Dynamics AE-001 Sharif University-Aerospace Dep. Fall 2004

Dynamics AE-001Dynamics AE-001

Sharif University-Aerospace Dep. Fall 2004

Rectangular ComponentsRectangular Components

Chapter 2 STATICS OF PARTICLES

Forces are vector quantities; they add according to theparallelogram law. The magnitude and direction of theresultant R of two forces P and Q can be determined eithergraphically or by trigonometry.

P

R

QA

A

Q

P

F

Any given force acting on a particle can be resolved into two or more components, i.e.., it can be replaced by two or more forces which have the same effect on the particle.

A force F can be resolvedinto two components Pand Q by drawing aparallelogram which hasF for its diagonal; thecomponents P and Qare then represented bythe two adjacent sidesof the parallelogramand can be determinedeither graphically or bytrigonometry.

x

y

Fx = Fx i

Fy = Fy j

F

i

j

A force F is said to have been resolved into two rectangularcomponents if its components are directed along the coordinateaxes. Introducing the unit vectors i and j along the x and y axes,

F = Fx i + Fy j

Fx = F cos Fy = F sin

tan = Fy

Fx

F = Fx + Fy2 2

When three or more coplanar forces act on a particle, the rectangular components of their resultant R can be obtained by adding algebraically the corresponding components of the given forces.

Rx = Rx Ry = Ry

The magnitude and direction of R can be determined from

tan = Ry

RxR = Rx + Ry

2 2

x

y

z

A

B

C

D

E

OFx

Fy

Fz

F

x

y

z

A

B

C

D

E

OFx

Fy

Fz

F

x

x

y

z

A

B

C

D

E

OFx

Fy

Fz

F

y

z

A force F in three-dimensional space can be resolved into components

Fx = F cos x Fy = F cos y

Fz = F cos z

x

y

z

Fy j

Fx i

Fz k

(Magnitude = 1)

cos x i

cos z k

cos y j F = F

The cosines ofx , y , and z

are known as thedirection cosines of the force F. Using the unit vectors i , j, and k, we write

F = Fx i + Fy j + Fz k

or

F = F (cosx i + cosy j + cosz k )

x

y

z

Fy j

Fx i

Fz k

(Magnitude = 1)

cos x i

cos z k

cos y j

F = F

= cosx i + cosy j + cosz k

Since the magnitude of is unity, we have

cos2x + cos2y

+ cos2z = 1

F = Fx + Fy + Fz2 2 2

cosx =Fx

Fcosy =

Fy

Fcosz =

Fz

F

In addition,

x

y

z

M (x1, y1, z1)

N (x2, y2, z2)

dx = x2 - x1

dz = z2 - z1 < 0

dy = y2 - y1

A force vector Fin three-dimensionsis defined by itsmagnitude F andtwo points M and N along its line ofaction. The vectorMN joining points and N is

MN = dx i + dy j + dz k

= = ( dx i + dy j + dz k ) MNMN

1d

F

The unit vector along the line of action of the force is

x

y

z

M (x1, y1, z1)

N (x2, y2, z2)

dx = x2 - x1

dz = z2 - z1 < 0

dy = y2 - y1

A force F is defined as theproduct of F and. Therefore,

F = F = ( dx i + dy j + dz k ) Fd

d = dx + dy + dz

22 2

From this it follows that

Fx =Fdx

dFy =

Fdy

dFz =

Fdz

d

When two or more forces act on a particle in three-dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces.

The particle is in equilibrium when the resultant of all forces acting on it is zero.

Rx = Fx

Ry = Fy

Rz = Fz

Vector AlgebraVector Algebra

q

V = P x Q

P

Q

The vector product of twovectors is defined as

V = P x QThe vector product of P and Q forms a vector which is

perpendicular to both P and Q, of magnitude

V = PQ sin This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through which brings vector P in line with vector Q. The three vectors P,Q, and V - taken in that order - form a right-hand triad. It follows that

Q x P = - (P x Q)

ik

j It follows from the definition of the vectorproduct of two vectors that the vectorproducts of unit vectors i, j, and k are

i x i = j x j = k x k = 0

i x j = k , j x k = i , k x i = j , i x k = - j , j x i = - k , k x j = - i

The rectangular components of the vector product V of twovectors P and Q are determined as follows: Given

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k

The determinant containing each component of P and Q isexpanded to define the vector V, as well as its scalar components

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k

V = P x Q =i

Px

Qx

jPy

Qy

kPz

Qz

= Vx i + Vy j + Vz k

where

Vx = Py Qz - Pz Qy

Vy = Pz Qx - Px Qz

Vz = Px Qy - Py Qx

Od A

F

Mo

r

The moment of force F about point O is defined as the vector product

MO = r x F

where r is the positionvector drawn from point O to the point of application of the force F. The angle between the lines of actionof r and F is .

