX. Kinetics of rigid bodies 125Figure4.4

z

y

x

B

e1

e2

e3

rB

ri

rCM

CMi

C

rBC

iBmi

M

F extMext

B

Figure IX.4: A three-dimensional rigid body B subjected to external forces F ext andexternal torques M ext. To derive the LMP and AMP, we partition B into infinitesimallysmall pieces of mass ∆mi. B denotes an arbitrary reference point and C = CM denotesthe center of mass.

Note: Since B is a three-dimensional rigid body, ρ is a volume density with units [ρ] = kg/m3.The mass M of B can be calculated as

M =

∫B

dm. (IX.4)

As we did before for systems of particles, we now define a center of mass C for the rigidbody B. We do this by simply taking the limit of ∆mi → 0, N → ∞ in the vector rCpointing to the location of center of mass for a system of particles (cf. V.13).

rC = limN→∞

∆mi→0

1

M

N∑i=1

ri∆mi =1

M

∫B

r dm.

By definition, therefore, the center of mass rC of the rigid body B is the point with respectto which the mass moment of B vanishes, i.e. ,∫

B

%C

dm = 0 , (IX.5)

with the notation %C

= r − rC .

Center of mass of a rigid body B

rC =1

M

∫B

r dm =1

M

∫B

r ρdV (IX.6)

126 IX.2. Linear momentum principle

rC . . . position vector of the center of massr . . . . . position vector to a general point of Bdm . . . infinitesimal mass elementρ . . . . . volume density of B, [ρ] = kg/m3

dV . . . infinitesimal volume element

Using (IX.6), we can write the linear momentum (IX.3) as follows,

P = M1

M

∫B

v dm = M1

M

∫B

d

dtr dm = M

d

dt

1

M

∫B

r dm

= Md

dtrC = MvC ,

where we have denoted the velocity of the center of mass by vC .

Note: We exploited the fact that the mass distribution of the rigid body B does not changein time, otherwise we could not have interchanged the order of integration and differenti-ation in the above derivation.

Definition. The linear momentum of a rigid body of mass M , whose center of massmoves with velocity vC , is defined as

P := MvC . (IX.7)

In the same manner, we can also derive the linear momentum principle for a rigidbody as the limit of the same principle for systems of particles. Indeed, taking the limit∆mi → 0, N →∞ as the LMP for such systems of particles gives:

The principle of linear momentum for a rigid body B

P = MaC = F ext . (IX.8)

P . . . . . linear momentum of BM . . . . mass of BaC . . . . acceleration of the center of mass of BF ext . . . resultant external force acting on B

Note: Conservation of linear momentum: if F ext ≡ 0 then linear momentum is conserved,P ≡ const.


IX.3 Angular momentum principle

We again consider a rigid body B of mass M , divided into N small pieces of mass ∆mi

at ri, as shown in Figure IX.4. The center of mass is located at rC and an arbitrary (andpotentially moving) point B is located at rB. The vector pointing from B to the masselement ∆mi is denoted by

%iB

= ri − rB .

To find the angular momentum HB w.r.t. the point B, we use the analogous formula fora system of N particles with mass ∆mi and take the limit N →∞ and ∆mi → 0:

HB = limN→∞

∆mi→0

N∑i=1

%iB× P = lim

N→∞∆mi→0

N∑i=1

%iB× (∆mivi) =

∫B

%B× v dm. (IX.9)

Taking then the same limit in the angular momentum principle for systems of particles,gives the following result:

The principle of angular momentum for a rigid body B

HB + vB × P = M extB . (IX.10)

HB . . . . angular momentum of B w.r.t. a point BvB . . . . . velocity of point BP . . . . . . linear momentum of body BM ext

B . . . resultant torque w.r.t. point B

Note: Conservation of angular momentum: if M extB = 0 and one of the three following

conditions is fulfilled:

• vB ≡ 0 or

• B ≡ C or

• vB ‖ vC ,

then the angular momentum of B is conserved, i.e., HB ≡ const.Recall that we can compute the resultant external torque w.r.t. a point B at rB as

M extB =

K∑k=1

rBk × F extk +J∑j=1

M j .

where F extk are external forces acting on the body at rk, and rBk is

rBk = rk − rB .

M j denotes torques directly exerted on the body.

128 IX.4. Moment of inertia

IX.4 Moment of inertia

We now reconsider formula (IX.9) to evaluate the general expression of the angular mo-mentum w.r.t. point B:

HB =

∫B

%B× v dm,

IX.4.1 Calculation of the angular momentum when Point B is fixed orcoincident with the center of mass C

We first express v using the velocity transfer formula w.r.t. the point B as

v = vB + ω × %B,

where %B

= r − rB , then insert this expression into the general formula (IX.9) for theangular momentum:

HB =

∫B

%B× (vB + ω × %

B) dm =

∫B

%B

dm× vB +

∫B

%B× (ω × %

B) dm.

Since either vB = 0 or B ≡ C is assumed, the first term of the equation above vanishes,i.e., ∫

B

%B

dm× vB = 0 .

The expression for the angular momentum then reduces to

HB =

∫B

%B× (ω × %

B) dm.

To simplify the calculations further, we shift the origin of the coordinate system to coincidewith the point B, i.e., let

%B

=

xyz

.

Recall that the cross product among three vectors a, b and c satisfies

a× (b× c) = (a · c)b− (a · b)c .

We use this identity to conclude

%B× (ω × %

B) = (%

B· %

B)ω − (%

B· ω)%

B.

Substituting the coordinates of %B

as well as

ω =

ω1

ω2

ω3


into the above expression gives

%B× (ω × %

B) =

(x2 + y2 + z2

)ω1

ω2

ω3

− (xω1 + yω2 + zω3)

xyz

=

(y2 + z2)ω1 − yxω2 − zxω3

(x2 + z2)ω2 − xyω1 − zyω3

(x2 + y2)ω3 − xzω1 − yzω2

=

y2 + z2 −xy −xz−xy x2 + z2 −yz−xz −yz x2 + y2

ω1

ω2

ω3

.

With this result, we may rewrite the angular momentum HB as

HB =

∫B

%B× (ω × %

B) dm = I B ω . (IX.11)

Here I B is the symmetric, positive definite moment of inertia tensor of the body B w.r.t.the point B, defined as

I B :=

Ixx Ixy IxzIxy Iyy IyzIxz Iyz Izz

, (IX.12)

with

Ixx =

∫B

(y2 + z2) dm, Ixy = −∫B

xy dm, Ixz = −∫B

xz dm,

Iyy =

∫B

(x2 + z2) dm, Iyz = −∫B

yz dm, Izz =

∫B

(x2 + y2) dm.

The angular momentum of a rigid body B w.r.t. a fixed point B orw.r.t. its center of mass B ≡ C

HB = I Bω (IX.13)

HB . . . angular momentum of B w.r.t. BI B . . . symmetric, positive definite moment of inertia tensor of B w.r.t. Bω . . . . . angular velocity of B

Note: If B ≡ C then we call I B the centroidal moment of inertia tensor.

IX.4.2 Calculation of the angular momentum when point B is arbitrary

So far, we have only seen how to compute the angular momentum w.r.t. the center ofmass or a fixed point. Unfortunately, there is no easy way to directly compute the angularmomentum w.r.t. an arbitrary point. However, we can relate the angular momenta w.r.t.

130 IX.4. Moment of inertia

two different points to each other. We can then first use equation (IX.13) and then usethe result below to find the angular momentum w.r.t. any desired point.

We first consider the angular momentum HB w.r.t. an arbitrary point B at rB, i.e.,

HB =

∫B

%B× v dm,

and substitute

%B

= rBC + %C.

We also use of the velocity transfer formula to obtain

v = vC + ω × %C.

In these formulas, C denotes the center of mass at rC ; rBC = rC − rB; and %C = r − rC .Using these definitions, we obtain

HB =

∫B

(rBC + %C

)× (vC + ω × %C

) dm

= rBC ×∫B

vC dm+ rBC × ω ×∫B

%C

dm+

∫B

%C× v dm. (IX.14)

(Note that we did not replace v in the last term.) The first factor

P =

∫B

vC dm

in the above equation is just the linear momentum of B. The second integral term is equalto 0 by definition of the center of mass as discussed before:∫

B

%C

dm = 0 .

The last integral is just the angular momentum of B w.r.t. C:

HC =

∫B

%C× v dm,

whose computation we already discussed. Therefore, (IX.14) becomes

HB = HC + P × rCB . (IX.15)

We can repeat the same argument for any other point A at rA and obtain

HA = HC + P × rCA (IX.16)

with rCA = rA − rC . Subtracting (IX.16) from (IX.15), we obtain

HB −HA = P × (rCB − rCA)


or

HB = HA + P × rAB,

where rAB = rCB − rCA = rB − rA. With this angular momentum transfer formula, wecan relate the angular momenta of some rigid body B w.r.t. two arbitrary points A, B toeach other.

The angular momentum transfer formula for a rigid body B

HB = HA + P × rAB (IX.17)

HB . . . angular momentum of B w.r.t. some arbitrary point BHA . . . angular momentum of B w.r.t. some arbitrary point AP . . . . . linear momentum of BrAB . . . vector pointing from point A to point B

IX.5 Properties of the moment of inertia tensor

In the following, we discuss various properties of the moment of inertia tensor that furthersimplify the computations of the angular momentum.

IX.5.1 Two-dimensional motion

Figure4.5

z

y

x x

z

y

!

Figure IX.5: A planar rigid body moving in the (x, y) - plane.

In general, we use (IX.13) to compute the angular momentum w.r.t. the center of mass ora fixed point:

HB = I Bω .

For planar rigid bodies (see Figure IX.5), we can simplify this equation further. As weknow from Chapter VIII, the angular velocity in the 2D-case only has a component inz-direction, i.e.

ω = ωez .

with this simplification, (IX.13) can be evaluated as

HB = (Ixz + Iyz + Izz)ωez = Izzωez = IBωez ,

132 IX.5. Properties of the moment of inertia tensor

because Ixz = Iyz = 0 for a planar rigid body. This yields a significantly simpler way tocompute the angular momentum for a planar rigid body:

Definition. The (scalar) mass moment of inertia w.r.t. a point B at rB for a planarrigid body B is defined as

IB := Izz =

∫B

(x2 + y2

)dm. (IX.18)

The angular momentum of a planar rigid body B w.r.t. a fixed pointB, or its center of mass B ≡ C

HB = IBωez (IX.19)

HB . . . angular momentum of B w.r.t. BIB . . . . mass moment of inertia of B w.r.t. Bω . . . . . angular velocity of B

Example IX.3: Centroidal moment of inertia of a diskFigure4.6

CR

y

x'

rM

drd'

rd'

(a) (b)

Figure IX.6: (a) A planar disk of thickness h. (b) An infinitesimal surface elementin cylindrical coordinates.

Consider a disk of mass M , radius R, thickness h and uniform density ρ, seeFigure IX.6 (a). The disk is constrained to move in the (x, y)-plane, and, hence, canbe viewed as a planar rigid body. Compute the corresponding (scalar) centroidalmoment of inertia IC of the disk.


Solution:

We use formula (IX.18) with the mass element

dm = ρ dx dy dz ,

expressed in cylindrical coordinates (see Figure IX.6 (b)):

dm = ρ rdϕ dr dz .

Noting that r2 = x2 + y2, we compute the centroidal moment of inertia as

IC =

∫B

(x2 + y2

)dm =

∫ h2

−h2

∫ R

0

∫ 2π

0

(x2 + y2

)ρr dϕ dr dz

= ρh

∫ R

0

∫ 2π

0r2r dϕ dr = ρh2π

R4

4=

1

2ρhR2πR2 ,

yielding the final result

IC =1

2MR2 ,

where we have we used the fact that M = ρR2πh.

IX.5.2 Additivity

The moment of inertia tensor is additive because integration is a linear operation. Thisadditivity property can be used for systems of several rigid bodies for which the momentof inertia tensor is either tabulated or much easier to calculate, as detailed below.

Superposition of moments of inertia tensorsOne can calculate the moment of inertia tensor I B of a rigid body B consisting ofseveral component-bodies Bi, i = 1, . . . , n with moments of inertia tensors I iB (aboutsome common point B and w.r.t. the same coordinate system) by the superpositionprinciple, i.e.,

I B =

n∑i=1

I iB. (IX.20)

Example IX.4: Moment of inertia of a Mercedes logoConsider the Mercedes logo consisting of a ring and 3 identical bars depicted inFigure IX.7. We know the moments of inertia IbarB about the center B for a bar andIringB for the ring, also about B. We now want to know what the moment of inertiaI logoB for the whole structure is.


Figure4.7

B B

B

+ 3 =

Figure IX.7: The moment of inertia of the Mercedes logo may be partitioned intothe moment of inertia of three bars and a ring.

To obtain I logoB we can simply use the superposition principle to obtain

I logoB = 3IbarB + IringB .

Example IX.5: Moment of inertia for a hollow cylinderConsider a hollow cylinder of mass M , inner radius R1 and outer radius R2. Whatis the moment of inertia IC of the cylinder w.r.t. its center? Hint : The moment ofinertia of a cylinder of radius Ri and mass Mi about its center is Ii = 1

2MiR2i .

Based on the superposition principle, the moment of inertia with respect to Ccan be found by subtracting the moment of inertia of a cylinder of radius R1 fromthe moment of inertia from a cylinder of radius R2,

IC =1

2M2R

22 −

1

2M1R

21 .

Letting M = M2 −M1 to be the mass of the cylinder, we find after some algebrathat

IC =1

2M(R2

1 +R22) .

Here we have used that

M2R21 −M1R

22 = ρhπR2

2R21 − ρhπR2

1R22 = 0 .

with ρ denoting the density and h denoting the length of the cylinder.

IX.5.3 Symmetries

A rigid body may exhibit various types of symmetry. Two of them will be discussed here indetail because they simplify the computations of the moment of inertia tensor. First, if oneof the axis is an axis of rotational symmetry, all off-diagonal terms of I B involving that axisvanish. For example, if the axis of rotational symmetry is the x-axis (see Figure IX.8 (a))then

Ixy = Ixz = 0 .

Chapter IX. Kinetics of rigid bodies 135Figure4.8

xz

y+y

y

B

xy

z

(a) (b)

B

B+

Figure IX.8: (a) The x-axis is an axis of rotational symmetry. (b) The z-axis is perpen-dicular to the plane of symmetry of the body.

Second, if an axis is perpendicular to a plane of symmetry then all off-diagonal termsof I B involving that axis vanish. For the z-axis, shown in Figure IX.8 (b), it follows thenthat

Iyz = Ixz = 0 .

IX.5.4 Parallel axes theorem (Steiner’s theorem)

We can relate the moment of inertia IB w.r.t. an arbitrary point B to the centroidalmoment of inertia IC via Steiner’s theorem. We will only state the result below withoutderivation (see for geometry Figure IX.9).

Figure4.9

z

x

y

M

CM

B

ab

c

x0

y0

z0

x

y

CM

b

a

y0

x0B

M

(a) (b)

Figure IX.9: When changing the reference point of the moment of inertia tensor from thecenter of mass C = CM to an arbitrary point B, Steiner’s theorem is applicable. (a)Three-dimensional case. (b) Two-dimensional case.


Parallel axis theorem (Steiner’s theorem)

I B = I C +M(|rBC |2 I − rBCrTBC

)(IX.21)

or

I B = I C +M

b2 + c2 −ab −ac−ab a2 + c2 −bc−ac −bc a2 + b2

(IX.22)

I B . . . moment of inertia tensor w.r.t. an arbitrary point BI C . . . moment of inertia tensor w.r.t. the center of mass CM . . . . mass of the bodyrBC . . . vector pointing from point B to center of mass C, rBC = (a, b, c).

Note: Steiner’s theorem is not valid between any two arbitrary points. Furthermore, therole of B and C cannot be interchanged in formula (IX.22).

Finally, Steiner’s theorem implies that the diagonal entries of the moment of inertiatensor are minimal w.r.t. the center of mass.

Parallel axis theorem for a planar rigid body (Steiner’s theorem)

IB = IC +M%2 = IC +M(a2 + b2) (IX.23)

IB . . . moment of inertia w.r.t. an arbitrary point BIC . . . moment of inertia w.r.t. the center of mass CM . . . mass of the body% . . . . distance between point B and center of mass C: %2 = a2 + b2.

Example IX.6: Moment of inertia of a disk about its point of contactwith a planeIn Example IX.3, we computed the centroidal moment of inertia of a disk of radiusR and mass m. Given that the centroidal moment of inertia is IC = 1

2mR2, what

is the moment of inertia IA w.r.t. the contact point A between the disk and thehorizontal plane, as shown in Figure IX.10?


Figure4.10

R

m

C

A

Figure IX.10: A disk of mass m and radius R.

Solution

We can directly apply Steiner’s theorem (IX.23) to compute IA:

IA = IC +m%2 =1

2mR2 +mR2 =

3

2mR2 .

