KINEMATICS OF

RIGID BODIES

Introduction

In rigid body kinematics, we use the relationships

governing the displacement, velocity and acceleration, but

must also account for the rotational motion of the body.

Description of the motion of rigid bodies is important for

two reasons:

1) To generate, transmit or control motions by using

cams, gears and linkages of various types and analyze the

displacement, velocity and acceleration of the motion to

determine the design geometry of the mechanical parts.

Furthermore, as a result of the motion generated, forces

may be developed which must be accounted for in the

design of the parts.

2) To determine the motion of a rigid body caused by the

forces applied to it. Calculation of the motion of a rocket

under the influence of its thrust and gravitational

attraction is an example of such a problem.

Rigid Body Assumption

A rigid body is a system of particles for which the

distances between the particles and the angle between

the lines remain unchanged. Thus, if each particle of such

a body is located by a position vector from reference axes

attached to and rotating with the body, there will be no

change in any position vector as measured from these

axes.

Of course this is an idealization since all solid materials

change shape to some extent when forces are applied to

them. Nevertheless, if the movements associated with the

changes in shape are very small compared with the

movements of the body as a whole, then the assumption of

rigidity is usually acceptable.

Plane Motion

All parts of the body move in parallel planes.

The plane motion of a rigid body is divided into several

categories:

1. Translation

2. Rotation

3. General Motion

1. TRANSLATION It is any motion in which every line in the body remains parallel to its original position at all times. In translation, there is no rotation of any line in the body. 1. Rectilinear Translation: All points in the body move in parallel straight lines.

Rocket test sled

2. Curvilinear Translation: All points move on congruent curves.

In each of the two cases of translation, the motion of the body is

completely specified by the motion of any point in the body, since all

the points have the same motion.

2. Fixed Axis Rotation

Rotation about a fixed axis is the angular motion about the axis. All

particles in a rigid body move in circular paths about the axis of rotation

and all lines in the body which are perpendicular to the axis of rotation

rotate through the same angle at the same time.

A

B

C

A

B

C

3. General Plane Motion It is the combination of translation and rotation.

A A

B

B

w O

Crank (Krank)

(Rotation)

Connecting rod (General Motion)

Piston

(Translation)

hinge

Rotation The rotation of a rigid body is described by its angular motion. The figure shows a rigid body which is rotating as it undergoes plane motion in the plane of the figure. The angular positions of any two lines 1 and 2 attached to the body are specified by q1 and q2 measured from any convenient fixed reference direction.

12 qq 12 qq

Because the angle b is invariant, the relation q2 = q1 + b upon differentiation with respect to time gives and during a finite interval, D q2 = D q1. All lines on a rigid body in its plane of motion have the same angular displacement, the same angular velocity and the same angular acceleration.

Angular Motion Relations

The angular velocity w and angular acceleration a of a rigid body in

plane rotation are, respectively, the first and second time derivatives

of the angular position coordinate q of any line in the plane of motion

of the body. These definitions give

dθdorαdθ ωdω

dt

dor

dt

d

dt

d

qqq

qq

aww

a

qq

w

2

2

For rotation with constant angular acceleration, the relationships become

200

02

02

0

2

1

2

tt

t

awqq

qqaww

aww

Rotation About a Fixed Axis

When a rigid body rotates about a fixed axis, all points other than those

on the axis move in concentric circles about the fixed axis. Thus, for

the rigid body in the figure rotating about a fixed axis normal to the

plane of the figure through O, any point such as A moves in a circle of

radius r. So the velocity and the acceleration of point A can be written

as

a

ww

w

ra

vrvra

rv

t

n

/22

These quantities may be expressed using the cross product

relationship of vector notation,

rrv

w

kk

aaww ,

n

t aarv

rrdt

rdr

dt

dr

dt

dv

dt

da

wwaww

w

wa

1. The angular velocity of a gear is controlled according to

w = 12 – 3t2, where w in rad/s and t is the time in seconds.

Find the net angular displacement Dq from the time t = 0 to

t = 3 s. Also find the total number of revolutions N through

which the gear turns during the three seconds.

