KINEMATICS OFKINEMATICS OF

RIGID BODIESRIGID BODIESRIGID BODIESRIGID 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 gravitationalunder 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.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:categories:

1. Translation

2. Rotation

3. General Motion

1. TRANSLATION

It is any motion in which every line in the body remains parallel to itsoriginal position at all times. In translation, there is no rotation of anyline in the body.

1. Rectilinear Translation: All points in the body move in parallelstraight 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

ωωωωO

Crank (Krank)

(Rotation)

Connecting rod (General Motion)

Piston

(Translation)

hingehinge

Rotation

The rotation of a rigid body is described byits angular motion. The figure shows a rigidbody which is rotating as it undergoes planemotion in the plane of the figure. The angularpositions of any two lines 1 and 2 attached tothe body are specified by θ1 and θ2 measuredfrom any convenient fixed referencedirection.direction.

12 θθ && = 12 θθ &&&& =Because the angle β is invariant, the relation θ2 = θ1 + β upondifferentiation with respect to time gives andduring a finite interval, ∆ θ2 = ∆ θ1.

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 ω and angular acceleration α of a rigid body in

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

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

of the body. These definitions give

dθ

dθdorαdθ ωdω

dt

dor

dt

d

dt

d

θθθ

θθ

αωω

α

θθ

ω

&&&&

&&&

&

==

====

==

2

2

For rotation with constant angular acceleration, the relationships become

( )020

2

0

2

t

θθαωω

αωω

−+=

+=

200

00

2

1tt αωθθ ++=

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

α

ωω

ω

ra

vrvra

rv

t

n

=

===

=

/22

These quantities may be expressed using the cross product

relationship of vector notation,

rrvvv&vv

×== ω

kkrrrr

ααωω == ,

( ) ( ){ {

{ ( )43421

rvvrrr

vrv

rvrr

rr

rvrv nt aa

rrdt

rdr

dt

dr

dt

dv

dt

da ××+×=×+×=×== ωωαω

ωω

{ {rvrv n

t aarv ×=ωα

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

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

Find the net angular displacement ∆θ 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

=∆

=−=−=−=

=⇒=

∫∫θ

θθ

ωθωθ

θ

SOLUTION

( ) ( ) radttdttd

statstopsitsttt

1622123

312 312

) 2 ( 2 312 0312

32

0

31

2

0

2

0

22

1

=−=−=⇒−=

====−=

∫∫ θθ

ω

θ

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

SOLUTION

( )

srevolutionNradsrevolution N

radrevolution

rad

radttdttd

66.3 23

2 1

23716

73

312 312

3

2

32

3

2

2

0

2

=⇒

=−+

−=−=⇒−= ∫∫

π

θθθ

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 α of the pulley and disk, (b) the

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

belt.

PROBLEMS

belt.

222

222

CB2

45

/456075

/6015.020

/ 20075.0

5.1

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

a

sma

smRa

sradr

v

smasmv

A

A

C

AC

t

n

=−=

===

===

=====

ω

ω

α

SOLUTION

2

222

222

2

2

/5.22075.0300

/5.37305.22/30075.020

/5.22075.0300

/30015.0

45

smra

smasmra

smra

sradR

a

C

B

B

B

A

n

t

t

=⋅=⋅=

=+=

=⋅=⋅=

=⋅=⋅=

===

α

ω

α

α

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 α which varies with time for a

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

t=6 s.

PROBLEMS

t=6 s.

SOLUTION

srad

revNst

B

B

/1060

2300

min/3002

ππ

ω =⋅=

=⇒=

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

( ) ( ) /202 sradbbvv ABABA =⇒=⇒= πωωω

( )

( ) ( )min/59.414

/415.4326

)6(/83.8622

20

22

6

2

2

6

220

revN

sradbbst

statsradtt

dttddt

dt

B

BBA

AA

t

AA

AA

A

=

=⇒=⇒=

==⇒+=−

+=⇒=⇒+= ∫∫=

ωωω

ωπω

ωω

ααω

π

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 definingthen proceed to take the time derivatives of the defining

geometric relations to obtain velocities and accelerations.

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

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.

BABA vvv /vvv

+=

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.

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

position AB to A´B´ during time ∆t. 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 ∆θ, 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 /

v∆

Brv

∆

BAr /v

∆

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

BABA rrr /vvv

∆∆∆ +=

Where has the magnitude r∆θ as ∆θ approaches zero.

Dividing the time interval ∆t and passing to the limit, we obtain the

relative velocity equation

BAr /v

∆

vvvvvv

+= BABA vvv /vvv

+=

The distance r between A

and B remains constant.

The magnitude of the relative velocity is thus seen to be

which, with becomes

Using to represent the vector , we may write the

dt

dr

t

r

t

rv

t

BA

tBA

θθ=

∆∆

=

∆

∆=

→∆→∆ 0

/

0/ limlim

v

θω &=

ωrv BA =/

BAr /v

rv

rv BAvvv

×=ω/

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

BAr /r

Therefore, the relative velocity equation becomes

rvv BAvvvv

×+= ω

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

ωv

We can better understand the relative velocity equation by visualizing

the translation and rotation components separately.

Translation Fixed axid rotation

In the figure, point B is chosen as the reference point and theIn 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=rω, 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 BAvvv

×=ω/

Bvv

θω &v=

Relative Acceleration

Equation of relative velocity is

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

relative acceleration equation, which is

BABA vvv /vvv

+=

BABA vvv /&v&v&v +=

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 aaa /vvv

+=

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,BAv /v

BAv /v

Where the magnitudes of the relative acceleration components are

( ) ( )tBAnBABA aaaa //

vvvv++=

( )( ) α

ω

rva

rrva

BAtBA

BAnBA

==

==

//

22// /

&

In vector notation the acceleration components are

( ) ( )( ) ra

ra

tBA

nBAvvv

vvvv

×=

××=

α

ωω

/

/

The relative acceleration equation, thus, becomes

( ) rraa BAvvvvvvv

×+××+= αωω ( ) rraa BAvvvv

×+××+= αωω

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. The triangular plate has a constant clockwise angular velocity of

3 rad/s. Determine the angular velocities and angular accelerations

of link BC for this instant.

PROBLEMS

3. 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

4. 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

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

v =45 cm/s a =90 cm/s

37o

A

O4 cm

4.5 cm

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

10 cm

6 cm

B

x

y

x=2 cm2

4

1xy =