IX.5.5 Principal axes of a body BConsider a basis [u1, u2, u3] in which the moment of inertia tensor becomes diagonal, i.e.,

I B =

I1 0 00 I2 00 0 I3

. (IX.24)

We call the axes [u1, u2, u3] the principal axes of the body, and I1, I2, I3 the principalmoments of inertia. We may find such an orthonormal basis [u1, u2, u3] by solving theeigenvalue problem

(I B − λI )u = 0 , (IX.25)

where we obtain the eigenvalues λi = Ii from the characteristic equation.

det(I B − λI

)= 0 . (IX.26)

(We have denoted by I the identity matrix.) Therefore, the principal axes form an (or-thonormal) eigenbasis of I B, and the principal moments of inertia are the correspondingeigenvalues. The eigenbasis can be selected orthonormal because I B is symmetric. Fur-ther, all eigenvalues are positive, as expected for the moments of inertia.

When we use a basis[ex, ey, ez

]for representing the inertia tensor

[IB

]xyz

, we find the

eigenbasis [u1, u2, u3] by solving the eigenvalue problem stated above. We may then trans-form any vector [r]xyz, represented in the

[ex, ey, ez

]basis, to its representation [r]123 in

the eigenbasis [u1, u2, u3] via

[r]xyz = U [r]123 ,

and

[r]123 = U T [r]xyz ,

where U = [u1, u2, u3] ∈ SO(3).


Note: When both the x and y-axis are axes of symmetry, we have

Ixz = Ixy = Iyz = 0 ,

i.e., all off-diagonal terms of I B become 0. Therefore, the moment of inertia tensor is inits principal form.

Example IX.7: Symmetric objectsConsider the two rigid bodies in Figure IX.11. Discuss for each of them how theirsymmetry properties can help in identifying the principal frame.

Figure4.11

B

e1

e2

e3

e02

e01 e1

e2

e3

(a) (b)

Figure IX.11: (a) A symmetric disk and two of its principal frames. (b) A blockwith e3 as principal axis. The full principal basis cannot be immediately determinedbased on the symmetry.

(a ) Disk

The disk shown in Figure IX.11 (a) has three planes of symmetry, each orthog-onal to one of the three axis [ e1, e2, e3], which therefore define a principal frame.Note that e3 is also an axis of rotational symmetry. Due to the high symmetry ofthe disk, rotating the [ e1, e2, e3] frame about e3 gives another principal frame,such as the frame [ e′1, e

′2, e

′3] shown in the figure.


(b ) Block

For the block shown in Figure IX.11 (b), only one plane of symmetry exists withunit normal e3. To find the other two principal axes e1 and e2 we need to solvethe eigenvalue problem (IX.25). The moment of inertia tensor takes the form

I B =

Ixx Ixy 0Ixy Iyy 00 0 I3

,

because e3 is normal to the plane of symmetry, and hence Izz = I3 is a principalmoment of inertia. We therefore only need to solve the reduced eigenvalue problemfor (

Ixx IxyIxy Iyy

),

in the (x,y) plane.

IX.5.6 Tables with moment of inertia tensors

For common bodies and constant density ρ = const, the moment of inertia tensors are tab-ulated in Figure IX.12. In all cases, the tensor is given in its principal basis. Figure IX.13lists the (scalar) moment of inertia for common two-dimensional rigid bodies.

Figure IX.12: Table of moment of inertia tensors for common 3D rigid bodies with constantdensity.

140 IX.6. Work-energy principle

Figure IX.13: Table of moments of inertia for the most common 2D rigid bodies.

IX.6 Work-energy principle

To derive the work-energy principle for a rigid body, we first define its kinetic energy. Wedivide the rigid body B of mass M into small pieces of mass ∆mi at ri, each with velocityvi (see Figure IX.14).

Figure4.14

B

x

y

i

j rC

ri

mi

%iC

Figure IX.14: The rigid body B is viewed as the union of small mass elements ∆mi.

When these mass elements are sufficiently small, they can be regarded as particles.We can then use expression (VI.14) for the kinetic energy of a particle system and takethe limit ∆mi → 0, N →∞ to obtain

T =1

2limN→∞

∆mi→0

N∑i=1

∆mi |vi|2 =1

2

∫B

|v|2 dm. (IX.27)

This is the general form of the kinetic energy of a rigid body. This form, however, is notdirectly applicable in specific examples, as the velocity distribution as a function of themass distribution is generally not know. To this end, we now relate the velocity v to the


velocity vB via the velocity transfer formula (VIII.41), where B either denotes the centerof mass C or is fixed (vB = 0). With %

B= r− rB and the angular velocity ω, the velocity

transfer formula states

v = vB + ω × %B.

Inserting this expression into (IX.27), we obtain

T =1

2

∫B

(vB + ω × %B

) · (vB + ω × %B

) dm

=1

2

∫B

|vB|2 dm+

∫B

vB · (ω × %B) dm+1

2

∫B

∣∣∣ω × %B

∣∣∣2 dm. (IX.28)

We now evaluate each of these integrals individually. For our choice of B, the secondintegral in (IX.28) always vanishes:

vB ·

ω × ∫B

%B

dm

= 0 .

This is because when B is fixed, then vB = 0. On the other hand, when B ≡ CM , then∫B

%B

dm = 0

holds by the definition of the center of mass. We now take a closer look at the integrandof the third term in (IX.28):∣∣∣ω × %

B

∣∣∣2 =(ω × %

B

)·(ω × %

B

).

We recall the following identity for three vectors a, b, c:

(a× b) · c = (b× c) · a .

When we apply this identity with a = ω, b = %B

and c =(ω × %

B

), we obtain that(

ω × %B

)·(ω × %

B

)=(%B× (ω × %

B))· ω .

Substituting this back into the third term of the kinetic energy in (IX.28) gives

1

2

∫B

∣∣∣ω × %B

∣∣∣2 dm =1

2

∫B

%B× (ω × %

B) dm · ω .

By (IX.11), this can be further evaluated as

1

2

∫B

%B× (ω × %

B) dm · ω =

1

2(I B ω) · ω =

1

2(I B ω)Tω =

1

2ωT I B ω .

Note that we switched from the dot product notation ("·") to matrix multiplication andused the fact that I B = I TB.

142 IX.6. Work-energy principle

The last term to be evaluated is the first integral in (IX.28), which simply yields

1

2

∫B

|vB|2 dm =1

2

∫B

dm |vB|2 =1

2M |vB|2 .

Combining all these results, we obtain the final form of (IX.28) as

T =1

2M |vB|2 +

1

2ωT I B ω

for the kinetic energy T, where B is either fixed (vB = 0) or B coincides with the centerof mass CM .

The kinetic energy T of a rigid body B

T =1

2M |vC |2 +

1

2ωT I C ω (IX.29)

or

T =1

2ωT I B ω (IX.30)

M . . . mass of BvC . . . velocity of the center of massI C . . . centroidal moment of inertia tensorω . . . . angular velocity of BI B . . . moment of inertia tensor w.r.t. a fixed point B (vB = 0).

Note: (IX.30) confirms, as stated before, that the moment of inertia tensor must be positivedefinite, since T ≥ 0 (T is a quadratic form of I B).Note: Recall that for a planar rigid body, both the angular velocity ω and the moment ofinertia tensor I B become scalars in formulas (IX.29)-(IX.30).

To derive the work-energy principle, we consider a rigid-body system of particles andapproximate the rigid body in the same manner as before, i.e., divide it into N pieces ofmass ∆mi, then take the limit ∆mi → 0, N →∞. This leads to the results below.

Work-energy principle for a rigid body B

W ext12 = T2 − T1 (IX.31)

W ext12 . . . work done by the external forces between states 1 and 2

Ti . . . . . kinetic energy of the system associated with state i

Note: As discussed before, the rigid body hypothesis simplifies the work-energy principlesignificantly. If the body is not rigid, the internal forces do work and computing their workbecomes complicated.


As for systems of particles, we can define the potential energy and the work-energyprinciple for conservative systems (i.e., systems with all external forces either being po-tential or doing no work).

Definition. The potential energy V (r) of a rigid body is defined as the sum of thepotentials Vk of all (external) potential forces F k acting on the system, k = 1, . . . ,K:

V (r) :=K∑k=1

Vk . (IX.32)

Work-energy principle for a conservative systemThe total mechanical energy

E := T + V

of a conservative system is constant along all motions, i.e.,

T1 + V1 = T2 + V2 (IX.33)

or, equivalently,

d

dtE(t) ≡ 0

holds for all times.E . . . total mechanical energyTi . . . kinetic energy of the rigid body at state iVi . . . potential energy of the rigid body at state i

IX.7 Rotating frames

In three-dimensional applications, it is often convenient to express mechanical quantitiesin a rotating frame attached to the body. For instance, the moment of inertia tensor maybe time-dependent unless expressed in the frame of the principal axes.Two main challenges arise in describing rigid-body motions in rotating frames. First, weneed to transform vectors from their representation in one frame into their representationin another frame (see Section IX.7.1). Second, the momentum principles only hold truein their classic form in inertial frames. In order to use the momentum principles withquantities expressed in a rotating frame, we need to know how to take derivatives of thesequantities relative to the inertial frame (see Section IX.7.2).

IX.7.1 Rotation transformation

Consider a frame [e1, e2, e3] and an arbitrary vector u expressed in this frame, i.e. [u]e.Next, we introduce another frame

[f

1, f

2, f

3

]that rotates w.r.t. the first frame [e1, e2, e3],

144 IX.7. Rotating framesFigure4.15

B

e1

e2

e3

f1

f2

f3

u

Figure IX.15: A vector u may be represented using the inertial frame [e1, e2, e3] or arotating frame

[f

1, f

2, f

3

].

as shown in Figure IX.15. Then there exists a transformation matrix R ∈ SO(3) thattransforms a vector from its representation [u]e in the e-frame into its representation [u]fin the f -frame:

[u]e = R [u]f .

The three columns of R are just the three unit vectors f1, f

2, f

3, respectively, expressed in

coordinates with respect to the frame [e1, e2, e3]. For example, when the coordinate framef is obtained by rotating the e-frame around the e1-axis by an angle α in the positivedirection (right-hand rule), then the transformation matrix R 1 ∈ SO(3) is given by

R 1 =

1 0 00 cosα sinα0 − sinα cosα

. (IX.34)

Note that in this case, e1 = f1. We can similarly identify the transformation matrices for

the case when the f -frame is obtained by rotating the e-frame about its e2- or e3-axis inthe positive direction, by an angle β or γ, respectively:

R 2 =

cosβ 0 − sinβ0 1 0

sinβ 0 cosβ

, (IX.35)

R 3 =

cos γ sin γ 0− sin γ cos γ 0

0 0 1

. (IX.36)

Note: We may put together any transformation matrix between two frames by applyingthe above three rotation matrices R 1, R 2, R 3 in a sequence, choosing the angles α, β andγ appropriately.

In case the rotation about an axis is negative (by the right hand rule), we simplyconstruct the rotation matrix as the transpose of the rotation matrix for the respectiveaxis in the positive direction. For example, for negative rotations about the e1-axis, wehave

R−1 = R T1 =

[R 1

]−1.


IX.7.2 Differentiation of quantities in rotating frames

Let us consider a vector quantity u represented in a rotating frame[f

1, f

2, f

3

], as illus-

trated in Figure IX.15. Assume that the e-frame is an inertial frame, with respect towhich the f -frame rotates with the angular velocity vector Ω. To express the momentumprinciples in the rotating frame f , we first write them down in the inertial e-frame, thentransform the resulting equations in the f -frame and carry out the time-differentiation. Tothis end, we first express an arbitrary vector u as

u =3∑i=1

uif i

in the rotating f -frame, with ui denoting the coordinates of u w.r.t. fi. Differentiating u

in time yields

u =d

dt

(3∑i=1

uif i

)=

3∑i=1

uif i +

3∑i=1

uif i .

The velocity fiof a rotating unit vector can be calculated using its associated angular

velocity Ω:

fi

= Ω× fi.

Therefore, the derivative u can be expressed as

u =3∑i=1

uif i +3∑i=1

uiΩ× f i .

We rewrite this result as

u = u+ Ω× u ,

denoting the time derivative of u in the moving frame by

u =

3∑i=1

uif i .

Differentiation of quantities in moving frames

u = u+ Ω× u (IX.37)

u . . . absolute time derivative of u in an inertial frameu . . . time derivative of u in the moving frameΩ . . . angular velocity of the rotating frame relative to the inertial frame

All quantities must be expressed in the moving frame!

146 IX.7. Rotating frames

Note: The above formula is only valid if the origin of the f -frame is fixed w.r.t. the inertiale-frame. Otherwise, we must consider that the origin of the f -frame moves as well.Note: For a body-attached frame, the angular velocity Ω of the frame is just the angularvelocity ω of the body:

Ω = ω .

Example IX.8: Bullet in a rotating barrel

Figure4.16

(a) (b)

Cx

y

z

B

xC

0

x

y f1

f2

f3

e1

e2

e3

xC

Figure IX.16: A bullet that moves inside a rotating barrel. (a) Side view of thesystem. (b) View from above.

Consider a bullet in a rotating barrel, as illustrated in Figure IX.16. We wantto apply the linear momentum principle in the rotating frame

[f

1, f

2, f

3

]that is

attached to the barrel by using (IX.37) to differentiate the quantities expressed inthe rotating frame f w.r.t. the inertial frame [e1, e2, e3]. We already derived theseresults before in Example IV.3 using a different approach.

Solution

The position of the bullet is denoted by

[xC ]f = xCf1

in the moving frame f . As aforementioned, we use (IX.37) to correctly differentiatethe position to obtain the absolute velocity and acceleration expressed in the f -frame:

[xC ]f = [vC ]f = xC + Ω× xC = xCf1+ ΩxCf2

,

[xC ]f = [aC ]f = vC + Ω× vC = xCf1+ 2ΩxCf2

− Ω2xCf1.

The angular velocity of the barrel (i.e., of the f -frame) is Ω = Ωf3. The linear

momentum principle states[P]f

= M [xC ]f =[F ext

]f,


with the differentiation carried out in the inertial e-frame. Substituting the expres-sion we have obtained for [xC ]f and rearranging some terms, we obtain

MxCf1=[F ext

]f

+MΩ2xCf1− 2MΩxCf2

.

As expected, this result agrees with that obtained in Example IV.3. Note thatMΩ2xCf1

denotes the centrifugal term and −2MΩxCf2the Coriolis term.

While the use of a non-inertial frame did alleviate the underlying computations,the true merit of formula (IX.37) becomes even more apparent for the angular mo-mentum principle in rotating frames. This is because the moment of inertia tensorI B is often time-dependent when rotating rigid bodies are described in inertialframes. In contrast, the same tensor is constant in its principal frame, which ismost of the time not an inertial frame. Hence, with the use of the results from thissection, we can express quantities in the rotating, principal frame, which in turnallows us to easily compute the angular momentum.

148 IX.7. Rotating frames


Prof. George Haller

Dynamics

X. Examples of rigid-bodydynamics in 2D




Chapter X

Examples of rigid-body dynamics in2D

In this section, we will apply the derived principles to concrete problems involving planarrigid bodies.

Example X.1: Falling stickA stick of mass m, length L and center of mass C at rC is falling to the ground due togravity, as seen in Figure X.1 (a). The point B denotes the contact point with the ground.We will assume that the stick is thin and, therefore, B remains in touch with the groundindependent of the angle ϕ, i.e.,

yB ≡ 0 .

Further, we assume that there is no friction between the ground and the stick. Note thatthe stick can slide along the surface.(a ) Draw a free-body diagram and decide on the necessary generalized coordinates to

describe the problem.(b ) Find the equation of motion by applying the appropriate principles.

(a ) FBD and generalized coordinates

From the free-body diagram in Figure X.1 (b) we see that the only forces acting onthe body are the normal force N attacking at B and the gravitational force mg. Per as-sumption, there is no friction force. The system is subjected to one constraint, i.e., yB = 0and hence DOF = 1× 3− 1 = 2. We choose the angle ϕ and the horizontal position xCof the center of mass as the generalized coordinates to describe the system.


(i) Linear momentum principleThe linear momentum principle for the system states

d

dtP = maC = mg +N . (X.1)

151

152

Figure4.17

i

j

C C

g

B0

xc

yc'

rC

m, l

smooth

B

N

mg

(a) (b)

Figure X.1: A falling stick of mass m and length L on a smooth surface. (a) The necessaryquantities are defined. (b) The free-body diagram of the system is depicted. Note thatthere is no friction force since the surface is assumed to be smooth.

For the x-component of (X.1), we obtain

mxC = 0

since there are no forces acting in x-direction. Integrating this twice yields

xC(t) = xC(0) + vC(0)t . (X.2)

When evaluating (X.1) in the y-direction, we see that this includes the unknown force N .To this end, we apply the angular momentum principle w.r.t. B, which does not includeN .