PROBLEMS

rad

radttdttd

dtddt

d

9

933123

312 , 312

33

0

33

0

2

0

D

q

qq

wqwq

q

SOLUTION

srevolutionNradsrevolution N

radrevolution

rad

radttdttd

radttdttd

statstopsitsttt

66.3 23

2 1

23716

73

312 312

1622123

312 312

) 2 ( 2 312 0312

3

2

32

3

2

2

0

32

0

31

2

0

2

0

22

2

1

qq

qq

w

q

q

Does the gear stop between t = 0 and t = 3 seconds?

SOLUTION

2. The belt-driven pulley and attached disk are rotating with increasing

angular velocity. At a certain instant the speed v of the belt is 1.5 m/s,

and the total acceleration of point A is 75 m/s2. For this instant

determine (a) the angular acceleration a of the pulley and disk, (b) the

total acceleration of point B, and (c) the acceleration of point C on the

belt.

PROBLEMS

2

222

222

2

2

222

222

CB2

/5.22075.0300

/5.37305.22/30075.020

/5.22075.0300

/30015.0

45

/456075

/6015.020

/ 20075.0

5.1

?ac)?ab)?a) / 75 / 5.1

smra

smasmra

smra

sradR

a

sma

smRa

sradr

v

smasmv

C

B

B

B

A

A

A

C

AC

n

t

t

t

n

a

w

a

a

w

w

a

SOLUTION

3. The design characteristics of a gear-reduction unit are under

review. Gear B is rotating clockwise (cw) with a speed of 300 rev/min

when a torque is applied to gear A at time t=2 s to give gear A a

counterclockwise (ccw) acceleration a which varies with time for a

duration of 4 seconds as shown. Determine the speed NB of gear B when

t=6 s.

PROBLEMS

SOLUTION

srad

revNst

B

B

/1060

2300

min/3002

w

The velocities of gears A and B are same at the contact point.

min/59.414

/415.4326

)6(/83.8622

20

22

/202

6

2

2

6

220

revN

sradbbst

statsradtt

dttddt

dt

sradbbvv

B

BBA

AA

t

AA

AA

ABABA

A

www

ww

ww

aa

www

w

Absolute Motion

In this approach, we make use of the geometric relations

which define the configuration of the body involved and

then proceed to take the time derivatives of the defining

geometric relations to obtain velocities and accelerations.

1) A wheel of radius r rolls on a flat surface without slipping. Determine the

angular motion of the wheel in terms of the linear motion of its center O. Also

determine the acceleration of a point on the rim of the wheel as the point

comes into contact with the surface on which the wheel rolls.

PROBLEM

2) Motion of the equilateral triangular plate ABC in its plane is controlled by the

hydraulic cylinder D. If the piston rod in the cylinder is moving upward at the

constant rate of 0.3 m/s during an interval of its motion, calculate for the

instant when q=30o the velocity and acceleration of the center of the roller B in

the horizontal guide and the angular velocity and angular acceleration of edge

CB.

PROBLEM

3) Derive an expression for the upward velocity v of the car hoist in terms of

q. The piston rod of the hydraulic cylinder is extending at the rate .

PROBLEM

s

4) Calculate the angular velocity w of the slender bar AB as a fuction of the

distance x and the constant angular velocity wo of the drum.

PROBLEM

Relative Motion

The second approach to rigid body kinematics uses the principles of

relative motion. In kinematics of particles for motion relative to

translating axes, we applied the relative velocity equation

to the motions of two particles A and B.

We now choose two points on the same rigid body for our two particles.

The consequence of this choice is that the motion of one point as seen by

an observer translating with the other point must be circular since the

radial distance to the observed point from the reference point does not

change.

BABA vvv /

The figure shows a rigid body moving in the plane of the figure from

position AB to A´B´ during time Dt. This movement may be visualized

as occurring in two parts. First, the body translates to the parallel

position A´´B´ with the displacement . Second, the body rotates

about B´ through the angle Dq, from the nonrotating reference axes

x´-y´ attached to the reference point B´, giving rise to the

displacement of A with respect to B. BAr /

D

BAr /

D

Br

D

With B as the reference point, the total displacement of A is

BABA rrr /

DDD

Where has the magnitude rDq as Dq approaches zero.

