MechanicsPhysics 151

Lecture 9Rigid Body Motion

(Chapter 4, 5)

What We Did Last Time

Discussed 3-dimensional rotationPreparation for rigid body motion

Movement in 3-d + Rotation in 3-d = 6 coordinates

Looked for ways to describe 3-d rotationEuler angles one of the many possibilitiesEuler’s theorem

Defined infinitesimal rotation dΩCommutative (unlike finite rotation)Behaves as an axial vector (like angular momentum)

d d= ×r r Ωd d= ΦΩ n

Goals For Today

Time derivatives in rotating coordinate systemTo calculate velocities and accelerationsCoriolis effectExpress angular velocity using Euler angles

Try to write down Lagrangian for rigid bodySeparate rotation from movement of CoMDefine inertia tensor

Will almost get to the equation of motion

Body Coordinates

Consider a rotating rigid bodyDefine body coordinates (x’, y’, z’)

Between t and t + dt, the bodycoordinates rotate by

The direction is CCW because weare talking about the coordinate axesObserved in the space coordinates,any point r on the body moves by

x x′

y

y′

zz′

d d d= Φ × = ×r n r Ω r

n

rdr

dΦ

d d= ΦΩ n

Sign opposite from last lecturebecause the rotation is CCW

General Vectors

Now consider a general vector GHow does it move in space/body coordinates?i.e. what’s the time derivative dG/dt ?

Movement dG differs in space and body coordinatesbecause of the rotation of the latter

If G is fixed to the body

( ) ( ) ( )space body rotd d d= +G G G Difference is due to rotation

( )body0d =G ( )space

d d= ×G Ω Gand

( )rotd d= ×G Ω G Generally true

Angular Velocity

For any vector G

ω = instantaneous angular velocityDirection = n = instantaneous axis of rotationMagnitude = dΦ/dt = instantaneous rate of rotation

Since this works for any vector, we can say

( ) ( )space bodyd d d= + ×G G Ω G

space body

d ddt dt

⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠G G ω G dt d d= = Φω Ω n

s r

d ddt dt

⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ω

space coordinates rotating coordinates

Coriolis Effect

Imagine a particle observed in a rotating systeme.g. watching an object’s motion on EarthVelocity:

Acceleration:

Newton’s equation works in the space (inertial) system,i.e.

s r

d ddt dt

⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ω

s r= + ×v v ω r

2 ( )

s ss s

s r

r r

d ddt dt

⎛ ⎞ ⎛ ⎞= = + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + × + × ×

v va ω v

a ω v ω ω r

sm=F a

eff 2 ( ) ( )r rm m m= = − × − × ×a F F ω v ω ω r

Object appears to move according to this force

Coriolis Effect

Last term is centrifugal forceMiddle term is Coriolis effect

Occurs when object is moving in therotating frame

Moving objects on Earth appears to deflect toward right in the northern hemisphere

Hurricane wind patternFoucault pendulum

eff 2 ( ) ( )r rm m m= = − × − × ×a F F ω v ω ω r ω

rv

r− ×ω v

L

Coriolis Effect

Free-falling particleDrop a ball and ignore air resistanceCoriolis effect is

Pointing east (out of screen) x axisIgnoring the centrifugal force

For sinθ = 1 and z = 100 m x = 2.2 cm

ω

θ

v

2 ( )m− ×ω vz

2 sinzmx m vmz mg

ω θ= −⎧⎨ = −⎩

3 2sin ,3 2

gt gtx zω θ= = −

3sin ( 2 )3

zxg

ω θ −=

Euler Angles

Use angular velocity ω to calculate particles’ velocitiesUse Euler angles to describe the rotation of rigid bodies

How are they connected?Infinitesimal rotations can be added like vectors

z

xy

ξ

η

ζ z

xy

ξ ′

η′ζ ′

φ

θz

xy

z′

x′

y′

ψ

z zξφ θ ψ′= + +ω n n n

zn

ξn z′n

Euler Angles

Let’s express ω in x’-y’-z’Doing this in x-y-z equally easy (or difficult)Must express nz, nξ, nz’ in x’-y’-z’nz Anz, nξ Bnξ (nz’ is OK)

z

xy

ξ

η

ζ z

xy

ξ ′

η′ζ ′

φ

θz

xy

z′

x′

y′

ψ

zn

ξn z′n

D C B

Euler Angles

From the last lecturecos sin 0sin cos 00 0 1

φ φφ φ

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

D1 0 00 cos sin0 sin cos

θ θθ θ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

Ccos sin 0sin cos 00 0 1

ψ ψψ ψ

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

B

cos cos cos sin sin cos sin cos cos sin sin sinsin cos cos sin cos sin sin cos cos cos cos sin

sin sin sin cos cos

ψ φ θ φ ψ ψ φ θ φ ψ ψ θψ φ θ φ ψ ψ φ θ φ ψ ψ θ

θ φ θ φ θ

− +⎡ ⎤⎢ ⎥= − − − +⎢ ⎥⎢ ⎥−⎣ ⎦

A

0 sin sin0 cos sin1 cos

z

ψ θψ θ

θ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

n A1 cos0 sin0 0

ξ

ψψ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

n B001

z′

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

n

Euler Angles

Remember: this is in the x’-y’-z’ system

Now we know how to express velocities in terms of time-derivatives of Euler angles