(ii) Angular momentum principle w.r.t. BWhen using the angular momentum principle w.r.t. B, we avoid having the unknownforce N in our equations (see Figure X.1). The angular momentum principle w.r.t. Bstates

HB + vB × P = MB . (X.3)

In the following we will compute all necessary terms in (X.3). We start with the linearmomentum

P = mrC .

From the geometry we obtain that

rC = xC i+l

2sinϕ j ,

which we then differentiate in time to obtain

rC = xC i+l

2ϕ cosϕ j .

Chapter X. Examples of rigid-body dynamics in 2D 153

This gives the linear momentum

P = m(xC i+l

2ϕ cosϕ j) .

We can also find rB from the geometry and then differentiate it in time to obtain

vB =d

dt(xC −

l

2cosϕ) i = (xC +

l

2ϕ sinϕ) i .

To compute HB, we first compute

HC = ICϕ k,

then use the angular momentum transfer formula

HB = HC + P × rCB .

Following this procedure, we obtain

HB = ICϕ k +m(xC i+l

2ϕ cosϕ j)× (− l

2cosϕ i− l

2sinϕ j)

=

((IC +m

l2

4cos2 ϕ

)ϕ−m l

2xC sinϕ

)k .

Differentiating this equation in time gives

HB =

((IC +m

l2

4cos2 ϕ

)ϕ− 1

4ml2 sin(2ϕ)ϕ2 −m l

2xC sinϕ−m l

2xCϕ cosϕ

)k

=

((IC +m

l2

4cos2 ϕ

)ϕ− 1

4ml2 sin(2ϕ)ϕ2 −m l

2xCϕ cosϕ

)k ,

where we used that xC = 0, which we obtained from the linear momentum principle (X.1).We also need to compute the vector product (vB × P ):

vB × P = m(xC +l

2ϕ sinϕ) i× (xC i+

l

2ϕ cosϕ j)

= ml

2(xC +

l

2ϕ sinϕ)ϕ cosϕ k .

The external torque w.r.t. B is

MB = −mg l2

cosϕ k .

We now insert all terms computed above into (X.3) to obtain the equation of motion

(IC +ml2

4cos2 ϕ)ϕ− 1

8ml2ϕ2 sin(2ϕ) +

1

2mgl cosϕ = 0 . (X.4)

We have not computed the centroidal moment of inertia IC yet. To compute it, we use

IC =

∫B

∣∣∣%C

∣∣∣2 dm

154

Figure4.18

C

m, lB

wx l

(thin stick)

Figure X.2: The stick is assumed to be thin, i.e., wx l.

with

dm = ρ dx dy =m

wxldx dy .

Here, wx is the width of the rod and l is its length (see Figure X.2). Since we assumedthe rod to be thin (wx l), we can write∣∣∣%

C

∣∣∣ = y2 ,

where y denotes the distance from the center of mass in the direction of l. So, we obtain

IC =

∫ l2

− l2

∫ wx2

−wx2

y2 m

wxldx dy .

Integration about y is symmetric, which simplifies the integral to

IC = 2

∫ l2

0

∫ wx2

−wx2

y2 m

wxldx dy .

We evaluate this integral to obtain

IC =2m

wxlwx

((l2

)33− 0

)=

1

12ml2 .

We could have also found the same result in look-up tables for moments of inertia. Finallywe plug this expression for IC into our equation of motion (X.4) to obtain the final result,

1

4ml2

(1

3+ cos2 ϕ

)ϕ− 1

8ml2ϕ2 sin(2ϕ) +

1

2mgl cosϕ = 0 . (X.5)

Together with (X.2), equation (X.5) gives a set of second-order ODEs that describe themotion of the system. Equation (X.2) was easily solved but (X.5) is nonlinear and canonly be solved numerically.Note: The two equations of motion are fully decoupled from each other, i.e. , there is oneseparate equation for each degree of freedom, ϕ and xC . This is usually not the case formulti-degree-of-freedom rigid bodies.


Example X.2: Falling hinged stickWe again consider a falling (thin) stick of mass m and length l, shown in Figure X.3 (a).This time, the contact point B is hinged and hence cannot slide. This system has only onedegree of freedom, prompting us to choose the angle ϕ as the generalized coordinate.(a ) What is the equation of motion for the system?(b ) Find the acceleration of the point A at the top of the stick. Then take its vertical

component for the moment the stick is released from rest and compare it to thegravitational acceleration g. Discuss the result.

Figure4.19

g

C

'B

A A

C

BRx

Ry

mg

(a) (b)

Figure X.3: (a) A falling hinged stick of mass m and length l, hinged at point B, andsubjected to gravity. (b) The corresponding free-body diagram.


To find the equation of motion, we apply the angular momentum principle w.r.t. thehinge point B:

HB + vB × P = MB . (X.6)

Since the reaction forces at the hinge do not produce a torque about B, we only have toinclude the torque arising from gravity,

MB = −mg l2

cosϕ k .

For this problem, we can compute the angular momentum directly because the point B isfixed:

HB = IBϕ k,

with IB = 13ml

2 obtained from a look-up table. Also note that vB = 0 and therefore(vB × P = 0). Inserting these quantities into (X.6), we find the equation of motion

1

3ml2ϕ = −mg l

2cosϕ ,

(X.7)

156

or, equivalently,

ϕ+3

2

g

lcosϕ = 0 . (X.8)

(b ) Acceleration of point A

To find the acceleration aA of the top point A, we differentiate the vector rBA twicein time to obtain

aA =d2

dt2rBA =

d2

dt2(l cosϕ i+ l sinϕ j)

= ld

dt(−ϕ sinϕ i+ ϕ cosϕ j)

= l(−ϕ2 cosϕ− ϕ sinϕ) i+ l(−ϕ2 sinϕ+ ϕ cosϕ) j .

We now take the vertical component avert = aA · j of the acceleration:

aA · j = l(−ϕ2 sinϕ+ ϕ cosϕ) .

In this expression, we express ϕ using (X.8):

aA · j = −3

2g cos2 ϕ− lϕ2 sinϕ .

Next, we determine the vertical acceleration of point A at the moment the stick is releasedi.e. , when ϕ(0) = 0:

avert(0) = −3

2g cos2 ϕ .

Notably, |avert(0)| > g when cos2 ϕ > 23 , i.e. for ϕ > 35. The end of the stick, therefore,

has an acceleration larger than the acceleration due to gravity!How can this be explained?In the hinge at B, there are reaction forces in both the x and the y-direction, as indicatedin Figure X.3 (b). These forces rotate the rigid body and add an extra component to thevertical acceleration of point A.

Example X.3: Sweet spot of baseball bats (center of percussion)An experienced baseball player is able to hit the ball on a distinguished spot of the bat,called the center of percussion. When the ball hits the bat at this spot, the player doesnot feel an impact force in direction parallel to the motion of the ball. The same is truefor racket sports, such as tennis. How can we find this sweet spot of a bat?

(a ) Modelling and free-body diagram

For our computation we model the bat as a 2D planar rigid body. The wrist of theplayer holding the bat is modeled as a smooth joint fixed in point B (see Figure X.4). Theend of the bat is hinged at point B. The distance between the center of mass C and the


Figure4.20

C

BRx

Ry

rBP

rBC

PF

ex

ey

Figure X.4: The baseball bat is modeled as a planar rigid body. The center of percussionis denoted by P . Also shown are the forces acting on the body. Note that F is impulsive.

point B is rBC . The unknown sweet spot is denoted by P , and the distance between Band P is by rBP . We will model the ball hitting the bat as an impulsive force F actingover the time interval [t−, t+], t+ → t−. F acts at the sweet spot P . The reaction forceat B is denoted by R = Rxex +Ryey.

(b ) Rotation of bat after impact

First, we need to find the angular velocity ω+ after the impact. We will apply the angularmomentum principle w.r.t. the point B:

HB = MB . (X.9)

Note that vB = 0, and therefore the term vB×P is zero in (X.9). The only force producinga torque about B is the impact force F , i.e. ,

MB = |F | rBP ez = FrBP ez .

Next, we integrate (X.9) over the interval [t−, t+]:

HB(t+)−HB(t−) =

∫ t+

t−MB dt =

∫ t+

t−F dt rBP ez . (X.10)

Although the time interval is infinitesimally small (t+ → t−), the integral of F does notvanish because F is impulsive, i.e. , not bounded. We denote this integral by

∆IF =

∫ t+

t−F dt .

Since B is fixed, we can compute the angular momentum using the moment of inertia IBabout B:

HB = IBωez .

158

The angular velocity ω− before the impact is equal to zero, ω− = 0, and hence HB(t−) =0. The angular momentum HB(t+) after the impact is equal to HB(t+) = IBω

+ez.substituting everything into (X.10) , we obtain

IBω+ k = ∆IF rBP ez ,

which can be solved for ω+ to yield

ω+ =∆IF rBPIB

. (X.11)

(c ) Sweet spot

We defined our sweet spot as the point of impact for which Rx = 0. To find this spot, weneed to relate the reaction forces R at B to rBP . To this end, we apply the angular mo-mentum principle w.r.t. the center of mass C and again integrate over the interval [t−, t+]to obtain

HC(t+)−HC(t−) =

∫ t+

t−MC dt . (X.12)

We find the torque MC about C to be

MC = F (rBP − rBC)ez +RxrBCez .

Note that in general both forces F and Rx are impulsive and hence cannot be neglectedeven over infinitesimally small time intervals. However, we are interested in finding thesweet spot for which Rx = 0. We then obtain∫ t+

t−MC dt =

∫ t+

t−F (rBP − rBC)ez dt = ∆IF (rBP − rBC)ez .

The angular momentumHC(t−) before the impact is zero, because ω− = 0 and the angularmomentum after the impact can be computed using the centroidal moment of inertia,

HC(t+) = ICω+ez = IC

∆IF rBPIB

ez .

Here we used (X.11) to replace ω+. Substituting everything into (X.12) yields

IC∆IF rBPIB

ez = ∆IF (rBP − rBC)ez ,

which we solve for rBP to obtain

rBP =IB

IB − ICrBC .

Using the parallel axis theorem, we relate IB to IC via

IB = IC +mr2BC .


Finally, we obtain

rBP =IB

m rBC. (X.13)

This is the location of the sweet spot, also called center of percussion in mechanicalterminology. When a hinged rigid body is hit at this point, no parallel reaction force isexperienced at the fixed axis of rotation of the body.

Thin rod When we assume the bat is a thin rod with IB = 13mL

2 and rBC = 12L, we

obtain

rBP =2

3L .

Hammer A hammer, modeled as point mass of mass m at L, with IB = mL2 andrBC = L, has its center of percussion at

rBP = L .

Example X.4: Chimney demolitionConsider an old chimney that is scheduled to be demolished. Explosives detonated at thebottom cause the chimney to bend over and fall to the ground (see Figure X.5). During thefall, the structure experiences positive normal stress ("pulling") that causes the structureto break. We are interested in finding the location of the point with the largest stress,i.e. , the location of the largest bending moment along the chimney. This will be the pointwhere the chimney is expected to break first.

160

Figure X.5: A chimney demolished by an explosion.

Figure X.6: The chimney is modeled as a falling stick hinged at a smooth joint at pointB. The corresponding free-body diagram is shown in the figure.


We will model the chimney as a falling hinged stick (rigid body) (see Example (X.2)),of mass m and length L, with constant cross section. This assumption is valid until thefirst break in the chimney which we are interested to determine. The system then has onedegree of freedom; we choose the angle θ between the vertical and the actual position ofthe chimney as our generalized coordinate. At the smooth joint B, a tangential force LBand a normal force NB are acting (see Figure X.6). Further, we denote the center of masswith C, where the force of gravity acts on the chimney.



We find the equation of motion to be the same as the equation of motion (X.8) for Exam-ple X.2. We only need to change the angular coordinate ϕ in (X.8) to θ by substitutingϕ = π

2 − θ, which gives

θ − 3

2

g

lsin θ = 0 . (X.14)

(c ) Normal reaction force NB

We first compute the reaction force NB acting at B normal to the chimney. The onlyforce producing a torque w.r.t. the center of mass C is NB, thus the angular momentumprinciple w.r.t. C is handy to use:

HC = MC (X.15)

with

HC = −IC θez , MC = −NBl

2ez .

The centroidal moment of inertia is IC = 112ml

2. By inserting the above expressionsinto (X.15), we obtain

−IC θez = −NBl

2ez .

We express θ from (X.14) in this equation and solve for NB to obtain

NB =2IClθ =

1

4mg sin θ . (X.16)

(d ) Distribution of the bending moment TA(x)

To find the bending moment at any location x of the chimney, we first consider an x−longsubset of the chimney, as shown in Figure X.7. Since we cut free the top part of thechimney, we have to introduce forces NA(x) and LA(x) and the bending moment TA(x),to account for the actual motion of the full chimney. Since we are only interested in TA,we apply the angular momentum principle about A to avoid involving the unknown forcesNA, LA, LB:

HA + vA × P = MA . (X.17)

Since the chimney rotates around B, we obtain vA ‖ vC , and therefore vA × P = 0. Wefind HA by applying the angular momentum transfer formula between A and C:

HA(x) = HC(x) + P (x)× rCA(x) , (X.18)

with all quantities involved depending on the length x. The partial mass mx can be

162

Figure X.7: We are interested in finding the bending moment of the chimney. (a) A sub-chimney of length x is cut free. (b) There is bending moment TA acting on the subchimneywhen it is cut free.

expressed as

mx =x

lm .

Similarly, we find

HC = −IC θez = − 1

12mxx

2θez = − 1

12mx3

lθez

and

P × rCA = mx vC |rCA| ez =(mx

l

)(x2θ) x

2ez .

Inserting all these terms into (X.18), we obtain

HA =1

6

m

lx3θez .

which gives

HA =1

6

m

lx3θez .

We express θ from the equation of motion (X.14) to obtain

HA =1

4

mg

l2x3 sin θez .

For the torque about A, we obtain

MA =(− TA −NBx+

(mx

l

)gx

2sin θ

)ez .

Combining all the above results together with (X.16) and substituting into (X.17), weobtain

1

4

mg

l2x3 sin θ = −TA −NBx+

(mx

l

)gx

2sin θ ,


which we can then solve for TA(x):

TA(x) = −mg4l2

sin θx(x− l)2 . (X.19)

We can see that for x ∈ [0, l], TA < 0, since for the considered range θ ∈ [0, π2 ], we havesin θ > 0, as seen in Figure X.8. This result is also visible from the acceleration θ of thechimney:

θ =3

2

g

lsin θ .

Because θ is inversely proportional to the length of the chimney a longer beam acceleratesslower than a shorter one. In other words, the lower portion of the chimney on its ownwould be rotating faster than the whole chimney but is kept back by the upper partresulting in a negative bending moment. You can also see this in Figure X.5 where thelower part of the chimney starts rotating more quickly after the first break.Since TA(0) = TA(L) = 0, and in between we have TA < 0, there must be a minimum forTA(x) at some point x ∈ [0, l]. We find this location by differentiating w.r.t. x,

d

dxTA(x) = −mg

4l2sin θ(x− l)(3x− l) ,

and setting this derivative equal to zero. From this calculation, we find the minimalbending moment (maximal in absolute value) to be at the location

x =l

3. (X.20)

This is the location at which the chimney is expected to break first. In practice, thechimney has a non-constant cross section, and hence the break-location will vary. StillFigure X.5 shows an actual case in which the chimney first broke close to the x = l

3 point.

Figure X.8: The distribution of the bending moment is qualitatively illustrated. Theminimum is at x = l

3 .

Example X.5: Dynamics of a towed wheelConsider a wheel that is towed with constant velocity u = const. , as illustrated in Fig-ure X.9. The wheel is linked to an inclined arm at the center of mass of the wheel. Thearm is hinged to the pivot axis at B. The center of mass C for the whole apparatus islocated at a horizontal distance a away from B. The distance between this center of massand the point of contact G is b. We want to understand the stability of such a wheel underdifferent towing directions(push or pull).

164

(a ) Model the given situation in 2D, find the degrees of freedom and then draw anappropriate free-body diagram.

(b ) Find the equation of motion.(c ) Discuss the stability of the dynamics. First, consider the special case where the arm

is not inclined (a=b=0). For the general case, linearize the equation of motion andthen discuss the stability based on the eigenvalues of the problem.

Figure4.25

C

G

B

u

a b

G

C

B

u = const

(a) (b)

ex

ey

ez

Figure X.9: A wheel towed with constant velocity u. (a) Side view of the wheel with therelevant geometric features. (b) The top view shows the direction in which the wheel istowed. We will model the motion as two-dimensional based on this view.


(i) Two-dimensional modelThe two-dimensional model we use corresponds to the top view of the three-dimensionalwheel, as seen in Figure X.9 (b). We will, therefore, not consider the kinematics of thewheel but simply focus on the moving contact point G and the center of mass C, bothof which exhibit planar motion. The constant towing velocity u is assumed to be in thehorizontal direction, i.e. negative ex direction. We denote the angle between the towingdirection and the orientation of the wheel as θ.