Dividing the time interval Dt and passing to the limit, we obtain the

relative velocity equation

BAr /

D

BABA vvv /

The distance r between A

and B remains constant.

The magnitude of the relative velocity is thus seen to be

which, with becomes

rv BA

w/

Using to represent the vector , we may write the relative velocity as the vector

dt

dr

t

r

t

rv

t

BA

tBA

qq

D

D

D

D

DD 0

/

0/ limlim

qw

wrv BA /

BAr /

r

Therefore, the relative velocity equation becomes

rvv BA

w

Here, is the angular velocity vector normal to the plane of the

motion in the sense determined by the right hand rule.

It should be noted that the direction of the relative velocity will

always be perpendicular to the line joining the points A and B.

Interpretation of the Relative Velocity Equation

We can better understand the relative velocity equation by visualizing

the translation and rotation components separately.

w

Translation Fixed axid rotation

In the figure, point B is chosen as the reference point and the

velocity of A is the vector sum of the translational portion , plus

the rotational portion , which has the magnitude

vA/B=rw, where , the absolute angular velocity of AB . The

relative linear velocity is always perpendicular to the line joining the

two points A and B.

rv BA

w/

Bv

qw

Relative Acceleration Equation of relative velocity is

By differentiating the equation with respect to time, we obtain the

relative acceleration equation, which is

or

This equation states that the acceleration of point A equals the vector

sum of the acceleration of point B and the acceleration which A appears

to have to a nonrotating observer moving with B.

BABA vvv /

BABA vvv /

BABA aaa /

If points A and B are located on the same rigid body, the distance r

between them remains constant. Because the relative motion is

circular, the relative acceleration term will have both a normal

component directed from A toward B due to the change of direction

of and a tangential component perpendicular to AB due to the

change in magnitude of . Thus, we may write,

Where the magnitudes of the relative acceleration components are

BAv /

BAv /

tBAnBABA aaaa //

a

w

rva

rrva

BAtBA

BAnBA

//

22// /

In vector notation the acceleration components are

ra

ra

tBA

nBA

a

ww

/

/

The relative acceleration equation, thus, becomes

rraa BA

aww

The figure shows the acceleration of A to be composed of two parts:

the acceleration of B and the acceleration of A with respect to B.

Solution of the Relative Acceleration Equation

As in the case of the relative velocity equation, the

relative acceleration equation may be carried out by

scalar or vector algebra or by graphical construction.

Because the normal acceleration components depend on

velocities, it is generally necessary to solve for the

velocities before the acceleration calculations can be

made.

1. The center O of the disk has the velocity and acceleration shown. If the

disk rolls without slipping on the horizontal surface, determine the velocity

of A and the acceleration of B for the instant represented.

PROBLEMS

2. If the velocity of point A is 3 m/s to the right and is constant for an

interval including the position shown, determine the tangential acceleration

of point B along its path and the angular acceleration of the bar AB.

PROBLEMS

3. The flexible band F attached to the sector at E is given a constant

velocity of 4 m/s as shown. For the instant when BD is perpendicular to OA,

determine the angular acceleration of BD.

PROBLEMS

4. A mechanism for pushing small boxes from an assembly line onto a conveyor

belt is shown with arm OD and crank CB in their vertical positions. For the

configuration shown, crank CB has a constant clockwise angular velocity of

rad/s. Determine the acceleration of E.

PROBLEMS

5. At a given instant, the gear has the angular motion shown. Determine the

acceleration of points A and B on the link and the link’s angular accelartion at

this instant.

PROBLEMS

6. The center O of the disk rolling without slipping on the horizontal surface

has the velocity and acceleration shown. Radius of the disk is 4.5 cm.

Calculate the velocity and acceleration of point B.

PROBLEMS

37o A

O 4 cm

4.5 cm

vo=45 cm/s ao=90 cm/s2

10 cm

6 cm

B

x

y

x=2 cm 2

4

1xy