5-1

Chapter 5 Kinematics of Rigid Bodies

5.1 Introduction Various types of rigid-body motion are grouped as follows:

1. Translation: A motion is said to be translation if any straight line inside the body keeps the same

direction during the motion.

- rectilinear translation (Fig. 15.1)

- curvilinear translation (Fig. 15.2)

2. Rotation about a fixed axis: In this motion, the particles forming

the rigid body move in parallel planes along circles centered on the

same fixed axis of rotation (Fig. 15.3). Because each particle

moves in a given plane, the rotation of a body about a fixed axis is

said to be a plane motion.

3. General plane motion: There are many other types of plane motion, i.e., motions in which all the

particles of the body move in parallel planes. Any plane motion which is neither a rotation not a

translation is referred as a general plane motion (Fig. 15.5).

5-2

4. Motion about a fixed point: the three-dimensional motion of a rigid body attached at a fixed point O,

e.g., the motion of a top on a rough floor (Fig. 15.6), is known as motion about a fixed point.

5. General motion: Any motion of a rigid body, which does not fall in any of the categories above, is

referred as a general motion.

5.2 Translation Consider a rigid body in translation, and let A and B be any two of its particles (Fig. 15.7). Denoting,

respectively, by rA and rB the position vectors of A and B with respect to a fixed frame of reference and by

rB/A the vector joining A and B, we write

Since rB/A is constant by the definition of rigid body translation,

Thus, when a rigid body is in translation, all the points of the body have the same velocity and the same

acceleration at any given instant.

ABAB rrr +=

AB

AB

aavv

==

5-3

5.3 Rotation about a fixed axis Fig. 15.8, 15.9

Consider a rigid body which rotates about a fixed axis AA’. Let P be a point of the body and r its

position vector with respect to a fixed frame of reference. For convenience, let us assume that the frame

is centered at point O on AA’ and that the z axis coincides with AA’ (Fig. 15.8). Let B be the projection

of P on AA’; since P must remain at a constant distance from B, it will describe a circle center B and of

radius rsin ϕ, where ϕ denotes the angle formed by r and AA’.

The position of P and of entire body is completely defined by the angle θ the line BP forms with the zx

plane. The angle θ is known as the angular coordinate of the body and is defined as positive when

viewed as counterclockwise from A’. Then the velocity of P is

The velocity v of P is a vector perpendicular to the plane containing AA’ and r, and of magnitude defined

above. We thus write

The vector

(15.6)

is directed along the axis of rotation, is called the angular velocity of the body. The acceleration a of the

particle P is determined as

φθ sin&rv =

rωrv ×==dtd

( )

vωrω

rωrω

rωva

×+×=

×+×=

×==

dtd

dtd

dtd

dtd

dtd

kkω θω &==

5-4

The vector dω/dt is denoted by α and is called the angular acceleration of the body. Then

Tangential + Normal components

Recall

(11.39)

Differentiating (15.6) and recalling that k is constant in magnitude and direction, we have

Rotation of a representative slab

Fig. 15.10 and Fig. 15.11

The rotation of a fixed axis can be defined by the motion of a representative slab in a reference plane

perpendicular to the axis of rotation. Let us choose the xy plane as the reference plane and assume that it

coincides with the plane of Fig. 15.10, with the z axis pointing out of the paper. Then

We express the velocity of any given point P of the slab as

The acceleration of point P is expressed as

( )rωωrαa ××+×=

kkkα θωα &&& ===

nt eeaρ

2vdtdv

+=

kω ω=

rkv ×= ω

rrka 2ωα −×=

5-5

Resolving a into tangential and normal components, we write

5.4 Equations defining the rotation of a rigid body about a fixed axis Recalling the relations

Two particular cases of rotation are frequently encountered:

1. Uniform rotation (ω = constant)

2. Uniformly accelerated rotation (α = constant)

Example 5.1) Load B is connected to a double pulley by one of

the two inextensible cables shown. The motion of the pulley is

controlled by cable C, which has a constant acceleration of 9

in./s2 and an initial velocity of 12 in./s, both directed to the right.