The magnitude of the moment of F about O can be expressed as

MO = rF sin = Fd

where d is the perpendicular distance from O to the line of actionof F.

x

y

z

Fx i

Fz k

Fy j

x i

y j

z k

O

A (x , y, z )

r

The rectangular components of themoment Mo of a force F are determined byexpanding thedeterminant of r x F.

Mo = r x F =ixFx

jyFy

kzFz

= Mx i + My j + Mzk

where Mx = y Fz - z Fy My = zFx - x Fz

Mz = x Fy - y Fx

x

y

Fx i

Fz k

Fy j

O

r

In the more general case of the moment about an arbitrary point B of a force F applied at A, we have

MB = rA/B x F =i

xA/B

Fx

jyA/B

Fy

kzA/B

Fz

where

z

B (x B, yB, z B)

A (x A, yA, z A)

rA/B = xA/B i + yA/B j + zA/B k

and xA/B = xA- xB yA/B = yA- yB zA/B = zA- zB

x

y

Fx i

FFy j

O

MB = (xA- xB )Fy + (yA- yB ) Fx

z

B

A

(xA - xB ) i

rA/B(yA - yB ) j

In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalar

MB = MB k

The right-hand rule is useful for defining the direction of themoment as either into or out of the plane (positive or negativek direction).

P

QThe scalar product of two vectors P and Q is denoted as P Q ,and is defined as

The scalar product of P and Q is expressed in terms of the rectangular components of the two vectors as

P Q = PQ cos

P Q = Px Qx + Py Qy + Pz Qz

where is the angle between the two vectors

x

y

z

O

L

A

xP

z

y

The projection of a vectorP on an axis OL can be obtained by forming the scalar product of P and the unit vectoralong OL.

POL = P

Using rectangular components,

POL = Px cos x + Py cos y + Pz cos z

The mixed triple product of three vectors S, P, and Q is

S (P x Q ) =Sx

Px

Qx

Sy

Py

Qy

Sz

Pz

Qz

The elements of the determinant are the rectangular components of the three vectors.

x

F

O

The moment of a force F about an axis OL is the projection OC on OL of the moment MO of the force F. This can be written as a mixed triple product.

z

A (x, y, z)

yMO

L

C

r

x

xFx

y

yFy

z

zFz

MOL =MO =(r x F) =

x, y , z = direction cosines of axis OLx, y , z = components of rFx, Fy , Fz = components of F

Kinematics of ParticlesKinematics of Particles

Chapter 11Chapter 11

x

PO

x

The velocity v of the particle is equal to the time derivative of the position coordinate x,

v =dxdt

and the acceleration a is obtained by differentiating v with respect to t,

a =dvdt

or a =d 2xdt 2

we can also express a as

a = vdvdx

Two types of motion are frequently encountered: uniform rectilinear motion, in which the velocity v of the particle is constant and

x = xo + vt

and uniformly accelerated rectilinear motion, in which the acceleration a of the particle is constant and

v = vo + atx = xo + vot + at21

2

v2 = vo + 2a(x - xo )2

x

O

xA

xB

xB/A

A B

When particles A and B move along the same straight line, the relative motion of B with respect to A can be considered. Denoting by xB/A the relative position coordinate of B with respect to A , we have

xB = xA + xB/A

Differentiating twice with respect to t, we obtain

vB = vA + vB/A aB = aA + aB/A

where vB/A and aB/A represent, respectively, the relative velocity

and the relative acceleration of B with respect to A.

A

B

C

xA

xB

xC

When several blocks are are connected by inextensible cords, it is possible to write a linear relation between their position coordinates. Similar relations can then be written between their velocities and their accelerations and can be used to analyze their motion.

x

yC

P

an = en

O

v 2

at = et

dvdt

It is sometimes convenient to resolve the velocity and acceleration of a particle P into components other than the rectangular x, y, and z components. For a particle P moving along a path confined to a plane, we attach to P the unit vectors et tangent to the path and en normal to the path and directed toward the center of curvature of the path.