(ii) Degrees of freedomSince we have one rigid body, the unconstrained system has three degrees of freedom.Point B is not allowed to move in the ey direction:

vB · ey = 0 .

This non-holonomic constraint can be integrated to a holonomic one. The towing velocityu is imposed, which is also non-holonomic constraint. When integrated, this constraint


still depends on the initial position, i.e. ,

xB(t) = xB(t0) + u(t− t0) .

From our discussion about non-holonomic constraints in Example IX.2, we know this isa debatable case. In this case we regard this constraint as holonomic and therefore thedegrees of freedom are reduced to one,

DOF = 3− 2 = 1 .

We now choose the angle θ as the generalized coordinate to describe the system.

(iii) Free-body diagramThe forces acting on the body are illustrated in Figure X.10 (a). The towing force F Tacts at B and the rolling friction force F roll acts in the direction of the wheel. Sincethe wheel does not purely roll in the direction where it points, there is also a third forcepresent: a lateral friction force F l resisting the lateral sliding. Note that this force onlyexists when a pneumatic tire is put on the wheel, otherwise the wheel can laterally slidewithout resistance. Furthermore, we will use the unit vectors

e = − cos θex − sin θey

pointing in the direction of the wheel and

n = − sin θex + cos θey

pointing in the direction perpendicular to the wheel.

Figure4.26

GC

B

(a) (b)

e

n

FT

F roll

F l

GF l

vG

vlat

↵

Figure X.10: (a) Free-body diagram for the towed wheel. (b) Model for the lateral frictionforce.

(iv) Analysis of the lateral friction force F lLet us take a closer look at the motion of the wheel and discuss how we can describe thelateral friction force F l shown in Figure X.10 (b). The velocity vG of the wheel consistsof a rolling component vroll in the direction of the wheel and a lateral component vlatnormal to the wheel:

vG = vrolle+ vlatn .

166

The slip angle α is the angle between the rolling direction and the actual direction ofmotion,

tanα =vlatvroll

.

Experiments show that the magnitude |F l| of the lateral friction force can be describedas the product between a so-called cornering coefficient Cα and the slip angle α:

|F l| = Cα α . (X.21)

Note that without lateral friction, steering a car would be impossible. We will now finda relation for α in terms of the geometry and the generalized coordinate θ, and expressthe lateral friction force F l accordingly. Using the angular velocity transfer formula, weobtain vG from vB = u in the form,

vG = vB + ω × rBG ,

where the angular velocity is

ω = θez ;

the velocity of point B is

vB = −uex ;

and the vector pointing from B to G is

rBG = (a+ b)(cos θex + sin θey) .

When we now project vG onto n and e, respectively, and insert the calculated terms, weobtain the lateral velocity vlat and the rolling velocity vroll in the form

vlat = vG · n = u sin θ + θ(a+ b) ,

vroll = vG · e = u cos θ .

We can now express α by its definition as

sinα = sin

(tan−1 vlat

vroll

)=vlat|vG|

=vlat√

v2lat + v2

roll

.

Substituting in all necessary expressions, we obtain

α = sin−1

u sin θ + θ(a+ b)√u2 + 2uθ(a+ b) sin θ + θ2(a+ b)2

.

Substituting this expression for α into (X.21) gives

F l = −Cα sin−1


n . (X.22)



To find the equation of motion, we apply the angular momentum principle w.r.t. B. Thischoice ensures that we exclude the unknown forces F T and F roll since they do not producea torque about B. The angular momentum principle about B states

HB + vB × P = MB . (X.23)

We already know the velocity of B, i.e.

vB = −uex .

Via the velocity transfer formula we also find the velocity vC of the center of mass:

vC = vB + ω × rBC= −uex + θez × (a cos θex + a sin θey)

= −(u+ aθ sin θ)ex + aθ cos θey .

The linear momentum P is then

P = mvC = m(− (u+ aθ sin θ)ex + aθ cos θey

).

We now compute the vector product vB × P using the above equations:

vB × P = −muaθ cos θez .

To compute the angular momentum about B, we first compute the angular momentumHC about the center of mass C, then use the angular momentum transfer formula. Theangular momentum about C can be computed using the centroidal moment of inertia IC ,i.e.

HC = IC θez .

The angular momentum about B is given by

HB = HC + P × rCB =(

(IC +ma2)θ +mua sin θ)ez

with the vector

rCB = (−a cos θex − a sin θey)

pointing from C to B. We also need to differentiate the angular momentum about B intime,

HB =(

(IC +ma2)θ +muaθ cos θ)ez .

Finally, the torque about B can be computed as

MB = rBG × F l= (a+ b)(cos θex + sin θey)× F l

= −Cα sin−1


(a+ b)ez .

168

We now substitute all the terms into (X.23) to obtain the equation of motion:

(IC +ma2)θ + Cα sin−1


(a+ b) = 0 . (X.24)

This is a 2nd order nonlinear ODE in θ for which we need initial conditions (θ(0), θ(0)) tobe able to solve it. The equilibrium for steady towing is θ(t) ≡ 0.

(c ) Stability

(i) Non-inclined wheelIn case of no inclination in the arm (i.e. , a = b = 0), we obtain the following simplifiedequation of motion from (X.24):

θ = 0 .

We can directly integrate this twice with the initial conditions θ(0), θ(0) and obtain

θ(t) = θ(0) + θ(0)t .

From this result we see that for an initial perturbation from the equilibrium position (i.e. ,θ(0) 6= 0, θ(0) 6= 0), the wheel never reaches the equilibrium, and hence the system isalways unstable. Therefore, only an inclined arm has a chance to stabilize the towing.We now examine what specific configurations make the system stable.

(ii) Stability in the general caseTo investigate the stability in the general case, we linearize the equation of motion (X.24)at the equilibrium θ = 0, which is the desired operational state for steady towing. Forthat, we Taylor-expand (X.24) around θ = 0, θ = 0 and only keep the linear terms. Theresult of this Taylor-expansion is

(IC +ma2)θ + Cα

((uθ + θ(a+ b) +O(θ3)

)( 1√u2

+O(θ2, θθ)))

(a+ b) = 0 .

Dropping the terms of higher order, we obtain

(IC +ma2)θ + Cα(a+ b)2

|u| θ + Cα sign(u)(a+ b)θ = 0 , (X.25)

the linearized equation of motion (Note that we substituted u√u2

= sign(u) in the equationabove). The general solution of (X.25) can be written as

θ(t) = K1eλ1t +K2e

λ2t (X.26)

for Kj , λj ∈ C, j = 1, 2. Here λj denotes the jth eigenvalue of the linearized equation ofmotion or the jth distinct root of the characteristic polynomial of the equation, i.e. ,

(IC +ma2)λ2 + Cα(a+ b)2

|u| λ+ Cα sign(u)(a+ b) = 0 .


When we solve the characteristic polynomial we obtain

λ1,2 =

−Cα(a+ b)2

|u| ±√C2α

(a+ b)4

|u|2− 4(IC +ma2)Cα sign(u)(a+ b)

2(IC +ma2). (X.27)

To continue, we need to differentiate two cases: the case of sign(u) > 0 and that ofsign(u) < 0.

Case 1: sign(u) > 0: towingIn this case, the real part of the eigenvalues are negative, i.e. , Re(λi) < 0. We can nowexpress the general solution (X.26) such that we split λj into its real and imaginary part:


λ2t

=2∑j=1

KjeRe(λj)teIm(λj)t

=2∑j=1

KjeRe(λj)t

(cos(Im(λj)t) + i sin(Im(λj)t)

).

We can see that

eRe(λj)t −−−→t→∞

0

in this case. The equilibrium of the towed wheel, therefore, is asymptotically stable. Todesign a robust wheel, we want the oscillations to die out fast. The larger Re(λj), thefaster the rate of decay eRe(λj)t, where

eRe(λj)t = exp

(−Cα

(a+ b)2t

2 |u| (IC +ma2)

).

For faster decay, therefore, the following design principles can be used:• Select a tire with a large Cα

• A larger inclination (a+b) of the wheel helps also.

• Make |u| smaller.

• Make the inertia term IC +ma2 small.While making such design decisions, one must also pay attention to Im(λj), as it deter-mines the frequency of wheel oscillation. We need to make sure to avoid resonances withother parts of the vehicle.

Case 2: sign(u) < 0: pushingIn this case, both eigenvalues are real, i.e. , λj ∈ R, j = 1, 2. One of them is positive andthe other one is negative: λ1 < 0 < λ2. The general solution then becomes


λ2t ,

170

and the term with the positive eigenvalue λ2 becomes unstable, i.e.

eRe(λ2)t −−−→t→∞

∞ .

Note that this is all valid within our linearized setting, as long as trajectories remain closeto the equilibrium. Of course, for an unstable solution approaching infinity, this closenessassumption is no longer valid. In reality, the wheel will turn by 180 and be towed again,therefore returning to the stable solution. This behavior is consistent with observationsof wheel flipping of carts in supermarkets.


Prof. George Haller

Dynamics

XI. Examples of rigid bodydynamics in 3D




Chapter XI

Examples of rigid body dynamics in3D

In this section, various problems involving three-dimensional rigid bodies are discussed.

Example XI.1: Pan mill (Kollergang)A pan mill, shown in Figure XI.1, is a device used to grind various granular materials,such as grains. We can mechanically model this grinder as two massless shafts that areperpendicular to each other and a disk of mass m and radius R. The horizontal shaft haslength a and is hinged at O to the vertical shaft, which rotates with angular velocity ωz.The system is subjected to gravity. The milling performance of the device is determinedby the magnitude of the normal reaction force N between the disk and the ground.What is the reaction force N?

Figure XI.1: (a) A model of a pan mill consisting of two perpendicular, massless shaftsand a disk. (b) Pan mills in various applications.

(a ) Coordinate system

Consider the inertial frame [ e1, e2, e3] as indicated in Figure XI.2 (a). In this frame,as in any other inertial frame, the disk has a time-dependent, non-diagonal moment of

173

174

inertia tensor. To avoid dealing with such a tensor, we pick the (non-inertial) frame[f ′

1, f ′

2, f ′

3

]based at O, co-moving with the horizontal bar. In this frame, the moment of

inertia tensor is diagonal and time-independent, which alleviates our computational effort.We will also use the frame

[f

1, f

2, f

3

], which is parallel to the f ′-frame, but is based at

the center of mass C of the disk.

Figure4.28

A

0

!

N

A

C

mgK

0

f 02

f 01

f 01

f 03

f 03

f3

f1

e1 e2

e3

(a) (b)

Figure XI.2: (a) The frame[f ′

1, f ′

2, f ′

3

], co-moving with the horizontal bar. (b) Free-

body diagram. The dashed line indicates (massless) cone whose motion is kinematicallyequivalent to that of the pan mill.

(b ) Kinematics

The angular velocity ω of the disk can be expressed as

ω = ωzf3− ω1f1

.

Since we assume rolling without slipping for the disk, the OA axis is the instantaneousaxis of rotation. Hence, we must have that

ωzω1

=R

a,

or, equivalently,

ω1 = ωza

R.

The total angular velocity of the disk is then

ω = ωzf3− ωz

a

Rf

1.

Note that the motion of the disk is kinematically equivalent 1 to the rolling motion of acone of radius R, hinged at O. We have seen this example before (Example VIII.4) and

1The kinetics of the cone and the disk, however, differ since their mass distributions are different.

Chapter XI. Examples of rigid body dynamics in 3D 175

we inferred the direction of the angular velocity ω here using the results from the rollingcone.

(c ) Free-body diagram

The free-body diagram of the system is depicted in Figure XI.2 (b). Beyond the forceof gravity G = mg = −mgf ′

3, there is a reaction force K at the hinge point O and a

normal reaction force N = Nf ′3at the contact point with the ground.

(d ) Angular momentum principle about O

We seek the reaction force N , but not the reaction force K. To this end, we applythe angular momentum principle w.r.t. point O to engage N but not K. As indicatedbefore, the f ′-frame is preferable over the inertial e−frame since the moment of inertiatensor is then in its principal form. The angular momentum principle w.r.t. O states that

HO + vO × P = M extO . (XI.1)

The term vO×P vanishes since vO = 0. The sum of the external torquesM extO is composed

of contributions from the gravitational force G and the normal force N :

M extO = (mga−Na)f ′

2. (XI.2)

The angular momentum can simply be computed as

HO = I Oω , (XI.3)

because O has zero velocity (vO = 0). Also note that, strictly speaking, O must alsobe part of the body B, which is not true in this case. Note, however, that though themassless cone sketched in Figure XI.2(b), we can extend the body B without changing thedynamics to include the point O as well. To find the moment of inertia tensor I O w.r.t.O, we apply Steiner’s theorem:

[I O]f ′

=[I C]f

+m

0 0 00 a2 00 0 a2

=

12mR

2 0 00 I2 +ma2 00 0 I2 +ma2

. (XI.4)

The components of the centroidal moment of inertia tensor[I C]f

= diag(

12mR

2, I2, I2

)were obtained from the table in Figure IX.13. Before we can compute the angular mo-mentum HO from formula (XI.3), we also need to convert the angular velocity vector ωinto its representation in the f ′-frame. This is simple in this case, since the axes of thef ′- and f -frame are parallel:

[ω]f ′ = [ω]f = ωzf′3− ωz

a

Rf ′

1. (XI.5)

Next, we compute the angular momentum in the f ′− frame using the expressions (XI.3)-(XI.5):

[HO]f ′ =[I C]f ′

[ω]f ′ = (I2 +ma2)ωzf′3− 1

2mRaωzf

′1.

176

The time derivative of HO in the inertial frame is (c.f. (IX.37))

HO = HO + Ω×HO = 0 + ωzf′3×(

(I2 +ma2)ωzf′3− 1

2mRaωzf

′1

),

=⇒ H0 = −1

2mRaω2

zf′2. (XI.6)

(Here we have used the fact that the angular velocity of the f ′-frame is Ω = ωzf′3).

Plugging both (XI.6) and (XI.2) into (XI.1) yields

−1

2mRaω2

z = mga−Na ,

or, equivalently,

N = mg +1

2mRω2

z .

From this result, we see that the extra term 12mRω

2z increases the milling force beyond

the weight mg of the disk. This increase is solely due to the circular motion of the disk.It does not arise for a disk moving with the same speed along a straight line.

Example XI.2: Eccentric and skewed disk on a rotating shaftConsider a disk of mass m, height h, radius R and center of mass at C, as illustrated inFigure XI.3. The disk is eccentrically mounted on a massless shaft at the point D, whichcoincides with the origin of the chosen inertial coordinate system [e1, e2, e3]. The eccen-tricity e measures the projection of the distance between D and C, onto the [e1, e2]-plane.Moreover, the disk is skewed at an angle δ relative to the horizontal. The distance betweenD and the floating bearing B of the shaft is l1; the distance between D and the fixedbearing A of the shaft is l2. The magnitude of the angular velocity of the shaft is denotedby ϕ. Gravity acts in the negative k - direction.What are the reaction forces on the bearings at points at A and B?

(a ) Degrees of freedom

The system (shaft and disk) has in its unconstrained state, six degrees of freedom. Thefixed bearing represents three constraints and the floating bearing adds an additional twoconstraints. We then obtain that the constrained system has one degree of freedom:

DOF = 6− 3− 2 = 1 .

We choose ϕ, i.e., the current rotation angle of the shaft, as a generalized coordinate todescribe the motion.

(b ) Free-body diagram

As shown in the free-body diagram of Figure XI.4, normal forces NA = NA er andNB = NB er arise at the two bearings. They result from the frictionless point contact


Figure4.29

C

D

D

e

C

'

'

m

h

R

B

A

l1

l2

g

D

e1 e2

e3

D

e1

e2

e3

(smooth joint)

(smooth joint)

masslessshaft

Figure XI.3: A skewed disk mounted eccentrically on a vertically rotating shaft.

between the bearings and the shaft, pointing in the direction of C in the [e1, e2]-projection.This is the direction of

er = cosϕe1 + sinϕe2 . (XI.7)

Since A is a fixed joint, there is also a vertical reaction force TA = TAez exerted in thedirection of the shaft. At C, the gravitational force mg acts upon the disk.

(c ) Linear momentum principle in the inertial frame [e1, e2, e3]

Since we are interested in finding the reaction forces, we first apply the linear momentumprinciple to the disk:

P = F ext . (XI.8)

To compute the quantities involved in this equation, we make use of the unit vector erdefined in (XI.7). We find the resultant external force to be

F ext = (NA +NB) er + (TA −mg)e3 ,

178

Figure4.30

C

D

D

C

g

D

e1 e2

e3

D

e1

e2

e3

B

A

NB

NA

TA

mg

NB

NA

Figure XI.4: The free-body diagram of the skewed and eccentric rotating shaft.

or, equivalently,

F ext = (NA +NB) cosϕe1 + (NA +NB) sinϕe2 + (TA −mg)e3 .