Determine (a) the number of revolutions executed by the pulley

in 2s, (b) the velocity and change in position of the load B after

2s, and (c) the acceleration of point D on the rim of the inner

pulley at t = 0.

ra

rka2ω

α

−=

×=

n

t

θωωθωα

θω

dd

dtd

dtddtd

===

=

2

2

tωθθ += 0

( )02

02

200

0

221

θθαωω

αωθθ

αωω

−+=

++=

+=

tt

t

5-6

Ans)

(a) Motion of Pulley

(b) Motion of Load B

(c) Acceleration of Point D at t=0

Homework

#15.10, 15.18, 15.25, 15.30

( ) ( )( )( ) ( )( ) ( )

( )( )

( )( ) ( )( )

rev23.2214 srevolution ofNumber

rad142srad3212srad4

21

srad102srad3srad4

srad3in3sin9

srad4in3sin12sin9

sin12

22200

20

2

0000

200

==

=+=++=

=+=+=

=→=→=

=→=→=

==

==

π

αωθθ

αωω

ααα

ωωω

sstt

st

ra

rv

tD

D

CtD

CD

aa

vv

( )( )( )( ) inry

rv

B

B

70rad14in5sin50srad10in5

===Δ===

θω

( )( ) ( )( ) ↓===

→==222

0

2

sin48srad4in3

sin9

ωDnD

CtD

ra

aa

5-7

5.5 General Plane Motion A general plane motion can always be considered as the sum of a translation and a rotation.

Fig. 15.12 and 15.13

Although the original motion differs from the combination of translation and rotation when these

motions are taken in succession, the original motion can be exactly duplicated by a combination of

simultaneous translation and rotation.

5-8

Fig. 15.14

We consider a small displacement, which brings two particles A and B of a representative slab,

respectively, from A1 and B1 into A2 and B2. This displacement can be divided into two parts: in one, the

particles move into A2 and B1’ while the line AB maintains the same direction; in the other, B moves into

B2 while A remains fixed. The first part of the motion is clearly a translation and the second part a

rotation about A.

5.6 Absolute and relative velocity in plane motion Fig. 15.15

Any plane motion of a slab can be replaced by a translation defined by the motion of an arbitrary

reference point A and a simultaneous rotation about A. The absolute velocity vB of a particle B of the

slab is obtained as

As an example, consider the rod AB of Fig. 15.16 or 15.17. Assuming that the velocity vA of end A is

known, find the velocity vB of end B and the angular velocity ω of the rod, in terms of the velocity vA, the

length l, and the angle θ.

ABA

ABAB

rkv

vvv

×+=

+=

ω

5-9

i) Choosing A as a reference point, we express that the given motion is equivalent to a translation with A

and a simultaneous rotation about A. The absolute velocity of B is equivalent to the vector sum

Solving for the magnitudes vB and ω, we write

Their directions are illustrated in Fig. 15.16.

ii) The same result can be obtained by using B as a point of reference. Resolving the given motion into a

translation with B and a simultaneous rotation about B, we write the equation

ABAB vvv +=

θω

θ

cos

tan

lv

lvvv

AAB

AB

==

=

BABA vvv +=

5-10

We note that vA/B and vB/A have the same magnitude lω but opposite sense. The sense of the relative

velocity depends, therefore, upon the point of reference, which has been selected and should be carefully

ascertained from the appropriate diagram. Note that the angular velocity ω of a rigid body in plane

motion is independent of the reference point.

Example 15.3) In the engine system shown, the crank AB has a constant clockwise angular velocity of

2000 rpm. For the crank position indicated, determine (a) the angular velocity of the connecting rod BD,

(b) the velocity of the piston P.