We can write down the Lagrangian

sin sin coscos sin sin

cosz zξ

φ ψ θ θ ψφ θ ψ φ ψ θ θ ψ

φ θ ψ′

⎡ ⎤+⎢ ⎥= + + = −⎢ ⎥⎢ ⎥+⎣ ⎦

ω n n n

Kinetic Energy

Kinetic energy of multi-particle system is

If we define the body axis from the center of mass

T’ depends only on the angular velocityMust be a 2nd order homogeneous function

2 21 12 2 i iT Mv m v′= +

Motion of CoM Motion around CoM

2 2 21 ( ) ( , , )2

T M x y z T φ θ ψ′= + + +

Remember Einstein convention

Potential Energy

Potential energy can often be separated as well

Uniform gravity gUniform magnetic field B and magnetic dipole moment M

Lagrangian can then be written as

1 2( , , ) ( , , )V V x y z V φ θ ψ= +

1V = − ⋅g r

2V = − ⋅M B

( , , , , , ) ( , , , , , )t rL L x y z x y z L φ θ ψ φ θ ψ= +

Translational Rotational

It is often possible to separate the translational and rotational motions by taking the center of mass as the origin of the body coordinate axes

Rotational Motion

We concentrate on the rotational partTranslational part same as a single particle Easy

Consider total angular momentumvi is given by the rotation ω as

i i im= ×L r v

i i= ×v ω r

2 2

2 2 2

2 2

( )

( )( ) ( )

( )

i i i

i i i i i i i i i

i i i i i i i i i i i i i

i i i i i i i i i

m

m r x m x y m x zm r m y x m r y m y z

m z x m z y m r z

= × ×

⎡ ⎤− − −⎢ ⎥⎡ ⎤= − ⋅ = − − −⎢ ⎥⎣ ⎦⎢ ⎥− − −⎣ ⎦

L r ω r

ω r r ω ω

Inertia tensor I

Inertia Tensor

Diagonal components are familiarmoment of inertia

What are the off-diagonal components?Iyx produces Ly when the object is turned around x axisImagine turning something like:

2 2 2 2( ) sinxx i i i i iI m r x m r= − = Θx

r

Θ

Balanced Unbalanced

Unbalanced one has non-zero off-diagonal components, which represents “wobbliness”of rotation

Inertia Tensor

Using

We can also deal with continuous mass distribution ρ(r)

1 2 3( , , ) ( , , )i i i i i ix y z x x x→

2 2

2 2

2 2

( )( )

( )

i i i i i i i i i

i i i i i i i i i

i i i i i i i i i

m r x m x y m x zm y x m r y m y zm z x m z y m r z

⎡ ⎤− − −⎢ ⎥= − − −⎢ ⎥⎢ ⎥− − −⎣ ⎦

I

( )2( )jk jk j kI r x x dρ δ= −∫ r r

( )2jk i i jk ij ikI m r x xδ= −

Kinetic Energy

Kinetic energy due to rotation isUsing

Using

Defining n as the unit vector in the direction of ω

NB: n moves with timeI = I(t) must change accordingly with time

12 i i iT m= ⋅v v

i i= ×v ω r

1 ( ) ( )2 2 2i i i i i iT m m= ⋅ × = ⋅ × = ⋅

ω ωv ω r r v L

=L Iω2

T ⋅=ω Iω

2 212 2

T Iω ω⋅= =n In

where 2 2( )i i iI m r⎡ ⎤= ⋅ = − ⋅⎣ ⎦n In r n

Moment of inertia about axis n

( )2jk i i jk ij ikI m r x xδ= −

ω=ω n

Shifting Origin

Origin of body axes does not have to be at the CoMIt’s convenient – Separates translational/rotational motion

If it isn’t, I can be easily translatedi i′= +r R rfrom origin from CoM

2 2

2 2

( ) [( ) ]

( ) ( ) 2 ( ) ( )i i i i

i i i i

I m m

M m m

′= × = + ×

′ ′= × + × + × ⋅ ×

r n R r n

R n r n R n r n

I from CoMI of CoM

Summary

Found the velocity due to rotationUsed it to find Coriolis effect

Connected ω with the Euler anglesLagrangian translational and rotational parts

Often possible if body axes are defined from the CoM

Defined the inertia tensorCalculated angular momentumand kinetic energy

Next: Equation of motion (finally!)

s r

d ddt dt

⎛ ⎞ ⎛ ⎞= + ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ω

( )2i i jk ij ikm r x xδ= −I

=L Iω 2 212 2

T Iω ω= =nIn