The velocity and acceleration are expressed in terms of tangential and normal components. The velocity of the particle is

v = vet

The acceleration is

a = et + en

v2

dvdt

v = vet

In these equations, v is the speed of the particle and is the radius of curvature of its path. The velocity vector v is directed along the tangent to the path. The acceleration vector a consists of a component at directed along the tangent to the path and a component an directed toward the center of curvature of the path,

a = et + en

v2

dvdt

x

yC

P

an = en

O

v 2

at = et

dvdt

x

P

O

e

r = r er

er

When the position of a particle moving in a plane is defined by its polar coordinates r and , it is convenient to use radial and transverse components directed, respectively, along the position vector r of the particle and in the direction obtained by rotating r through 90o counterclockwise. Unit

vectors er and e are attached to P and are directed in the radial and transverse directions. The velocity and acceleration of the particle in terms of radial and transverse components is

v = rer + re

. .

a = (r - r2)er + (r + 2r)e

... .. . .

x

P

O

e

r = r er

er v = rer + re

. .

a = (r - r2)er + (r + 2r)e

... .. . .

In these equations the dots represent differentiation with respect to time. The scalar components of of the velocity and acceleration in the radial and transverse directions are therefore

vr = r v = r. .

ar = r - r2 a = r + 2r... .. . .

It is important to note that ar is not equal to the time derivative of vr, and that a is not equal to the time derivative of v.

Newton’s Second LawNewton’s Second Law


To solve a problem involving the motion of a particle, F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have

Fx = max Fy = may Fz = maz

x

y

P

an

O

at

x

y

z

ax

ay

az

P

x

P

a

O

ar

Using tangential and normal components,

Ft = mat = mdvdtv2

Using radial and transverse components,

...

.. . .Fr = mar= m(r - r2)

r

Fn = man = m

F = ma = m(r + 2r)

0

mv0

r0

Or

P0

Pmv When the only force acting on a

particle P is a force F directed toward or away from a fixed point O, the particle is said to be moving under a central force. Since MO = 0 at any given instant, it follows that HO = 0 for all values of t, and

HO = constantWe conclude that the angular momentum of a particle moving under a central force is constant, both in magnitude and direction, and that the particle moves in a plane perpendicular to HO .

.

0

mv0

r0

Or

P0

Pmv

rmv sin = romvo sin o

Using polar coordinates and recalling that v = r and HO = mr2, we have

r2 = h.

where h is a constant representing the angular momentum per unit mass Ho/m, of the particle.

Recalling that HO = rmv sin , we have, for points PO and P

..

for the motion of any particle under a central force.

Work EnergyWork Energy


Chapter 13 KINETICS OF PARTICLES: ENERGY AND MOMENTUM METHODS

s2

A1

A2

A

s1

s

drF

ds

Consider a force F acting on a particle A. The work of F corresponding to the small displacement dr is defined as

dU = F drRecalling the definition of scalar product of two vectors,

dU = F ds cos

where is the angle between F and dr.

s2

A1

A2

A

s1

s

drF

ds

dU = F dr = F ds cos

The work of F during a finite displacement from A1 to A2 , denoted by U1 2 , is obtained

U1 2 = F drA1

A2

by integrating along the path described by the particle.

For a force defined by its rectangular components, we write

U1 2 = (Fxdx + Fydy + Fzdz) A1

A2

U1 2 = - Wdy = Wy1 - Wy2

y1

y2

y

dy

W

A1

A2

The work of the weight W of a body as its center of gravity moves from an elevation y1 to y2 is obtained by setting Fx = Fz = 0 and Fy = - W .

y1

y2

The work is negative when the elevation increases, and positive when the elevation decreases.

A

U1 2 = - k x dxx1

x2

= kx1 - kx2

B

B x1A1

A2

A

AO

F

B x2

x

The work of the force F exerted by a spring on a body A during a finite displacement of the body from A1 (x = x1) to A2 (x = x2) is obtained by writing

2 212

12

The work is positive when the spring is returning to its undeformed position.

.

spring undeformed

dU = -Fdx = -kx dx

The kinetic energy of a particle of mass m moving with a velocity v is defined as the scalar quantity

T = mv212

From Newton’s second law the principle of work and energy is derived. This principle states that the kinetic energy of a particle at A2 can be obtained by adding to its kinetic energy at A1 the work done during the displacement from A1 to A2 by the force F exerted on the particle:

T1 + U1 2 = T2

The power developed by a machine is defined as the time rate at which work is done:

Power = = F vdUdt

where F is the force exerted on the particle and v is the velocity of the particle. The mechanical efficiency, denoted by , is expressed as

=power outputpower input

When the work of a force F is independent of the path followed, the force F is said to be a conservative force, and its work is equal to minus the change in the potential energy V associated with F :

U1 2 = V1 - V2

The potential energy associated with each force considered earlier is

Force of gravity (weight):

Gravitational force:

Elastic force exerted by a spring:

Vg = Wy

Vg = - GMmr

Ve = kx212

U1 2 = V1 - V2

T1 + V 1 = T2 + V 1

This relationship between work and potential energy, when combined with the relationship between work and kinetic energy (T1 + U1 2 = T2) results in

This is the principle of conservation of energy, which states that when a particle moves under the action of conservative forces, the sum of its kinetic and potential energies remains constant. The application of this principle facilitates the solution of problems involving only conservative forces.