We also need the acceleration of the center of mass C. We find the position of C from thegeometry to be:

rC = e (er − sin δe3) = e (cosϕe1 + sinϕe2 − sin δe3) .

Differentiating this expression for rC twice in time yields

rC = eϕ (− sinϕe1 + cosϕe2) ,

rC = eϕ (− sinϕe1 + cosϕe2) + eϕ2 (− cosϕe1 − sinϕe2)

= −e(ϕ sinϕ+ ϕ2 cosϕ

)e1 + e

(ϕ cosϕ− ϕ2 sinϕ

)e2 .

Substituting all the above quantities into (XI.8), we obtain the following three component


equations:

−em(ϕ sinϕ+ ϕ2 cosϕ

)= (NA +NB) cosϕ , (XI.9)

em(ϕ cosϕ− ϕ2 sinϕ

)= (NA +NB) sinϕ , (XI.10)

0 = TA −mg . (XI.11)

Combining (XI.9) and (XI.10), we obtain

−emϕ2 = NA +NB , (XI.12)

and from (XI.11), we find that

TA = mg . (XI.13)

From (XI.12), we see that the force (NA + NB) points away from the center of mass C.(NA +NB) acts, therefore, as a centripetal force, keeping C on a circular orbit.

(d ) Angular momentum principle in principal frame (non-inertial!)

Just from the linear momentum principle worked out above, we do not have enoughequations to determine the reaction forces NA, NB. Therefore, we also need to applythe angular momentum principle. We choose to apply it w.r.t. C and use the principalframe

[f

1, f

2, f

3

], shown in Figure XI.5. This choice of the frame is motivated by the

simple form of the moment of the moment of inertia tensor I C in this frame. Indeed,[I C ]f is diagonal, time-independent and readily available from the table in Figure IX.12.Note: In the following, all column vectors and matrices are represented in the principalframe!

Since vC ‖ P , the angular momentum principle w.r.t. C states that

HC = M extC . (XI.14)

Since C is the center of mass of the disk, the angular momentum w.r.t. C can be computedas

HC = I Cω .

Since the shaft is the axis of rotation, the angular velocity ω points along this axis and ithas the magnitude ϕ:

ω = ϕe3 .

In the principal f−frame, ω can be expressed as

ω = −ϕ sin δf1

+ ϕ cos δf3

=

−ϕ sin δ0

ϕ cos δ

,

where the coordinates of the last vector are determined w.r.t. the[f

1, f

2, f

3

]-basis, as

noted before.

180Figure4.31

C

D

D

C

g

D

e1 e2

e3

D

e1

e2

e3

B

A

NB

NA

TA

mg

NB

NA

f1

f2

f1

f3

!

f1

f2

f3

C

m

↵

(a) (b)

Figure XI.5: The angular momentum principle is applied in the principal frame[f

1, f

2, f

3

].

(a) The orientation of the principal frame is shown. (b) The principal frame coincides withthe axes of symmetry of the disk.

Note: We can transform any vector u represented in the inertial e-frame as [u]e into itsrepresentation [u]f in the principal frame f by the transformation matrix

R fe = R 2R 3 .

We decomposed the transformation into two parts to describe the rotation of f . First,we rotate around the positive 3-axis by ϕ. Then, from the frame arising from the firstrotation, we rotate around the new positive 2-axis by δ. Therefore,

R fe =

cos δ 0 − sin δ0 1 0

sin δ 0 cos δ

cosϕ sinϕ 0− sinϕ cosϕ 0

0 0 1

.


Using this transformation matrix, we obtain

[ω]f = R fe [ω]e =

−ϕ sin δ0

ϕ cos δ

.

We now come back to computing the angular momentum HC . From the lookup table inFigure IX.12, we find the moment of inertia tensor about C in the principal frame:

I C =

Iα 0 00 Iβ 00 0 Iγ

,

with (c.f. Figure XI.5(b))

Iα = Iβ =1

4m

(R2 +

h2

3

),

Iγ =1

2mR2 .

Finally, we obtain

HC = I Cω =

Iα 0 00 Iβ 00 0 Iγ

−ϕ sin δ0

ϕ cos δ

=

−Iαϕ sin δ0

Iγϕ cos δ

.

To differentiate HC in the inertial e−frame, we must apply formula (IX.37) since HC isexpressed in the moving f -frame. We then obtain

HC = HC + Ω×HC ,

where Ω denotes the angular velocity of the body-attached frame, and hence

Ω = ω .

Substitution into the formula for HC gives

HC =

−Iαϕ sin δ0

Iγϕ cos δ

+

−ϕ sin δ0

ϕ cos δ

×−Iαϕ sin δ

0Iγϕ cos δ

=

−Iαϕ sin δ12 ϕ

2 sin(2δ)(Iγ − Iα)Iγϕ cos δ

.

For the torque M extC about C, we find

M extC =

0NB(l1 + e sin δ)−NA(l2 − e sin δ) + TAe cos δ

0

.

Substituting everything into the angular momentum principle (XI.14) yields −Iαϕ sin δ12 ϕ

2 sin(2δ)(Iγ − Iα)Iγϕ cos δ

=

0NB(l1 + e sin δ)−NA(l2 − e sin δ) + TAe cos δ

0

. (XI.15)

182

From the first and third component of this equation, we obtain

ϕ = 0

and hence

ϕ = const. = ω0 .

We could have anticipated this result since the angular momentum in the e3-direction isconserved about C. The second component of (XI.15) yields

1

2ϕ2 sin(2δ)(Iγ − Iα) = NB(l1 + e sin δ)−NA(l2 − l sin δ) + TAe cos δ . (XI.16)

When we combine (XI.16) with (XI.12), we can solve this for NA, NB to finally obtain

NB = −m(

18R

2 − 112h

2)

sin(2δ)ω20 + e

(g cos δ + ω2

0(l2 − sin δ))

l1 + l2,

NA = m

(18R

2 − 112h

2)

sin(2δ)ω20 + e

(g cos δ − ω2

0(l1 + sin δ))

l1 + l2.

Note that the first term with "sin(2δ)" denotes the effect of the skewedness and the secondterm with "e" represents the effect of the eccentricity.Note: For e = δ = 0 (unskewed disk without eccentricity), both reaction forces disappear:NA = NB = 0. However, when e, δ 6= 0, then NA, NB ∼ ω2

0 grow (!) quadratically withincreasing angular velocity. In practice, this implies that high-speed turbo machinery mustbe carefully balanced.


Prof. George Haller

Dynamics

XII. GyroscopesContents

XII.1 Euler equations for gyroscopes . . . . . . . . . . . . . . . . . . . . 185




Chapter XII

Gyroscopes

XII.1 Euler equations for gyroscopes

' '

'

x

z = z1

yy

z = z1

xy1

x1x1 = x2

y2

y1

z2

(a) (b) (c)

y

z = z1

x

x1 = x2

y2

y1

z2 = z3

x3

y3

''

! = e

x1 = x2

x3y

y2

y1

y3

z = z1z2 = z3

x

!' = 'e'

! = e

f1

f3

f2

g

(d)

C

Figure XII.1: A spinning top hinged at the origin O. The degrees of freedom are indicatedby three angles (Euler angles).

Consider a freely rotating rigid body hinged at the origin O of the inertial coordinatesystem [x, y, z], as illustrated in Figure XII.1. The system may be subjected to external

185

186 XII.1. Euler equations for gyroscopes

forces and torques. The center of mass of the body is located at C and the axis rOC is aprincipal axis of the body. For such systems, we can simplify the general angular momen-tum principle to Euler’s equations for gyroscopes (or spinning tops). We will derive theseequations by applying the angular momentum principle w.r.t. O in the principal frame ofthe body.

Degrees of freedom

The fixed joint at O can be described by three independent holonomic constraints(nodisplacement in 3 directions). Therefore, the system has three degrees of freedom:

DOF = 6− 3 = 3 .

Euler’s idea was to use three angles as generalized coordinates for the system. Thosethree angles are constructed by a "3-1-3" rotation sequence from the inertial frame to theprincipal frame of the body, as illustrated in Figure XII.1. This "3-1-3" sequence is asfollows:

1. The first rotation is a rotation by ϕ about the axis z ≡ z1 to the coordinate system[x1, y1, z1]. The corresponding transformation is given by

[u]1 =


0 0 1

[u]xyz .

We call ϕ the precession angle, with corresponding angular velocity component

ωϕ = ϕ eϕ = ϕ ez = ϕ ez1 .

2. Next we rotate by the nutation angle ν about the axis x1 ≡ x2 to the coordinateframe [x2, y2, z2] such that z2 is aligned with the principal axis rOC of the body. Thissecond transformation is given by

[u]2 =

1 0 00 cos ν sin ν0 − sin ν cos ν

[u]1 ,

with the associated angular velocity component

ων = ν eν = ν ex1 = ν ex2 .

Note that the angular velocity of the frame is now Ω2 = ων + ωϕ.

3. Finally, we rotate about z2 ≡ z3 by the angle ψ such that the resulting frame[x3, y3, z3] corresponds to a principal frame of the spinning top. The transforma-tion is

[u]3 =

cosψ sinψ 0− sinψ cosψ 0

0 0 1

[u]2 ,

Chapter XII. Gyroscopes 187

or, equivalently,

[u]3 =

cosψ sinψ 0− sinψ cosψ 0

0 0 1

1 0 00 cos ν sin ν0 − sin ν cos ν


0 0 1

[u]xyz ,

when expressed in the original inertial frame [x, y, z]. The angular velocity componentcorresponding to ψ is

ωψ = ψ eψ = ψ ez2 = ψ ez3 .

The angle ψ is called spin.

The generalized coordinates (ϕ, ν, ψ) are commonly referred to as Euler angles. Othervariants of Euler angles also exist, but the present one is the most common choice.Note: The spin ψ comes from the fast-spinning of the top around its principal axis. Theprecession ϕ describes the change of orientation of the spinning axis, i.e. , the top precessesaround the inertial axis z. The nutation ν describes small oscillations of the spinning axisduring the precession.

Free-body diagram

The spinning top may be subjected to a resultant force F ext, as well as the gravitationalforce mg, and a resultant torque M ext

C , leading to a reaction force N of general orientationat the joint at O (see Figure XII.2).

g

Figure4.33

N

C

mg

F ext

MextC

Figure XII.2: The free-body diagram of the spinning top.

Angular velocity

The total angular velocity of the body is

ω = ωϕ + ων + ωψ ,

which is also the angular velocity of the principal frame. We express each summand of ωin the principal frame [x3, y3, z3] using the "3-1-3" transformation described above thenadd them up. We then obtain

ω =

ω1

ω2

ω3

=

ϕ sin ν sinψ + ν cosψϕ sin ν cosψ − ν sinψ

ϕ cos ν + ψ

. (XII.1)


Angular momentum w.r.t. O

Since we are in a principal frame, the moment of inertia tensor is diagonal and time-independent. We then find the angular momentum to be

HO = I Oω =

I1 0 00 I2 00 0 I3

ω1

ω2

ω3

=

I1ω1

I2ω2

I3ω3

. (XII.2)

Recall that the angular velocity Ω3 = ω of the principal frame [x3, y3, z3] coincides withthe angular velocity of the body.

Angular momentum principle w.r.t. O in the principal frame

The angular momentum principle w.r.t. O states that

HO = M extC + rOC × (mg) , (XII.3)

because vc × P = 0. The resultant torque in this equation is

MO =

M1

M2

M3

= M extC + rOC × (mg) ,

and hence the constraint force N does not appear in (XII.3). To compute HO, we useformula (IX.37) to obtain

HO = HO + ω ×HO ,

=

I1ω1

I2ω2

I3ω3

+

ω1

ω2

ω3

×I1ω1

I2ω2

I3ω3

.

Substituting into (XII.3) gives the following set of Euler’s equations for a spinning top inits principal frame:

Euler’s equations of motion for a spinning top

I1ω1 + (I3 − I2)ω3ω2 = M1 ,

I2ω2 + (I1 − I3)ω1ω3 = M2 , (XII.4)I3ω3 + (I2 − I1)ω2ω1 = M3 .

Ii : principal moments of inertiaωi : ith component of the angular velocity expressed in the principal frameMi: ith component of the resultant torque about O expressed in the principal

frame


Example XII.1: Steady precession of an axisymmetric gyroscopeConsider an axisymmetric gyroscope hinged at its tip at O, as shown in Figure XII.3.We are going to use the notation and coordinate frames we have just introducedfor a general spinning top (see above). In this case, since the body is axisymmetric,the frame [x2, y2, z2] is already a principal frame even though it is not attached tothe body! We can, therefore, also apply the angular momentum principle with allquantities expressed in the [x2, y2, z2]-frame. We are interested in in the case ofsteady precession, i.e. ν = ν0 = const., ϕ = ϕ0 = const. and ψ = ψ0 = const.

''

! = e

x1 = x2y

y2

y1

z = z1z2 = z3

x

!' = 'e'! = e g

Figure4.34

CC

Figure XII.3: An axisymmetric gyroscope. Note that the frame [x2, y2, z2] is al-ready a principal frame since the gyroscope is axisymmetric and hence the thirdtransformation to Euler’s angles is not necessary.

(a ) Angular velocity

The angular velocity of the [x2, y2, z2]-frame is

Ω = ϕ eϕ + ν eν =

νϕ sin νϕ cos ν

=

0ϕ0 sin ν0

ϕ0 cos ν0

,

and the angular velocity of the body is

ω = ϕ eϕ + ν eν + ψ eψ =

νϕ sin ν

ϕ cos ν + ψ

=

0ϕ0 sin ν0

ϕ0 cos ν0 + ψ0

.

All quantities are expressed w.r.t. the [x2, y2, z2]-frame. We used the fact that weconsider steady precession, therefore ν = 0. Note that Ω 6= ω since the frame isnot attached to the body.

(b ) Angular momentum principle w.r.t. O in the [x2, y2, z2]-frame


The angular momentum principle w.r.t. O is again

HO = M extO . (XII.5)

with

HO =

I1ω1

I2ω2

I3ω3

,

as in the case of the general spinning top, but now with I1 = I2. The time derivativeof HO in the absolute frame is

HO = HO + Ω×HO ,

as noted already for a general gyroscope. In case of steady precession, the timederivative in the moving frame is zero:

HO = 0 ,

and hence we obtain

HO = Ω×HO .

Substituting this expression into the angular momentum principle (XII.5) gives

Ω×HO = M extO . (XII.6)

Evaluating the left-hand side of (XII.6) for steady precession gives

Ω×HO =

I3Ω2ω3 − I1Ω3ω2



=

I3ϕ0 sin ν0(ϕ0 cos ν0 + ψ0)− I1ϕ20

sin(2ν)2

00

.

Note that ω1 = Ω1 = 0 in our case. Therefore, in steady precession, the externaltorque is parallel to the nutation axis x1 ≡ x2 and its magnitude balances (I3Ω2ω3−I1Ω3ω2).

(c ) Geometry of steady precession

Consider, for example, an axisymmetric gyroscope that spins with ψ0 and is slightlytilted (see Figure XII.4). The gravitational force mg will create a torqueMO aboutO. Instead of making the gyroscope fall, this torque causes the gyro to develop aϕ0 6= 0 and therefore a Ω 6= 0 to maintain the steady precession condition (XII.6).In other words, the gyroscope reacts to mg by rotating about mg! An equilibriumν = ν0 is reached when the external torque MO precisely balances the gyro-actiontorque (Ω ×HO). Then the gyroscope steadily precesses around mg. Gyroscopesare often used for navigation because of their stable orientation relative to thedirection of gravity.

Chapter XII. Gyroscopes 191Figure4.35

C

mg

O

|rOC | · sin 0

0

! = 0ez2 !' = '0ez1

! = !' + !

x2

y2z2

Figure XII.4: When an axisymmetric gyroscope spins fast its angular momentumHC is almost parallel to ωψ.

(d ) Fast gyroscope

For a fast-rotating gyroscope in steady precession, we have

I3ω3 I1ω2 = I1ϕ0 sin ν0 .

Hence, we can approximate the angular momentum HO as

H0 ≈ I3ωψ ,

and the angular velocity Ω of the frame as

Ω ≈ ωϕ .

Originally, the angular momentum principle stated

Ω×HO = M extO ,

which we can now approximate as

ωϕ × ωψ ≈1

I3M ext

O

in the case of a fast-spinning gyroscope. From this, we see that the gyroscopewill react to an external torque by developing a rotational component, i.e. , by aprecession with angular velocity ωϕ normal to both M ext

O and to the spin axis ωψ.This can be memorized by the following "TSP"-rule:

1. The Torque must be aligned with the middle finger


2. The Spin must be aligned with the index finger3. The Precession axis must be aligned with the thumb.We sketch this TSP-rule in Figure XII.5.

Figure XII.5: "TSP"-rule to determine the direction of the precession axis.