Ans)

Motion of Crank AB

Motion of connecting Rod BD

Resolving the motion of BD into a translation with B and a rotation about B, we obtain

( )

( ) ( )( )o50sin3.628

sin3.628srad4.209in3

srad4.20960

rad22000

∠=

===

=⎟⎠⎞

⎜⎝⎛=

B

ABB

AB

ABvs

v

ω

πω

o

o

95.13in3

sinin840sin

=

=

β

β

5-11

Homework

15.57, 5.58, 15.58, 15.60, 15.61

15.7 Instantaneous Center of Rotation in Plane Motion Fig. 15.18, 15.19

At any instant the velocities of the various particles of the slab are the same as if the slab were

rotating about a certain axis of perpendicular to the plane of the slab, called the instantaneous axis of

rotation. Thus axis intersects the plane of the slab at a point C, called the instantaneous center of

rotation of the slab.

We first recall that the plane motion of a slab can always be replaced by a translation defined by the

motion of an arbitrary reference point A and by a rotation about A. As far as the velocities are

concerned, the translation is characterized by the velocity vA of the reference point A and the

rotation is characterized by the angular velocity ω of the slab. The instantaneous center C locates

on the perpendicular to vA at a distance r = vA /ω.

↑=

====

==

+=

rad/s0.62

sin9.495sin4.523

05.76sinsin3.628

50sin95.53sin

BD

BDBD

DP

BDD

lvvv

vv

ω

ω

ooo

BDBD vvv

5-12

Example (Fig. 15.20)

As far as the velocities are concerned, the slab seems to rotate about the instantaneous center C at

the instant considered.

In general, the particle C does not have zero acceleration and, therefore, that the accelerations of the

various particles of the slab cannot be determined as if the slab were rotating about C.

15.8 Absolute and Relative Acceleration in Plane Motion Fig. 15.22

( )( )

ABABAB

ABnAB

ABtAB

ABAB

rrkaa

ra

rka

aaa

2

2

ωα

ω

α

−×+=

−=

×=

+=

5-13

Fig.15.23 & 15.24

As an example, consider the rod AB whose extremities slide,

respectively, along a horizontal and a vertical track. Assuming that

the velocity vA and the acceleration aA of A are known, determine the

acceleration aB of B and the angular acceleration α of the rod.

Ans)

Four different vector polygons can be obtained, depending upon the

sense of aA and the relative magnitude of aA and (aB/A)n. If we are to

determine aB and α from one of 4 diagrams in Fig. 15.24, we must

know not only aA and θ but also ω; In the case of polygon a, we write

and solve for aB and α.

( ) ( )nABtABAB

ABAB

aaaa

aaa

++=

+=

θαθω

θαθω

sincos

cossin02

2

lla

lla

B

A

−−=−

−+=

5-14

Example 15.7) Crank AB of the engine system of Sample Pro. 15.3 has a constant clockwise angular

velocity of 2000rpm. For the crank position shown, determine the angular acceleration of the

connecting rod BD and the acceleration of point D.

Ans)

Motion of Crank AB

Motion of the connecting Rod BD.

From the result of prob. 15.3

The motion of BD is resolved into a translation with B and a rotation about B. The relative acceleration

aD/B is resolved into normal and tangential components:

( )o40sft962,10

sft962,10

0srad4.2092000

2

22

∠=

==

===

B

ABB

AB

AB

ra

rpm

a

ω

αω

o95.13

srad0.62

=

↑=

β

ω BD

( ) ( ) ( )( )( ) ( )( ) o

o

05.766667.0

6667.0

95.13sft2563

sft2563srad0.62ft128

2

222

∠=

==

∠=

=⎟⎠⎞

⎜⎝⎛==

BDnBD

BDBDtBD

nBD

BDnBD

BDa

BDa

α

αα

ω

a

a

5-15

While (aD/B )t must be perpendicular to BD, its sense is not known. Noting that the acceleration aD

must be horizontal, we write

Referring to the force diagram,

x components:

y components:

Solving the equations simultaneously, we obtain

Homework

15.111, 15.117, 15.125, 15.137

5.9 Rate of Change of a Vector with respect to a Rotating Frame Fig. 15.26

Consider two frames of reference centered at O, a fixed frame OXYZ and a frame Oxyz, which rotates

about the fixed axis OA; let Σ denote the angular velocity of the frame Oxyz at a given instant. Consider

now a vector function Q(t) represented by the vector Q attached at O; as the time t varies, both the

( ) ( )tBDnBDBBDBD aaaaaa ++=+=

ooo 95.13sin6667.095.13cos256340cos962,10 BDDa α+−−=−

ooo 95.13cos6667.095.13sin256340sin962,100 BDα++−=

←=

↑=2

2

sft9290

srad9940

D

BD

a

α

5-16

direction and the magnitude of Q change. Let us resolve the vector Q into components along the x, y

and z axes of the rotating frame.

With respect to the rotating frame Oxyz:

With respect to the fixed frame OXYZ:

The last three terms in the above equation represent the velocity of a particle located at the tip of Q and

fixed to the rotating frame with an angular velocity Ω:

Thus

We conclude that the rate of change of the vector Q with respect to the fixed frame OXYZ is made of two

parts: The first part represents the rate of change of Q with respect to the rotating frame Oxyz; the second

part, Ω x Q, is induced by the rotation of the frame Oxyz.

5.10 Plane Motion of a Particle Relative to a Rotating Frame. Coriolis

Acceleration The absolute velocity vp of the particle in Fig. 15.27 is defined as

the velocity observed from the fixed frame OXY and is equal to the

rate of change of r with respect to that frame. We can

also express vp in terms of the rate of change observed

from the rotating frame. Denoting by Ω the angular velocity of the

frame Oxy with respect to OXY at the instant considered, we write

Denoting the rotating frame by F for short, we represent the velocity

of P relative to the rotating frame by vp/F . Let us assume that a rigid slab

has been attached to the rotating frame. Then vp/F represents the velocity

of P along the path that it describes on the slab(Fig. 15.28), and the term

kjiQ zyx QQQ ++=

( ) kjiQ zyxOxyz QQQ &&&& ++=

( ) kjikjiQdtdQ

dtdQ

dtdQQQQ zyxzyxOXYZ +++++= &&&&

QΩkji ×=++dtdQ

dtdQ

dtdQ zyx

( ) ( ) QΩQQ ×+= OxyzOXYZ&&

( )OXYr&( )Oxyr&

( ) ( )OxyOXYp rrΩrv && +×==

( )Oxyr&

5-17

Ω x r represents the velocity vp’ of the point P’ of the slab –or rotating frame-which coincides with P at

the instant considered.

Thus, we have

where vp = absolute velocity of particle P

vp’ = velocity of point P’ of moving frame F coinciding with P,

vp/F = velocity of P relative to moving frame F,

The absolute acceleration ap of the particle is defined as

where ap = absolute acceleration of particle P

ap’ = acceleration of point P’ of moving frame F coinciding with P

ap/F = acceleration of P relative to moving frame F,

ac = complementary, or Coriolis, acceleration,

Fig. 15.29 (See the direction of the Coriolis effect)

Homework

15.172, 15.179

FPpp vvv += ′

( )[ ]

( )[ ] ( ) ( )

( )( ) ( ) ( )( ) ( ) ( )

cpp aaa

rrΩrΩΩrΩ

rΩrrrΩΩrΩa

rΩrr

rrΩrΩva

++=

+×+××+×=

×+++××+×=

×+=

+×+×==

′ F

OxyOxy

OxyOxyOxyp

OxyOxyOxy

Oxypp

dtd

dtd

&&&&

&&&&&

&&&&

&&&&

2

rΩ×( )Oxyr&

( )rΩΩrΩ ××+×= &

( )OxyrΩ &×2

( )Oxyr&&