Impulse & MomentumImpulse & Momentum


The linear momentum of a particle is defined as the product mv of the mass m of the particle and its velocity v. From Newton’s second law, F = ma, we derive the relation

mv1 + F dt = mv2t1

t2

where mv1 and mv2 represent the momentum of the particle at a time t1 and a time t2 , respectively, and where the integral defines the linear impulse of the force F during the corresponding time interval. Therefore,

mv1 + Imp1 2 = mv2

which expresses the principle of impulse and momentum for a particle.

When the particle considered is subjected to several forces, the sum of the impulses of these forces should be used;

Since vector quantities are involved, it is necessary to consider their x and y components separately.

mv1 + Imp1 2 = mv2

The method of impulse and momentum is effective in the study of impulsive motion of a particle, when very large forces, called impulsive forces, are applied for a very short interval of time t, since this method involves impulses Ft of the forces, rather than the forces themselves. Neglecting the impulse of any nonimpulsive force, we write

mv1 + Ft = mv2

mv1 + Ft = mv2

In the case of the impulsive motion of several particles, we write

where the second term involves only impulsive, external forces.

In the particular case when the sum of the impulses of the external forces is zero, the equation above reduces to

mv1 = mv2

that is, the total momentum of the particles is conserved.

In the case of direct central impact, two colliding bodies A and B move along the line of impact with velocities vA and vB , respectively. Two equations can be used to determine their velocities v’A and v’B after the impact. The first represents the conservation of the total momentum of the two bodies,

mAvA + mBvB = mAv’A + mBv’B

A

B

vA

vB

Line of Impact

A

B

v’A

v’B

Before Impact

After Impact

mAvA + mBvB = mAv’A + mBv’B

The second equation relates the relative velocities of the two bodies before and after impact,

v’B - v’A = e (vA - vB )

The constant e is known as the coefficient of restitution; its value lies between 0 and 1 and depends on the material involved. When e = 0, the impact is termed perfectly plastic; when e = 1 , the impact is termed perfectly elastic.

A

B

vA

vB

Line of Impact

A

B

v’A

v’B

Before Impact

After Impact

A

B

vA

vB

In the case of oblique central impact, the velocities of the two colliding bodies before and after impact are resolved into n components along the line of impact and t components along the common tangent to the surfaces in contact. In the t direction,

mA (vA)n + mB (vB)n = mA (v’A)n + mB (v’B)n

Line of Impact

A

Bv’A

v’B

Before Impact

After Impact

n

t

vA

vB

(vA)t = (v’A)t (vB)t = (v’B)t

(v’B)n - (v’A)n = e [(vA)n - (vB)n]

while in the n directionn

t

A

B

vA

vB mA (vA)n + mB (vB)n = mA (v’A)n + mB (v’B)n

Line of Impact

A

Bv’A

v’B

Before Impact

After Impact

n

t

vA

vB

(vA)t = (v’A)t (vB)t = (v’B)t

(v’B)n - (v’A)n = e [(vA)n - (vB)n]

n

tAlthough this method was developed for bodies moving freely before and after impact, it could be extended to the case when one or both of the colliding bodies is constrained in its motion.

System of ParticleSystem of Particle


The theory in this chapter serves as a support to Chapter 16.

Chapter 14 SYSTEMS OF PARTICLES

The effective force of a particle Pi of a given system is the product miai of its mass mi and its acceleration ai with respect to a newtonian frame of reference centered at O. The system of the external forces acting on the particles and the system of the effective forces of the particles are equipollent; i.e., both systems have the same resultant and the same moment resultant about O :

Fi =miaii =1

n

i =1

n

(ri x Fi ) =(ri x miai)i =1

n

i =1

n

The linear momentum L and the angular momentum Ho about point O are defined as

L =mivii =1

n

Ho =(ri x mivi)i =1

n

It can be shown that

F = L Mo = Ho

. .

This expresses that the resultant and the moment resultant about O of the external forces are, respectively, equal to the rates of change of the linear momentum and of the angular momentum about O of the system of particles.