Example XII.2: Pushing a gyroscopeA fast-spinning gyroscope is pushed with a finger, as shown in Figure XII.6. Theforce exerted on the gyroscope results in an external torque M ext

C . Note that, inthis case, the center of mass C coincides with the hinge point O. As we have seen,the gyroscope will react to the torque by precessing around it. The precession axiscan be determined by the "TSP"-rule, as shown in Figure XII.6. A resistance isfelt because the gyroscope does not move in the direction of the push, but, rather,rotates around it.

Figure4.37

!'

!

MextC

Figure XII.6: A gyroscope pushed upwards.


Example XII.3: Spinning bicycle-wheel suspended on a ropeConsider bicycle wheel of mass m and moments of inertia I3 I1, suspended on arope, as illustrated in Figure XII.7 (a). The wheel is spinning with angular velocityωψ. The wheel is initially held steady, then released.What happens to the wheel when it is released?

Figure4.38

g

x

z

0! 0

R

mg

(a) (b)

m, I3 I1

Figure XII.7: (a) A bicycle-wheel suspended on a rope, spinning fast about itsdominant principle axis of inertia. (b) Free-body diagram of the same wheel.

Answer:

From the free-body diagram in Figure XII.7 (b), we see that there is an exter-nal force G = mg acting on the wheel, which in turn creates a torque about thesuspension point O. The gyroscope must react to G by turning around the force G.To find the direction of precession, we can again apply the "TSP"-rule, as shownin Figure XII.8. Note that the rope cannot be perfectly vertical, since there is ahorizontal force needed to maintain the circular motion of the wheel. This leads toa rotation of the wheel about the vertical axis in the direction shown in Figure XII.7(b).

Figure4.39

P

S

T!'

!

zy

x

MextO = rOC (mg)

Figure XII.8: The "TSP" - rule is applied to determine the direction of the preces-sion axis for the spinning bicycle wheel.



Prof. George Haller

Dynamics

XIII Single-DOF vibrationsContents

XIII.1 One-DOF vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 199XIII.1.1 Free vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 200XIII.1.2 Forced vibrations . . . . . . . . . . . . . . . . . . . . . . . . 207




Chapter XIII

Vibrations

In this chapter, we study the motion of mechanical systems near equilibria. This topic iscentral to engineering dynamics since most engineered systems are designed to operate atan equilibrium point. The question is how the system responds to its desired operatingconditions. We call the resulting perturbed motion vibration.

Definition. Vibrations are small oscillations around a stable equilibrium.

Since we focus on small oscillations around a stable equilibrium, we will approximatethe motion by linearizing the equation of motion around the equilibrium. This step allowsus to use a set of analytical tools only valid for linear systems, which greatly simplifies ouranalysis.Note: Linearization, however, is only justified as long as the oscillations remain small.For motions leaving a neighborhood of an equilibrium, the linearized equations of motionbecome inaccurate.

We start out with one of the simplest example of nonlinear vibrations, i.e., the motionof a forced-damped pendulum. On this example we illustrate the derivation of the linearizedequations of motion starting from the momentum principles. Subsequently, we analyzethe general equation for vibrations of a one-degree-of-freedom system (see Section XIII.1).Afterwards, we extend our discussion to multi-degree-of-freedom vibrations in Section XIV.

Example XIII.1: Periodically forced and damped pendulumConsider a pendulum of mass m connected to a massless bar of length l, hinged in asmooth joint at O, as shown in Figure XIII.1 (a). The smooth joint moves accordingto yB = a sinωt in the ey-direction. We describe the motion with the generalizedcoordinate ϕ denoting the deflection of the bar from its vertical position. Gravityacts upon the system, as does a linear, angular damper, exerting a damping torqueMd = −cϕex.(a ) Derive the equation of motion of the system.(b ) Linearize the equation of motion, first for a = 0 and then for 0 < a 1.

197

198Figure5.1

yB(t)B

l

mc

ex

ey

ez

'

massless

stable unstable

(a) (b)

B

angularviscousdamper

g 'O = 0 'O =

Figure XIII.1: (a) A damped and forced pendulum. (b) The stable and the unstableequilibrium of the pendulum.


We recall the resulting equation of motion (II.2)

ml2ϕ+ cϕ+mlyB cosϕ+mgl sinϕ = 0

from Example II.2, which was concerned with the same forced-damped pendulum.Next, we substitute the specific forcing term yB = −aω2 sinωt to obtain the equa-tion of motion

ml2ϕ+ cϕ−malω2 sinωt cosϕ+mgl sinϕ = 0 . (XIII.1)

This is a 2nd order, nonlinear, non-autonomous ODE for the unknown motion ϕ(t).

(b ) Linearization of the equation of motion

(i) Unforced vibration (a = 0)Equilibria of system (XIII.1) are motions satisfying ϕ(t) ≡ ϕ0 = const. A directsubstitution of this relation into (XIII.1) gives the two equilibria ϕ0 = 0 and ϕ0 =π, sketched in Figure XIII.1 (b). We now introduce a new variable x = ϕ−ϕ0 thatdescribes a small deviation (or perturbation) from the equilibrium. Substitutingϕ = x+ ϕ0 into (XIII.1) and setting a = 0, we obtain

ml2x+ cx+mgl sin(x+ ϕ0) = 0 .

We linearize this equation by Taylor-expanding it to first order in x at x = 0 andneglecting higher order terms in the expansion. We then obtain the linearizedequation of motion

Mx+ Cx+Kx = 0 , (XIII.2)

where we introduced the parameters M = ml2, C = c and K = mgl cosϕ0. Equa-tion (XIII.2) describes free vibrations in a neighborhood of the equilibrium. It is

Chapter XIII. Vibrations 199

a 2nd-order homogeneous, linear ODE.

For a stable equilibrium and for small initial perturbation (|x(0)|, |x(0)| small),the linearized equation of motion remains valid for all times since the perturbationfrom the equilibrium will remain small. As expected physically, ϕ0 = 0 is a stableequilibrium. For this choice of ϕ0, we obtain K = mgl, and hence the quantitiesM , C, K > 0 are all positive. We will see that this is true for any linearizationaround a stable equilibrium. In contrast, ϕ0 = π marks an unstable equilibriumpoint (cf. Figure XIII.1 (b)). For this choice of ϕ0, we have K = −mgl < 0, whichresults in exponentially growing solutions for the ODE (XIII.2).

(ii) Harmonically forced vibration (0 < a 1)For small a 1, the linear approximation (XIII.2) to the equation of motion isexpected to remain valid, at least near the unforced equilibrium ϕ0 = 0. We needto add, however, the force arising from the motion of the joint to our linearization.We then obtain

Mx+ Cx+Kx = F0 sinωt , (XIII.3)

with F0 = malω2 cosϕ0. We call such an externally excited vibration a forced vi-bration. The above equation is a 2nd order inhomogeneous, linear, non-autonomousODE.

XIII.1 One-DOF vibrations

We now consider a general one-degree-of-freedom mechanical system, oscillating aroundits stable equilibrium. As illustrated in the previous example, the linearized equation ofmotion for such systems, near sinusoidal forcing, can be written as

Mx+ Cx+Kx = F0 sinωt , (XIII.4)

with the scalar variable x denoting the perturbation from the stable equilibrium. One canthink of this system as a mass attached to a wall via a linear spring and a viscous lineardamper (see Figure XIII.2). The mass is excited by the external force F0 sinωt, whichcauses oscillations around the equilibrium x = 0 position.

Figure5.2

c

k

x

F0 sin(!t)M

Figure XIII.2: A block of mass M attached via a linear spring and a viscous damper tothe wall, while excited by an external force. The system represents the linearization of atypical 1-DOF mechanical system at a stable equilibrium point.

In order to work with normalized variables, we define the characteristic damping

δ :=C

2M,

200 XIII.1. One-DOF vibrations

the natural frequency

ω0 :=

√K

M,

and the normalized force amplitude

f0 =F0

M.

We use these parameters to rewrite equation (XIII.4) of one DOF vibrations as follows:

The general equation for vibrations with one degree of freedom

x+ 2δx+ ω20x = f0 sinωt (XIII.5)

x(t) . . . displacement from the stable equilibriumδ . . . . . characteristic damping [1/s]

ω0 . . . . natural frequency [1/s]

f0 . . . . amplitude of normalized excitation forceω . . . . . frequency of excitation

Sometimes, we also use the nondimensionalized parameter D, called Lehr’s damping,defined as

D :=δ

ω0=

C

2√KM

. (XIII.6)

XIII.1.1 Free vibrations

Let us first consider the case of f0 = 0, i.e. a free vibration. Then (XIII.5) becomes

x+ 2δx+ ω20x = 0 . (XIII.7)

This equation can also be converted into a 2 × 2 system of 1st order linear, homogeneousODEs. For this purpose, we introduce the state vector x ∈ R2, defined as

x =

(x1

x2

)=

(xx

).

We call the space of possible x values the phase space. Hence, we can represent a motion(x1, x2) in the phase space as a trajectory x(t), i.e., as a curve parameterized by time. Inthe phase space, (XIII.7) is represented by the system of ODEs

x = Ax =

(0 1−ω2

0 −2δ

)x . (XIII.8)

We can find the general solution of (XIII.7) via the ansatz

x(t) = Aeλt,


with A, λ ∈ C , A 6= 0. Substituting this ansatz into (XIII.7), we obtain

Aeλt(λ2 + 2δλ+ ω20) = 0 .

Solving this equation for λ amounts to finding the roots λ1,2 of the characteristic polynomial

p(λ) = λ2 + 2δλ+ ω20 ,

given that A 6= 0 is assumed. Solving for these roots, we obtain

λ1,2 = −δ ±√δ2 − ω2

0, (XIII.9)

the eigenvalues of the matrix A . We now need to distinguish two fundamentally differentcases:

Case 1: Distinct roots: λ1 6= λ2

For two distinct roots λ1 6= λ2, the general solution of (XIII.7) is of the form

x(t) = A1eλ1t +A2e

λ2t , (XIII.10)

with Aj ∈ C, whenever λj ∈ C. Using the initial conditions x(0) = x0, x(0) = v0, we canexpress A1 and A2 from (XIII.10). The solution can then also be represented in the phasespace as

x(t) = A1s1eλ1t +A2s2e

λ2t ,

where sj = (1, λj). The vector sj represents the eigenvector of A corresponding to theeigenvalue λj , i.e., we have Asj = λjsj . The initial condition for the trajectory x(t) in thephase space is given by

x0 =

(x0

v0

)∈ R2 .

Note: Although Aj , λj ∈ C, the solution x(t) is always real, as the position x(t) and thevelocity x(t) is real by definition.

General solution for free vibrations (one DOF) with distinct roots λ1 6=λ2

The solution in the time domain is

x(t) = A1eλ1t +A2e

λ2t , (XIII.11)

while in the phase space it is given by the trajectory

x(t) = A1s1eλ1t +A2s2e

λ2t , (XIII.12)

where λ1,2 = −δ ±√δ2 − ω2

0 and sj = (1, λj). The coefficients A1, A2 ∈ C can bedetermined from the initial conditions (x0, v0).


Case 2: Repeated roots: λ1 = λ2

For this case, the discriminant of (XIII.9) vanishes, which is only possible if δ = ω0. Thegeneral solution of (XIII.7) is then of the following form:

General solution for free vibrations (one DOF) with a repeated eigen-value λ = λ1 = λ2

The solution in the time domain is

x(t) = A1eλt +A2te

λt, (XIII.13)

and in the phase space, it is given by the trajectory

x(t) = (A1s1 +A2s1)eλ1t +A2s1teλ2t , (XIII.14)

where λ1,2 = −δ = −ω0, s1 = (1, λ) and s1 = (0, 1). The vector s1 is the singleeigenvector that can be found for the eigenvalue λ of A , defined by the equationAs1 = λs1. The coefficients A1, A2 ∈ R can be determined from the initial conditions(x0, v0). Note that this only occurs when δ = ω0.

Physical sub-cases

Writing the above two fundamental cases further sub-cases arise, as listed below, dependingon the properties of the eigenvalues λ1,2 = −δ ±

√δ2 − ω2

0.

Overdamped vibrations Overdamped vibrations arise when

δ > ω0

and hence λ1,2 = −δ ±√δ2 − ω2

0 are real, distinct and negative:

λ1 < λ2 < 0, λ1, λ2 ∈ R .

From (XIII.11), we find the solution in the time domain in the form

x(t) = A1 exp

((−δ −

√δ2 − ω2

0

)t

)+A2 exp

((−δ +

√δ2 − ω2

0

)t

), (XIII.15)

and from (XIII.12), we obtain the trajectory in the phase space as

x(t) = A1s1 exp

((−δ −

√δ2 − ω2

0

)t

)+A2s2 exp

((−δ +

√δ2 − ω2

0

)t

). (XIII.16)

Note that A1, A2 ∈ R and

s1 =

(1

−δ −√δ2 − ω2

0

), s2 =

(1

−δ +√δ2 − ω2

0

).


This case is called overdamped because damping is large enough to prevent oscillationsaround the equilibrium. Indeed the solution x(t) decays exponentially from the initial po-sition towards the equilibrium position, without any oscillation, as shown in Figure XIII.3.A nonzero initial velocity v0 6= 0 will cause the system to either overshoot (v0 > 0) orundershoot (v0 < 0) before approaching the equilibrium. In the phase space, solutionsdecay to an invariant line (blue, spanned by the eigenvector s2), then creep towards theequilibrium in a close vicinity of this line. There is another invariant line (red, spannedby the eigenvector s1), along which solutions decay to the origin at a fast rate, withoutany creeping. Note that two trajectories cannot ever cross in the phase space due totheir uniqueness. As a consequence, the invariant lines separate the phase space into fourregions. This type of equilibrium is called a stable node.

Figure5.3

x

x

t0 50 100 150

x(t)

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

v0 > 0 v0 = 0 v0 < 0 v0 < 0

x

t

x

x

(a) (b)

00

s2

s1

Figure XIII.3: The motions in an overdamped vibration are qualitatively shown in thetime domain and the phase space. (a) The solution decays exponentially towards theequilibrium. Depending on the initial velocity v0, there is an overshoot or undershoot.(b) In the phase space, typical trajectories converge towards the equilibrium along theinvariant line. The equilibrium is called a node.

Critically damped vibrations The case of repeated roots λ1 = λ2 = λ, for which

δ = ω0 ,

is called critically damped. The solutions in this case can be seen from (XIII.13) in thetime domain, and from (XIII.14) in the phase space. The solution in the time domain looksqualitatively the same as in the overdamped case, as illustrated in Figure XIII.4 (a). Inthe phase space, however, the phase portrait is somewhat different, compared to the caseof overdamped vibrations. The two invariant lines from the case of overdamped vibrationshave merged into a single invariant line in the direction of the eigenvector

s1 =

(1−ω0

).

This type of equilibrium is called a degenerate (stable) node, sketched in Figure XIII.4 (b).


-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x

t0 2 4 6 8 10 12 14 16 18 20

x(t)

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

v0 > 0 v0 = 0 v0 < 0 v0 < 0

Figure5.4

x

t

(a) (b)

x

x

00

s1

Figure XIII.4: Motions in a critically damped vibration are qualitatively shown in the timedomain and in the phase space. (a) The solution decays exponentially towards the equilib-rium. Depending on the initial velocity v0, there is either an overshoot or an undershoot.(b) In the phase space, trajectories converge to the equilibrium along the invariant linespanned by the single eigenvector s1. Such an equilibrium is called a degenerate stablenode.

Underdamped vibrations Underdamped vibrations occur when the damping is weakenough to allow for oscillations around the equilibrium. In this case, we have

δ < ω0 .

The roots of the characteristic polynomial then satisfy

λ1,2 = −δ ± i√ω2

0 − δ2 ,

i.e., they have nonzero imaginary parts. Substitution into (XIII.11) then gives

x(t) = e−δt(A1e

i√ω20−δ2t +A2e

−i√ω20−δ2t

)for the solution in the time domain. Because x(t) ∈ R, the constants A1 and A2 must becomplex conjugate, i.e., A2 = A1. With the help of two real constants C1, C2 ∈ R, we cantherefore write:

A1 =1

2(C1 − iC2) ,

A2 =1

2(C1 + iC2) .

Using Euler’s formula, eiu = cosu+ i sinu, we obtain the solution in the time domain as

x(t) = e−δt[C1 cos

(√ω2

0 − δ2 t

)+ C2 sin

(√ω2

0 − δ2 t

)]. (XIII.17)

This equation describes oscillations around the stable equilibrium x = 0 with damped nat-ural frequency ωd =

√ω2

0 − δ2, and with an amplitude decaying as e−δt (see Figure XIII.5).


For the solution in the phase space, we use (XIII.14) and the same real constants C1, C2 ∈ Ras above to obtain

x(t) =

(C1

−C1δ + C2ωd

)e−δt cos(ωdt) +

(C2

−C2δ − C1ωd

)e−δt sin(ωdt) . (XIII.18)

The phase portrait in this case is clearly distinguished from the previously discussed phaseportraits. There is no invariant line anymore. Solutions form spirals converging towardsa stable focus-like equilibrium, as shown in Figure XIII.5. Remember that solutions areunique and therefore trajectories cannot intersect in the phase space.

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40 50 60 70 80 90 100

t-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

x(t)

Figure5.5

t

Td =2

!d> T =

2

!0

(a) (b)

0

x

x

x

0

Figure XIII.5: Underdamped vibrations are shown in the time domain and in the phasespace. (a) The solution in the time domain oscillates around the stable equilibrium with(pseudo-) period Td and exponentially decaying amplitude e−δt. (b) In the phase space,there are no invariant lines anymore. The solutions converge as non-intersecting spiralstowards the equilibrium.

Undamped vibrations Undamped vibrations arise in the presence of zero damping,i.e., for

δ = 0 .

The roots of the characteristic polynomial p(λ) are then purely imaginary:

λ1,2 = ±iω0 .

To obtain the specific form of solutions, we simply set δ = 0, ωd = ω0 in formula (XIII.17)for the underdamped case. In the time domain, we then obtain

x(t) = C1 cos(ω0t) + C2 sin(ω0t)

from (XIII.17). We can also rewrite this solution as

x(t) =√C2

1 + C22

(C1√

C21 + C2

2

cos(ω0t) +C2√

C21 + C2

2

sin(ω0t)

). (XIII.19)

We define now the amplitude constant C0 and the phase angle ϕ0 via

C0 =√C2

1 + C22 , sinϕ0 =

C1√C2

1 + C22

, cosϕ0 =C2√

C21 + C2

2

.


With these constants, we can express the undamped solution (XIII.19) as

x(t) = C0 sin(ω0t+ ϕ0) . (XIII.20)

The constants C0 and ϕ0 can be uniquely determined from C1 and C2, which in turn areobtained from the initial conditions (x0, v0). The solution (XIII.20) describes a sinusoidaloscillation with phase shift ϕ0 and natural frequency ω0 (see Figure XIII.6). This solutionwill never converge to the equilibrium; instead, it will oscillate without decay around theequilibrium. To identify the corresponding trajectory in the phase space, we set δ = 0 andωd = ω0 in (XIII.18) to obtain

x(t) =

(C1

C2ω0

)cos(ω0t) +

(C2

C1ω0

)sin(ω0t) . (XIII.21)

This formula defines elliptic trajectories that encircle the origin (see Figure XIII.6). Thistype of equilibrium is called a center.Note: The phase portrait is structurally unstable. The addition of the slightest amount ofdamping (δ > 0) causes all solutions to converge to the origin (underdamped case).

-1.5 -1 -0.5 0 0.5 1 1.5-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0 10 20 30 40 50 60 70 80

t-1.5

-1

-0.5

0

0.5

1

1.5

x(t)

Figure5.6

x

t

T =2

!0

(a) (b)

x

x

0

0

Figure XIII.6: Undamped vibrations in the time domain and in the phase space. (a) Thesolutions in the time domain oscillate around the stable equilibrium with period T andconstant amplitude C0. (b) The corresponding trajectories in the phase space form ellipsesencircling the equilibrium.


XIII.1.2 Forced vibrations

In this section, we consider oscillations under nonzero external forcing (f0 6= 0). We firstrecall the general equation for a one DOF vibration system (XIII.5) is

x+ 2δx+ ω20x = f0 sinωt . (XIII.22)

We can express the solution of this inhomogeneous linear ODE as the superposition of thegeneral solution of the associated homogeneous ODE and one particular solution of thefull inhomogeneous ODE. Specifically, we can write

x(t) = xh(t) + xp(t) ,

where xh(t) denotes the general solution of the homogeneous ODE and xp(t) is a particularsolution of the inhomogeneous ODE.

General solution of a forced vibration with one degree of freedomThe general solution x(t) to a forced vibration with one degree of freedom can bewritten as

x(t) = xh(t) + xp(t) , (XIII.23)

where xh(t) denotes the general solution to the associated free vibration (with f0 = 0)and xp(t) denotes a particular solution to the forced vibration (XIII.22).

We have already found the general solution for the homogeneous case, listed in (XIII.11)and (XIII.13). We are now interested in finding the particular solution of the inhomoge-neous ODE (XIII.22). For damped systems (δ 6= 0), the homogeneous solution xh(t)converges to 0, thus, we can approximate the general solution (XIII.23) for longer times as

x(t) ≈ xp(t) .

This underlines the importance of finding xp(t) explicitly. The ansatz we choose tofind the particular solution xp(t) is

xp(t) = A sin(ωt− ϕ) , (XIII.24)

where ω is the frequency of the external forcing. Substituting this ansatz into (XIII.22),we obtain

A(−ω2 + ω20) sin(ωt− ϕ) + 2δAω cos(ωt− ϕ) = f0 sin(ωt) .

Using the trigonometric identities

sin(ωt− ϕ) = sin(ωt) cosϕ− cos(ωt) sinϕ ,

and

cos(ωt− ϕ) = cos(ωt) cosϕ+ sin(ωt) sinϕ ,


we obtain

sin(ωt)(A(ω2

0 − ω2) cosϕ+ 2δAω sinϕ− f0

)=

cos(ωt)(A(ω2

0 − ω2) sinϕ− 2δAω cosϕ).

This equation can only hold for all times t if both sides are equal to zero, i.e.,

A(ω20 − ω2) cosϕ+ 2δAω sinϕ = f0 , A(ω2

0 − ω2) sinϕ− 2δAω cosϕ = 0 .

Combining these last two equations gives

A(ω20 − ω2) = f0 cosϕ , 2δAω = f0 sinϕ .

Solving these last two equations for the amplitude A of the particular solution, we obtain

A =f0√

(ω20 − ω2)2 + 4δ2ω2

. (XIII.25)

Similarly, we find the phase shift ϕ to be

ϕ = tan−1

(2δω

ω20 − ω2

). (XIII.26)

To analyze the response to forcing in more detail, we introduce the magnification factor

V =A

(f0/ω20), (XIII.27)

the ratio between the amplitude A generated by the forcing f0 sin(ωt) and the amplitudeof the static response under constant forcing f0. We find the static response amplitude(f0/ω

20) from the forced vibration equation (XIII.5) by seeking a steady-state solution x

(i.e. setting ¨x = ˙x = 0):

¨x+ 2δ ˙x+ ω20x = f0 ⇒ x =

f0

ω20

. (XIII.28)

With the frequency ratio defined as

η :=ω

ω0,

and Lehr’s damping defined as

D :=δ

ω0,

the amplification factor V defined in (XIII.27) can be written as

V (η,D) =1√

(1− η2)2 + (2Dη)2. (XIII.29)


We can also rewrite the phase shift ϕ in (XIII.26) in terms of η and D to obtain

ϕ = tan−1

(2Dη

1− η2

). (XIII.30)

The undamped vibration amplitude A can be expressed as A = f0ω20V = MF0

KM V = F0K V as

well. Finally, we state the particular solution (XIII.24) in terms of the amplification factorV :

The steady-state solution (particular solution) for a one-DOF forcedvibration

xp(t) =F0

KV (η,D) sin (ωt− ϕ(η,D)) (XIII.31)

xp(t) . . . . . steady state solutionF0 . . . . . . . . amplitude of external periodic forceK . . . . . . . . stiffness from (XIII.4)V (η,D) . . . amplification factorω . . . . . . . . . frequency of forcingϕ(η,D) . . . phase shift of steady state solution

We now further discuss the amplification factor V (η,D) and the phase shift ϕ(η,D),both plotted in Figure XIII.7. For each listed damping value D, there exists a maximumof V along the curve. For D = 0 (no damping), V tends to infinity when the excitationfrequency ω approaches the natural frequency ω0. This phenomenon is called resonance,arising when a system is excited precisely at its natural frequency ω0. The less dampingthere is, the closer the maximum amplification is to the natural frequency ω0, and thelarger is the maximum of the amplification factor V .For the phase shift ϕ(η,D), we see that η < 1 yields ϕ < π

2 . In this case, we say that thesystem oscillates in phase. For η > 1, the phase shift satisfies ϕ > π

2 . Such oscillations arecalled out of phase. In this case, the exciting frequency is larger than the natural frequencyand the system lacks the ability to keep up with the oscillation. When D → 0, the phaseshift jumps at η = 1 from 0 to π

2 . Furthermore, we can interpret the general solution

x(t) = xh(t) + xp(t)

as one that invariably settles on the particular solution xp(t). In contrast, the dampedoscillations xh(t) of the homogeneous system die out, as shown in Figure XIII.8.

210 XIII.1. One-DOF vibrationsFigure5.7

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

2

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

V

D = 0D = 0.15D = 0.2D = 0.3D = 0.5

D =p

0:5D = 1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

2

0

:/2

:

'

D = 0D = 0.15D = 0.2D = 0.3D = 0.5

D =p

0:5D = 1

in phase in counter phase

Figure XIII.7: The amplification factor V (η,D) and the phase shift ϕ(η,D), plotted asfunctions of the frequency ratio η, for various damping values D. The dotted-red lineindicates the curve of the maximum amplification for select values of D. Note that, forD = 0, the phase shift jumps at the natural frequency ω0.

0 2 4 6 8 10 12 14 16 18 20-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

xP(t) x(t)

Figure5.8

t

x

Figure XIII.8: The general solution x(t) = xh(t) + xp(t), indicated in red, consists of thehomogeneous solution xh(t) and the particular solution xp(t), with the latter indicatedin blue. For nonzero damping, the homogeneous solution dies out and hence the generalsolution x(t) approaches the particular solution xp(t).


Prof. George Haller

Dynamics

XIV Multi-DOF vibrationsContents

XIV.1 Free, undamped vibrations . . . . . . . . . . . . . . . . . . . . . . 214XIV.1.1 Case 1: Purely imaginary pair of eigenvalues (λ = ±iωj) . 216XIV.1.2 Case 2: A pair of zero eigenvalues (λ = 02) . . . . . . . . . 216XIV.1.3 General solution . . . . . . . . . . . . . . . . . . . . . . . . 217

XIV.2 Modal decomposition of free, undamped vibrations . . . . . . . . 221XIV.3 Multi-DOF forced, damped vibrations . . . . . . . . . . . . . . . 223




Chapter XIV

Multi-DOF vibrations

In this section, we consider vibrations of mechanical systems with multiple degrees offreedom, such as the system shown in Figure XIV.1. This system has n = 2 degrees offreedom and oscillates around its equilibrium. From our previous discussion of multi-DOFsystems, we know that we need n independent generalized coordinates to describe such anoscillatory system.

Figure5.9

m1 m2

c1 c2 c3

k1 k2 k3

x1 x2

Figure XIV.1: Two carts of mass m1 and m2 are attached to each other and each to a wallvia linear spring-damper elements.

In general, let

r(t) ∈ Rn

be the vector of generalized coordinates for an n-degree-of-freedom system, and let

r0 ∈ Rn

be a stable equilibrium (constant solution of the equations of motion) of the same system.We denote the deviation from the equilibrium r0 by

x(t) = r(t)− r0 .

For small motions with small |x(t)| (i.e., for vibrations), we can linearize the system ofequations of motion at the equilibrium r0. The resulting linear system is typically of theform

M x+ C x+K x = F (t) ,

with the positive definite mass matrix

M = M T ∈ Rn×n ,

213

214 XIV.1. Free, undamped vibrations

the positive semidefinite damping matrix

C = C T ∈ Rn×n ,

and the positive semidefinite stiffness matrix

K = K T ∈ Rn×n .

The vector F (t) ∈ Rn contains the external forces acting on the system. Recall that apositive definite matrix A ∈ Rn×n is defined as a matrix whose associated quadratic formxTAx satisfies

xTAx > 0 , x 6= 0 .

Similarly, a positive semidefinite matrix B ∈ Rn×n is a matrix whose associated quadraticform satisfies

xTB x ≥ 0 , x 6= 0 .

Equation for vibrations with n degrees of freedom

M x+ C x+K x = F (t) (XIV.1)

x(t) . . . deviation from the stable equilibrium; x(t) ∈ Rn

M . . . . positive definite mass matrix; M = M T ∈ Rn×n

C . . . . . positive semidefinite damping matrix; C = C T ∈ Rn×n

K . . . . positive semidefinite stiffness matrix; K = K T ∈ Rn×n

F (t) . . . external force acting on the system; F (t) ∈ Rn

In the most general case, the linearization of an arbitrary mechanical system aroundits equilibrium yields

M x+ (C +G )x+ (K +N )x = F (t) ,

where we call the skew-symmetric matrix G = −G T the gyro matrix and the likewise skew-symmetric matrix N = −N T the follower force matrix. We will not treat this completelygeneral setting here.

XIV.1 Free, undamped vibrations

Let us first consider the case of a free and undamped multi-DOF vibration. Equa-tion (XIV.1) then becomes

M x+K x = 0 . (XIV.2)

We use the ansatz

x(t) = ueλt (XIV.3)

Chapter XIV. Multi-DOF vibrations 215

with u ∈ Cn \ 0, λ ∈ C, to seek solutions of (XIV.2). When we substitute this ansatzinto (XIV.2), we obtain

(λ2M +K )ueλt = 0 , ⇒ (λ2M +K )u = 0 . (XIV.4)

Equation (XIV.4) is called the generalized eigenvalue problem for the matrices M and K .Indeed, rewriting the equation as

K u = −λ2M u ,

multiplying with M −1 from the left and defining B := M−1K and Λ = −λ2 lead to theclassic eigenvalue problem

B u = Λu .

We now multiply (XIV.4) by uT from the left to obtain

uTM uλ2 + uTK u = 0 .

Substituting u = v + iw, where v, w ∈ Rn, we obtain

(v − iw)TM (v + iw)λ2 + (v − iw)TK (v + iw) = 0 ,

or, equivalently,

(vTM v + wTM w)λ2 + (vTK v + wTK w) = 0 , (XIV.5)

where we have used the relations wTM v = vTM w and wTK v = vTK w. We nowintroduce the two parameters α, β as

α = vTM v + wTM w ,

and

β = vTK v + wTK w .

Since M is positive definite, it follows that α > 0. Similarly, because K is positivesemidefinite β ≥ 0. The eigenvalues λ of (XIV.4) can therefore be written, using (XIV.5),as

λ2 = −βα

= −ω2

or

λ = ±iω (XIV.6)

for some ω ≥ 0. To find the eigenvalues λ, we use the fact that the matrix (λ2M + K )must be singular according to (XIV.4). This implies that

det(λ2M +K

)= 0 . (XIV.7)

Equation (XIV.7) is called the characteristic equation of (XIV.2), which is an 2nth - orderpolynomial in λ. The roots of this polynomial define the 2n eigenvalues λ = ±iωj , j =1, . . . , n, of the oscillatory system (XIV.2).Note: To find the eigenvector uj corresponding to λj = ±iωj , one needs to solve (XIV.4)with λj = ±iωj substituted. Since this generalized eigenvalue problem is quadratic in λ,the eigenvalue pair λj = ±iωj share the same purely real eigenvector uj ∈ Rn.

From (XIV.6), we see that the eigenvalues may only arise in purely imaginary pairsor zero pairs. We can find two independent solutions to (XIV.2) for any such eigenvaluepair. We will distinguish two cases: ω = 0 and ω > 0.


XIV.1.1 Case 1: Purely imaginary pair of eigenvalues (λ = ±iωj)In this case, the solution (XIV.3) is of the specific form

xj(t) = uj(t)(pjeiωjt + qje

−iωjt) (XIV.8)

for some constants pj , qj ∈ C. This follows by superposing the two independent solutionspjuj(t)e

iωjt and qjuj(t)e−iωjt, obtained from the general ansatz (XIV.3) after substitutingthe two eigenvalues λ = ±iωj , and the corresponding eigenvector uj . From the undampedcase of a one-DOF system, we already know that qj = pj must hold since xj(t) ∈ R. UsingEuler’s formula e±iωt = cos(ωt)± i sin(ωt), we can rewrite (XIV.8) as

xj(t) = uj

((pj + qj) cosωt+ i(pj − qj) sinωt

),

or, equivalently,

xj(t) = uj

√a2j + b2j

aj√a2j + b2j

cosωt+bj√a2j + b2j

sinωt

, (XIV.9)

where we have introduced the notation aj = pj+qj ∈ R and bj = i(pj−qj) ∈ R. Finally, weintroduce the amplitude Aj =

√a2j + b2j and the phase shift ϕj ∈ [0, 2π] via the formulae

sinϕj =aj√a2j + b2j

, cosϕj =bj√a2j + b2j

to obtain the solution (XIV.9) in the final form

xj(t) = ujAj sin(ωjt+ ϕj) . (XIV.10)

Both of these constants can be determined from the initial conditions, which can be, e.g.,xj(t0) = x0

j and xj(t0) = x0j .