The mass center G of a system of particles is defined by a position vector r which satisfies the equation

mr =mirii =1

n

where m represents the total mass mi. Differentiating both

F = ma

.

i =1

n

members twice with respect to t, we obtain

L = mv L = ma.

where v and a are the velocity and acceleration of the mass center G. Since F = L, we obtain

Therefore, the mass center of a system of particles moves as if the entire mass of the system and all the external forces were concentrated at that point.

miv’i The kinetic energy T of a system of particles is defined as the sum of the kinetic energies of the particles.

T = mivi2

i = 1

n12

x

y

z

Ox’

y’

z’

Gr’i Pi

Using the centroidal reference frame Gx’y’z’ we note that the

kinetic energy of the system can also be obtained by adding the kinetic energy mv2 associated with the motion of the mass center G and the kinetic energy of the system in its motion relative to the frame Gx’y’z’ :

12

T = mv 2 + miv’ii = 1

n12

212

miv’i

x

y

z

Ox’

y’

z’

Gr’i Pi

T = mv 2 + miv’ii = 1

n12

212

The principle of work and energy can be applied to a system of particles as well as to individual particles

T1 + U1 2 = T2

where U1 2 represents the work of all the forces acting on the particles of the system, internal and external.

If all the forces acting on the particles of the system are conservative, the principle of conservation of energy can be applied to the system of particles

T1 + V1 = T2 + V2

x

y

O

(mAvA)1

(mBvB)1

(mCvC)1

x

y

O

(mAvA)2

(mBvB)2

(mCvC)2

x

y

t1

t2

Fdt

t1

t2

MOdtO

The principle of impulse and momentum for a system of particles can be expressed graphically as shown above. The momenta of the particles at time t1 and the impulses of the external forces from t1 to t2 form a system of vectors equipollent to the system of the momenta of the particles at time t2 .

x

y

O

(mAvA)1

(mBvB)1

(mCvC)1

x

y

O

(mAvA)2

(mBvB)2

(mCvC)2

If no external forces act on the system of particles, the systems of momenta shown above are equipollent and we have

L1 = L2 (HO)1 = (HO)2

Many problems involving the motion of systems of particles can be solved by applying simultaneously the principle of impulse and momentum and the principle of conservation of energy or by expressing that the linear momentum, angular momentum, and energy of the system are conserved.

(m)vA

mivi

AS

B

(m)vB

mivi

AS

B

For variable systems of particles, first consider a steady stream of particles, such as a stream of water diverted by a fixed vane or the flow of air through a jet engine. The principle of impulse and momentum is applied to a system S of particles during a time interval t, including particles which enter the system at A during that time interval and those (of the same mass m) which leave the system at B. The system formed by the momentum (m)vA of the particles entering S in the time t and the impulses of the forces exerted on S during that time is equipollent to the momentum (m)vB of the particles leaving S in the same time t.

S F t

M t

(m)vA

mivi

AS

B

(m)vB

mivi

AS

B

Equating the x components, y components, and moments about a fixed point of the vectors involved, we could obtain as many as three equations, which could be solved for the desired unknowns. From this result, we can derive the expression

F = (vB - vA)dmdt

where vB - vA represents the difference between the vectors vB and vA and where dm/dt is the mass rate of flow of the stream.

S F t

M t

Kinematics of Rigid BodiesKinematics of Rigid Bodies


Chapter 15 KINEMATICS OF RIGID BODIES

In rigid body translation, all points of the body have the same velocity and the same acceleration at any given instant.

Considering the rotation of a rigid body about a fixed axis, the position of the body is defined by the angle that the line BP, drawn from the axis of rotation to a point P of the body,

x

yA

rP

O

A’

B

z

v = = r sin

forms with a fixed plane. The magnitude of the velocity of P is

dsdt

.

where is the time derivative of ..

v = = r sin dsdt

.

The velocity of P is expressed as

v = = x rdrdt

where the vector

= k = k.

is directed along the fixed axis of rotation and represents the angular velocity of the body.

x

yA

rP

O

A’

B

z

v = = x rdrdt

The vector represents the angular acceleration of the body and is directed along the fixed axis of rotation.