XIV.1.2 Case 2: A pair of zero eigenvalues (λ = 02)

One solution for this case can be found directly by substituting the ansatz (XIV.3) intothe linearized equation of motion (XIV.2). This means substituting

x = pje0·tuj = pjuj ,

into (XIV.2), which gives

K x = 0 ,

or

K uj = 0 . (XIV.11)

This implies that uj is an eigenvector of K with eigenvalue λj = 0. To find a secondindependent solution for the eigenvalue pair λ = 02, we use the ansatz

x(t) = qj(uj + ujt) .


When we substitute this into the vibration equation (XIV.2), we obtain

K (uj + ujt) = 0 .

By (XIV.11), this is the same as

K uj = 0 .

We are free to choose any uj to satisfy this last equation. By (XIV.11), we can actuallychoose

uj = uj ,

so the general solution for the eigenvalue pair λj = 02, becomes

xj(t) = pjuj + qjuj(1 + t) ,

or, with two different constants Bj , Cj ∈ R,

xj(t) = uj(Bj + Cjt) . (XIV.12)

Physically, this corresponds to a rigid-body translation with uniform velocity. In otherwords, all degrees of freedom move with some constant velocity, without any oscillation.

XIV.1.3 General solution

Finally, we can find the general solution of the full linearized system as the superpositionof all individual solutions xj , j = 1, . . . , n:

x(t) =

n∑j=1

xj(t) .

Depending on whether the jth eigenvalue pair is purely imaginary or zero, we use ei-ther (XIV.10) or (XIV.12) to compute the jth solution component xj(t). This leads to thefollowing general result.

General solution to a free, undamped vibration of n degrees of freedomThe general solution x(t) to (XIV.2) is

x(t) = u1A1 sin(ω1t+ ϕ1) + u2A2 sin(ω2t+ ϕ2) + . . .+ ulAl sin(ωlt+ ϕl)

+ ul+1(Bl+1 + Cl+1t) + ul+2(Bl+2 + Cl+2t) + . . .+ un(Bn + Cnt)(XIV.13)

with l ≤ n denoting the number of purely imaginary eigenvalue pairs, and n denotingthe total number of eigenvalue pairs. The parameters (Aj , ϕj), j = 1, . . . , l, and(Bk, Ck), k = l + 1, . . . , n, can be determined using 2n initial conditions.

To find the eigenvector uj , commonly called the jth mode shape, the general-ized eigenvalue problem

(λ2jM +K )uj = 0 (XIV.14)


for each eigenvalue pair λj = ±iωj must be solved. The eigenvalues, also called modalfrequencies, are determined by the characteristic equation

det(λ2M +K

)= 0 . (XIV.15)

Note that each purely imaginary eigenvalue pair λj = ±iωj adds an oscillatory term

xj = ujAj sin(ωjt+ ϕj)

to the general solution x(t), whereas a zero eigenvalue pair (λ = 02) adds a rigid bodytranslation

xj = uj(Bj + Cjt) .

M . . . mass matrixK . . . stiffness matrix

Note: One usually determines the 2n parameters (Aj , ϕj) and (Bk, Ck) via the initialposition x(0) = x0 ∈ Rn and the initial velocity x(0) = v0 ∈ Rn. Substituting these initialconditions into the general solution (XIV.13) gives precisely 2n scalar equations for the 2nunknown parameters.

Example XIV.1: Two masses connected via a linear springConsider a system of two masses m1 and m2, connected to each other via a spring withspring constant k, as illustrated in Figure XIV.2. Both masses can slide on the groundwithout friction. The system has two degrees of freedom (n = 2). We choose the positionx1 of the center of mass m1 and the position x2 of the center of mass m2 as generalizedcoordinates. For x1 = x2 = 0, the system is in an unstretched, stable equilibrium. There-fore, xj denotes the deviation of mj from this equilibrium.

(a ) Find the linearized equations of motion.

(b ) What are the corresponding eigenvalues and mode shapes of the vibration?

(c ) Determine the general solution for this vibration and interpret it physically.

Figure5.10

m1 m2

x1 x2k

frictionless

Figure XIV.2: Two masses, m1 and m2, are attached to each other via a spring,but can otherwise move freely.


(a ) Equation of vibrations

Applying the linear momentum principle to each mass individually, we obtain

m1x1 = k(x2 − x1) ,

m2x2 = −k(x2 − x1) .

These equations are already linear and describe the deviation from the equilibrium (0, 0).With the state vector

x =

(x1

x2

),

we can formulate the equations of motion in the matrix form(m1 00 m2

)(x1

x2

)+

(k −k−k k

)(x1

x2

)= 0 .

Note: For the mass matrix,

M =

(m1 00 m2

)(XIV.16)

the associated quadratic form satisfies

xTM x = m1x21 +m2x

22 > 0 ,

therefore the mass matrix M is positive definite, as assumed earlier in our general discus-sion. Similarly, the quadratic form associated with the stiffness matrix

K =

(k −k−k k

)(XIV.17)

satisfies

xTK x = k(x1 − x2)2 ≥ 0 ,

and hence the stiffness matrix K is positive semidefinite, as we generally assumed.

(b ) Eigenvalues and mode shapes

(i) EigenvaluesTo find the eigenvalues of this vibration problem, we need to solve the characteristicequation

det(λ2M +K

)= 0 .

Substitution of (XIV.16) and (XIV.17) into this last equation gives

det

(λ2m1 + k −k−k λ2m2 + k

)= 0 ,


or, equivalently,

λ2(m1m2λ

2 + k(m1 +m2))

= 0 .

As roots to this equation, we obtain the two eigenvalue pairs

λ1,2 = ±i√k(m1 +m2)

m1m2= ±iω1 , λ3,4 = 0 , (XIV.18)

i.e., one purely imaginary pair and one zero pair.

(ii) Mode shapesWe find the mode shapes u1 and u2 by solving

(λ21,2M +K )u1 = 0, (XIV.19)

(λ23,4M +K )u2 = 0, (XIV.20)

respectively. Substituting λ1,2 from (XIV.18) into (XIV.19) gives(−ω2

1m1 + k −k−k −ω2

1m2 + k

)(u11

u21

)= 0 .

Solving this equation for the mode shape u1, we obtain

u1 =

(u11

u12

)=

(m2

−m1

).

Similarly, we find the mode shape u2 for the eigenvalue pair λ3,4 in the form

u2 =

(11

).

(c ) General solution

Using (XIV.13) we can write the general solution of this vibration problem in the form

x(t) = u1A1 sin(ω1t+ ϕ1) + u2(B2 + C2t) .

Substituting the expressions obtained above for the eigenvalues and the mode shapes, weobtain

x(t) =

(m2

−m1

)A1 sin

√k(m1 +m2)

m1m2t+ ϕ1

+

(11

)(B2 + C2t) ,

The first term xA in this solution corresponds to an oscillation around the overall centerof mass due to the spring force between the masses. This oscillation leaves the center ofmass fixed. The second term xB describes the rigid-body translation of the overall centerof mass. Note that for this system, the total linear momentum is conserved (no externalforces act on the system in the direction of the motion), as illustrated in Figure XIV.3.


Figure5.10

m1 m2

(a) (b)

m1 m2

v v

Figure XIV.3: (a) For the oscillation mode xA(t), we have m1xA1 (t) +m2x

A2 (t) ≡ 0.

When differentiating this w.r.t. time, we recover the conservation of linear momen-tum for the system: m1v

A1 +m2v

A2 ≡ 0. (b) In the rigid-body mode, both masses

move with the same constant velocity C2, i.e. vB1 = vB2 = C2. Linear momentum is,therefore, also conserved for this mode.

XIV.2 Modal decomposition of free, undamped vibrations

We again consider the free, undamped vibration equation

M x+K x = 0 , (XIV.21)

withM ,K ∈ Rn×n and x ∈ Rn. As previously established, the general solution to (XIV.21)is the sum of the solution components xj = Cjuje

iωjt, j = 1, . . . , n. We refer to these so-lution components as normal modes. From (XIV.14), we observe that

−ω2jM uj +K uj = 0 , (XIV.22)

−ω2kM uk +K uk = 0 (XIV.23)

for any two distinct eigenvalue pairs with ωj , ωk with k 6= j. Combining these equationsas

uTk · (XIV.22)− uTj · (XIV.23),

we obtain

(ω2k − ω2

j )uTj M uk = 0 .

Since (ω2k − ω2

j ) 6= 0 for k 6= j, it follows that

uTj M uk = 0 (XIV.24)

Let us also combine (XIV.22) and (XIV.23) as follows:

uTk · (XIV.22) + uTj · (XIV.23) .

This gives us

−(ω2k + ω2

j )uTj M uk + 2ujK uk = 0 ,

or, equivalently,

ujK uk = 0 , (XIV.25)

222 XIV.2. Modal decomposition of free, undamped vibrations

where we have used (XIV.24). The relations (XIV.24) and (XIV.25) show that distinctuj and uk vectors are orthogonal with respect to the metric generated by the mass andstiffness matrices. This orthogonality relation motivates the term "normal modes".

These results will imply that we can decompose (XIV.21) into n independent, 1-DOFvibration equations that are completely decoupled from each other. To obtain this decom-position, we first define the (nonsingular) modal matrix U as

U =[u1 u2 . . . un

]∈ Rn×n ,

∣∣uj∣∣ = 1 ,

from the corresponding mode shapes uj , normalizing them to∣∣uj∣∣ = 1. From (XIV.24),

we observe that

U TM U = diag(m1, m2, . . . , mn

), (XIV.26)

a diagonal matrix with entries mj = uTj M uj , j = 1, . . . , n. Similarly,

U TK U = diag(k1, k2, . . . , kn

)(XIV.27)

is also a diagonal matrix with entries kj = uTj K uj , j = 1, . . . , n.Note: The constants mj and kj depend on our choice for uj . Normalizing all eigenvectors(no unit) to∣∣uj∣∣ = 1

results in the correct mass units for mj and stiffness units for kj .We now introduce the modal coordinates y by letting

x(t) = U y(t) =n∑k=1

ykuk . (XIV.28)

Since U is non-singular, we can also write

y(t) = U −1x(t) .

When we now substitute x(t) = U y(t) into (XIV.21) and then multiply by U T from theleft, we obtain

U TM U y + U TK U y = 0 . (XIV.29)

By the expressions (XIV.26) and (XIV.27), both (U TM U ) and (U TK U ) are diagonalmatrices. The ODE (XIV.29), therefore, decouples into n 2nd order homogeneous linearODEs for the modal coordinates y:

mj yj + kjyj = 0, j = 1, . . . , n .

We can interpret the general solution (XIV.28) as a weighted superposition of its individualmodes, as shown, e.g., in Figure XIV.4.


Modal decomposition of free, undamped vibrationsA free, undamped vibratory system of n degrees of freedom,

M x+K x = 0 ,

can be decoupled into its mode shapes using the modal coordinates y(t) defined as

x(t) = U y(t) , (XIV.30)

where each coordinate represents one mode shape . The matrix

U =(u1 u2 . . . un

)∈ Rn×n (XIV.31)

is the corresponding modal matrix of the system with |uj | = 1, j = 1, 2, . . . , n denotingthe normalized eigenvectors of the system. In these modal coordinates, the equationsof motion become

U TM U y + U TK U y = 0 , (XIV.32)

with

U TM U = diag(m1, m2, . . . , mn

), (XIV.33)

U TK U = diag(k1, k2, . . . , kn

), (XIV.34)

or, equivalently,

mj yj + kjyj = 0, j = 1, . . . , n . (XIV.35)

Note: The eigenfrequency/natural frequency of each mode, i.e., the imaginary part of thecorresponding eigenvalue pair, is easily computed from the modal decomposition as

ωj =

√kjmj≥ 0, j = 1, . . . , n .

XIV.3 Multi-DOF forced, damped vibrations

Finally, we consider the general equations of forced and damped vibrations for a linearsystem of n degrees of freedom:

M x+ C x+K x = F (t) . (XIV.36)

For simplicity we assume structural damping, which means that the damping matrix C isa linear combination of the mass matrix M and the stiffness matrix K :

C = aM + bK , a, b ∈ R+ . (XIV.37)

We will show that under this assumption, the damped-forced system (XIV.36) can still bedecoupled into n one-DOF damped-forced oscillations using the modal coordinates

y(t) = U −1x(t)

224 XIV.3. Multi-DOF forced, damped vibrations

of the corresponding free, undamped system. As before, we substitute the modal coordi-nates into our vibration equation (XIV.36), then multiply by U T from the left to obtain

U TM U y + U TC U y + U TK U y = U TF (t) . (XIV.38)

From (XIV.33), we already know that

U TM U = diag(m1, m2, . . . , mn

)and from (XIV.34), we have

U TK U = diag(k1, k2, . . . , kn

).

By the structural damping hypothesis (XIV.37) we have

U TC U = U T (aM + bK )U = aU TM U + b U TK U ,

or, equivalently,

U TC U = a diag(m1, m2, . . . , mn

)+ b diag

(k1, k2, . . . , kn

).

We, therefore, obtain that for structural damping, C also becomes a diagonal matrix

U TC U = diag(c1, c2, . . . , cn

),

with

cj = amj + bkj ≥ 0, j = 1, . . . , n .

Substituting into the system (XIV.38) then yields n forced, damped one-DOF vibrationequations in the form

mj yj + cj yj + kjyj = fj(t), j = 1, . . . , n ,

with

fj = (U TF )j .

Solution to n-DOF forced, damped vibrations with structural dampingA forced, damped vibration system of n degrees of freedom

M x+ C x+K x = F (t)

with structural damping

C = aM + bK

can be decoupled using the modal coordinates y = U −1x of the corresponding un-damped, free vibration system. Each mode is then described by a forced, dampedone-DOF vibration of the form

mj yj + cj yj + kjyj = fj(t), j = 1, . . . , n (XIV.39)


with (U TM U

)jj

= mj , (XIV.40)(U TK U

)jj

= kj , (XIV.41)(U TC U

)jj

= cj = amj + bkj , (XIV.42)(U TF (t)

)j

= fj(t) . (XIV.43)

For each decoupled equation, our previous results for forced, damped one-DOF vibra-tions are applicable (see Section XIII.1).

Our earlier results on forced-damped 1-DOF oscillations specifically imply the follow-ing:

• A mode may be underdamped, overdamped, critically damped or undamped withthe respective consequences for the nature of the solution.

• The resonance frequency for the mode j is given by

ωj =

√kjmj

.

When the frequency of the external force fj also falls near the resonance frequency,the mode is excited at its resonance and hence its amplitude rises.

• When the external force is given by

F (t) = F 0 sin(ωt) ,

we can use the particular solution (XIII.31) of a single DOF vibratory system to findthe particular solution for each mode. In particular, the jth mode is excited by theforce component

fj(t) = f0j sin(ωt) =

(U TF 0

)j

sin(ωt) ,

giving the particular solution yp(t) (steady-state solution)

(yp(t))j

= APj sin(ωt+ ϕj) , (XIV.44)

with

APj =f0j√

(ω2j − ω2)2 + c2

jω2

and

ϕj = tan−1

(cjω

ω2j − ω2

).


Transforming the particular solution back into the original x coordinates yields

xp = U yp(t) =

(u1 u2 . . . un

)AP1 sin(ωt+ ϕ1)

...APn sin(ωt+ ϕn)

,

or, equivalently,

xp(t) =n∑k=1

APk uk sin(ωt+ ϕk) (XIV.45)

With the amplitude

Bj(ω) =

n∑k=1

APk (uk)j

of the particular response of the jth degree of freedom, we can define the amplificationfactor

Vj(ω) =|Bj ||Bstatic

j |

with Bstaticj = Bj(0). This amplification factor expresses the ratio between the forced

and the static response of the jth degree of freedom as a function of the forcing frequencyωj . It has local maxima at all natural frequencies ω1, . . . , ωn of the system, as shownin Figure XIV.5. We also observe that between two adjacent maxima, there is a pointfor which the amplification factor becomes zero and hence the jth degree of freedom isnot excited at all, even though an external sinusoidal force is applied to the system. Thisphenomenon is called anti-resonance. This is exploited in practice for vibration eliminationby choosing system parameters for which an anti-resonance of a degree of freedom falls neara known external forcing frequency. Just as in the case of single degree of freedom systems,the amplification factor approaches infinity at the resonances when there is no dampingpresent:

limω→ωj

Vj(ω) =∞, when C = 0 .


Figure5.12

Mq + Kq = p

p(t)

Figure XIV.4: The frame of a car is excited by the external force p(t), as shown. Theequations of the linearized deformation of the car can be decomposed into the individualmodes of the system, which are graphically displayed (a two-mode model is assumed forthe car – Courtesy: M. Geradin, D. Rixen, Mechanical Vibrations, Wiley).


Figure5.13

!1 !2 !3

!...

ResonancesAnti Resonances

|Bj ||Bstatic

j |Vj(!)

Figure XIV.5: The amplification factor Vj(ω) for the jth degree of freedom is shown qualita-tively for a forced, damped vibratory system. We see that the maxima of the amplificationfactor lie at the natural frequencies of the system and that there is a minimum betweeneach adjacent maxima where the considered degree of freedom is not excited.

Bibliography

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