Denoting by the derivative d/dt of the angular velocity, we express the acceleration of P as

a = x r + x ( x r)

= k = k.

differentiating and recalling that k is constant in magnitude and direction, we find that

= k = k = k. ..

x

yA

rP

O

A’

B

z

x

y

O

= k

v = k x r

r

Consider the motion of a representative slab located in a plane perpendicualr to the axis of rotation of the body. The angular velocity is perpendicular to the slab, so the velocity of point P of the slab is

x

y

= k

v = k x rwhere v is contained in the plane of the slab. The acceleration of point P can be resolved into tangential and normal components respectively equal to

at = k x r at = r

an= -2 r an = r2 = k

at = k x r

O an= -2 r

P

P

The angular velocity and angular acceleration of the slab can be expressed as

=ddt

= =ddt

d2dt2

= dd

or

Two particular cases of rotation are frequently encountered: uniform rotation and uniformly accelerated rotation. Problems involving either of these motions can be solved by using equations similar to those for uniform rectilinear motion and uniformly accelerated rectilinear motion of a particle, where x, v, and a are replaced by , , and .

A

B

vA

vB

Plane motion = Translation with A + Rotation about A

A

B

vA

vA

The most general plane motion of a rigid slab can be considered as the sum of a translation and a rotation. The slab shown can be assumed to translate with point A, while simultaneously rotating about A. It follows that the velocity of any point B of the slab can be expressed as

vB = vA + vB/A

where vA is the velocity of A and vB/A is the relative velocity of B with respect to A.

B

y’

x’

vB/A

rB/AA

(fixed)k

B

y’

x’

vB/A

rB/AA

(fixed)k

A

B

vA

vB


A

B

vA

vA vA vB

vB/A

vB = vA + vB/A

Denoting by rB/A the position of B relative to A, we note that

vB/A = k x rB/A vB/A = (rB/A )= r

The fundamental equation relating the absolute velocities of points A and B and the relative velocity of B with respect to A can be expressed in the form of a vector diagram and used to solve problems involving the motion of various types of mechanisms.

C

A

B

vA

vB

vA

vB

CAnother approach to the solution of problems involving the velocities of the points of a rigid slab in plane motion is based on determination of the instantaneous center of rotation C of the slab.


aB = aA + aB/A

The fact that any plane motion of a rigid slab can be considered the sum of a translation of the slab with reference to point A and a rotation about A is used to relate the absolute accelerations of any two points A and B of the slab and the relative acceleration of B with respect to A.

A

B

aA

aBA

B

aA

aA

A

B

y’

x’

(aB/A)n

kk

(aB/A)t

aB/A

where aB/A consists of a normal component (aB/A )n of magnitude r2 directed toward A, and a tangential component (aB/A )t of magnitude r perpendicular to the line AB.


aB = aA + aB/A

The fundamental equation relating the absolute accelerations of points A and B and the relative acceleration of B with respect to A can be expressed in the form of a vector diagram and used to determine the accelerations of given points of various mechanisms.

A

B

aA

aBA

B

aA

aA

A

B

y’

x’

(aB/A)n

kk

(aB/A)t

aB/A

(aB/A)n

(aB/A)t

aA

aB aB/A

The instantaneous center of rotation C cannot be used for the determination of accelerations, since point C , in general, does not have zero acceleration.


aB = aA + aB/A

A

B

aA

aBA

B

aA

aA

A

B

y’

x’

(aB/A)n

kk

(aB/A)t

aB/A

(aB/A)n

(aB/A)t

aA

aB aB/A

Inertia Inertia

Chapter 9Chapter 9

It is unlikely that you will calculate the mass moment of inertia. But the theory behind it has to be covered.

Chapter 9 DISTRIBUTED FORCES: MOMENTS OF INERTIA

x

y

y

dx

x

The rectangular moments of inertia Ix and Iy of an area are defined as

Ix = y 2dA Iy = x 2dA These computations are reduced to single integrations by choosing dA to be a thin strip parallel to one of the coordinate axes. The result is

dIx = y 3dx dIy = x 2ydx13

y

x

The polar moment of inertia of an area A with respect to the pole O is defined as

JO = r 2dA

The distance from O to the element of area dA is r. Observingthat r 2 =x 2 + y 2 , we established the relation

JO = Ix + Iy

x

yr

A

dA

O

yThe radius of gyration of an area A with respect to the x axis is defined asthe distance kx, whereIx = kx A. With similardefinitions for the radii ofgyration of A with respectto the y axis and with respect to O, we have

kx =

x

kx2

O

Ix

Aky =

Iy

AkO =

JO

A

A

The parallel-axis theorem states that the moment of inertia I of an area with respect to any given axis AA’ is equal to the moment of inertia I of the area with respect to the centroidal

I = I + Ad 2

A

B’

A’

B

d

c

axis BB’ that is parallel to AA’ plus the product of the area A and the square of the distance d between the two axes:

This expression can also be used to determine I when the moment of inertia with respect to AA’ is known:

I = I - Ad 2

A similar theorem can be used with the polar moment of inertia. The polar moment of inertiaJO of an area about O and the polar moment of inertia JC of the area about its

JO = JC + Ad 2

d

c

The parallel-axis theorem is used very effectively to compute the moment of inertia of a composite area with respect to a given axis.

o

centroid are related to the distance d between points C and Oby the relationship

r1

r2

m1

m2

r3m3

A

A’

Moments of inertia of mass areencountered in dynamics. Theyinvolve the rotation of a rigid bodyabout an axis. The mass moment of inertia of a body with respectto an axis AA’ is defined as

I = r 2dm

where r is the distance from AA’to the element of mass.

The radius of gyration of the body is defined as

k =Im

The moments of inertia of mass with respect to the coordinate axes are

Ix = (y 2 + z 2 ) dm

Iy = (z 2 + x 2 ) dm

Iz = (x 2 + y 2 ) dm

A

A’

B

B’d

G

The parallel-axis theorem also applies to mass moments of inertia.

I = I + d 2mI is the mass moment of inertia with respect to the centroidal BB’ axis, which is parallel to the AA’ axis. Themass of the body is m.

A

A’

B

B’

CC’

t

b

a

A

A’

B

B’

C

C’

tr

The moments of inertia of thin platescan be readily obtained from themoments of inertia of their areas. For a rectangular plate, the moments ofinertia are

IAA’ = ma 2 IBB’ = mb 2112

112

ICC’ = IAA’ + IBB’ = m (a 2 + b 2)112

For a circular plate they are

IAA’ = IBB’ = mr 214

ICC’ = IAA’ + IBB’ = mr 212

PLANE MOTION OF RIGID BODIES:PLANE MOTION OF RIGID BODIES:FORCES AND ACCELERATIONSFORCES AND ACCELERATIONS


Chapter 16 PLANE MOTION OF RIGID BODIES:FORCES AND ACCELERATIONS

The relations existing between the forces acting on a rigid body, the shape and mass of the body, and the motion produced are studied as the kinetics of rigid bodies. In general, our analysis is restricted to the plane motion of

G

F1

F2

F3

F4

HG

ma

G

.

rigid slabs and rigid bodies symmetrical with respect to the reference plane.

The two equations for the motion of a system of particles apply to the most general case of the motion of a rigid body. The first equation defines the motion of the mass center G of the body.

G

F1

F2

F3

F4

HG

ma

G

.

F = ma

where m is the mass of the body, and a the acceleration of G. The second is related to the motion of the body relative to a centroidal frame of reference.

MG = HG

.

G

F1

F2

F3

F4

HG

ma

G

.F = ma

MG = HG

.

where HG is the rate of change of the angular momentum HG of the body about its mass center G.

These equations express that the system of the external forces is equipollent to the system consisting of the vector ma attached at G and the couple of moment HG.

.

.

.

G

F1

F2

F3

F4

HG

ma

G

.

HG = I

For the plane motion of rigid slabs and rigid bodies symmetrical with respect to the reference plane, the angular momentum of the body is expressed as

where I is the moment of inertia of the body about a centroidal axis perpendicular to the reference plane and is the angular velocity of the body. Differentiating both members of this equation

HG = I= I. .

G

F1

F2

F3

F4For the restricted case considered here, the rate of change of the angular momentum of the rigid body can be represented by a vector of the same direction as (i.e.

The plane motion of a rigid body symmetrical with respect to the reference plane is defined by the three scalar equations

maG

I

perpendicular to the plane of reference) and of magnitude I.

Fx = max Fy = may MG = I

The external forces acting on a rigid body are actually equivalent to the effective forces of the various particles forming the body. This statement is known as d’Alembert’s principle.

G

F1

F2

F3

F4 d’Alembert’s principle can be expressed in the form of a vector diagram, where the effective forces are represented by a vector ma attached at G and a couple I. In the case of a slab in translation, the effective forces (part b of the figure) reduce to a

(a) (b)

single vector ma ; while in the particular case of a slab in centroidal rotation, they reduce to the single couple I ; in any other case of plane motion, both the vector ma and I should be included.

maG

I

G

F1

F2

F3

F4 Any problem involving the plane motion of a rigid slab may be solved by drawing a free-body-diagram equation similar to that shown. Three equations of of motion can then be obtained by equating the x components,

y components, and moments about an arbitrary point A, of the forces and vectors involved.

This method can be used to solve problems involving the plane motion of several connected rigid bodies.

Some problems, such as noncentroidal rotation of rods and plates, the rolling motion of spheres and wheels, and the plane motion of various types of linkages, which move under constraints, must be supplemented by kinematic analysis.

maG

I

PLANE MOTION OF RIGID BODIES:PLANE MOTION OF RIGID BODIES:ENERGY AND MOMENTUM ENERGY AND MOMENTUM

METHODSMETHODS


Chapter 17 PLANE MOTION OF RIGID BODIES:ENERGY AND MOMENTUM METHODS

The principle of work and energy for a rigid body is expressed in the form

T1 + U1 2 = T2

where T1 and T2 represent the initial and final values of the kinetic energy of the rigid body and U1 2 the work of the external forces acting on the rigid body.The work of a force F applied at a point A is

U1 2 = (F cos ) ds s1

s2

where F is the magnitude of the force, the angle it forms with the direction of motion of A, and s the variable of integration measuring the distance traveled by A along its path.

U1 2 = M ds 1

2

The work of a couple of moment M applied to a rigid body during a rotation in of the rigid body is

The kinetic energy of a rigid body in plane motion is

T = mv 2 + I212

12

where v is the velocity of the mass center G of the body, the angular velocity of the body, and I its moment of inertia about an axis through G perpendicular to the plane of reference.

G

v

T = mv 2 + I212

12

The kinetic energy of a rigid body in plane motion may be separated into two parts: (1) the kinetic energy mv 2 associated with the motion of the mass center G of the

12

12body, and (2) the kinetic energy I2 associated with the rotation

of the body about G.

For a rigid body rotating about a fixed axis through O with an angular velocity ,

T = IO212

O

where IO is the moment of inertia of the body about the fixed axis.

G

v

When a rigid body, or a system of rigid bodies, moves under the action of conservative forces, the principle of work and energy may be expressed in the form

which is referred to as the principle of conservation of energy. This principle may be used to solve problems involving conservative forces such as the force of gravity or the force exerted by a spring.

T1 + V1 = T2 + V2

The concept of power is extended to a rotating body subjected to a couple

Power = = = MdUdt

M ddt

where M is the magnitude of the couple and is the angular velocity of the body.

(m)v

The principle of impulse and momentum derived for a system of particles can be applied to the motion of a rigid body.

Syst Momenta1 + Syst Ext Imp1 2 = Syst Momenta2

Pmv

I

For a rigid slab or a rigid body symmetrical with respect to the reference plane, the system of the momenta of the particles forming the body is equivalent to a vector mv attached to the mass center G of the body and a couple I. The vector mv is associated with translation of the body with G and represents the linear momentum of the body, while the couple Icorresponds

to the rotation of the body about G and represents the angular momentum of the body about an axis through G.

x

y

O

I1

mv1

The principle of impulse and momentum can be expressed graphically by drawing three diagrams representing respectively the system of initial momenta of the body, the impulses of the external forces acting on it, and the system of the final momenta of the body. Summing and equating respectively the x components, the y components, and the moments about any given point of the vectors shown in the figure, we obtain three equations of motion which may be solved for the desired unknowns.

x

y

O x

y

O

I2

mv2Fdt

GG

x

y

O

I1

mv1

x

y

O x

y

O

I2

mv2Fdt

In problems dealing with several connected rigid bodies each body may be considered separately or, if no more than three unknowns are involved, the principles of impulse and momentum may be applied to the entire system, considering the impulses of the external forces only.

When the lines of action of all the external forces acting on a system of rigid bodies pass through a given point O, the angular momentum of the system about O is conserved.

G G

The eccentric impact of two rigid bodies is defined as an impact in which the mass centers of the colliding bodies are not located on the line of impact. In such a situation a relation for the impact involving the coefficient of restitution e holds, and the velocities of points A and B where contact occurs during the impact should be used.

n

A

vA

vB

nB

(a) Before impact

(v’B)n - (v’A)n = e[(vA)n - (vB)n] n

A

v’A

v’B

n

(b) After impact

B

(a) Before impact (b) After impact

(v’B)n - (v’A)n = e[(vA)n - (vB)n]

where (vA)n and (vB)n are the components along the line of impact of the velocities of A and B before impact, and (v’A)n and (v’B)n their components after impact. This equation is applicable not only when the colliding bodies move freely after impact but also when the bodies are partially constrained in their motion.

n

A

vA

vB

nB

n

A

v’A

v’B

n

B

Documents

Dynamics AE-001 Sharif University-Aerospace Dep. Fall 2004