1. No category

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
Crank (Krank)
(Rotation)
Piston
(Translation)
O
w
Connecting rod (General Motion)
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.
Because the angle b is invariant, the relation q2 = q1 + b upon
differentiation with respect to time gives q2  q1 and q2  q1
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
dq 
w
q
dt
dw
a
 w
dt
or
ωdω  αdθ
or
a
d 2q
 q
dt 2
qdq  qdθ
For rotation with constant angular acceleration, the relationships
become
w  w 0  at
w 2  w0 2  2a q  q 0 
1 2
q  q 0  w 0 t  at
2
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
v  rw
an  rw 2  v 2 / r  vw
at  ra
These quantities may be expressed using the cross product
 

relationship of vector notation, w  wk , a  ak

   
v  r  wr

    
 d 
d   dw   dr
a  v   w  r  
r w
a
 r  w  w  r 




dt
dt
dt
dt

at




a
  
v wr
an
PROBLEMS
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.
SOLUTION
dq
w
dt
q

dq 
0
dq  wdt

 12  3t dt
3
2
0
Dq  9 rad
3
,
3 3
q  12t  t  123  33  9 rad
3 0
SOLUTION
Does the gear stop between t = 0 and t = 3 seconds?
w  12  3t 2  0
q1

dq 
0
12  3t 2 t  2 s (it stops at t  2 s )
 12  3t dt
2
2
2

3
q1  12t  t 3  122  23  16 rad
3 0

3
q 2  12t  t 3
3
0
q2

0
dq 
 12  3t dt
3
2
2
3
 7 rad
2
16   7  23 rad
1 revolution
2 rad
N revolutions
23 rad

N  3.66 revolutions
PROBLEMS
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.
SOLUTION
vC  1.5 m / s
w
a A  75 m / s 2
a) a  ?
b) a B  ?
c) a C  ?
vC
1.5

 20 rad / s
r
0.075
a An  w 2 R  20 2 0.15  60 m / s 2
a At  75 2  60 2  45 m / s 2
a
a At
R

45
 300 rad / s 2
0.15
a Bt  a  r  300  0.075  22.5 m / s 2
a Bn  w 2  r  20 2  0.075  30 m / s 2
aC  a  r  300  0.075  22.5 m / s 2


2
2
2
 a B  22.5  30  37.5 m / s

PROBLEMS
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.
SOLUTION
t2 s 
N B  300 rev / min
w B  300 
2
 10 rad / s
60
The velocities of gears A and B are same at the contact point.
w A b   w B 2b  
v A  vB 
aA  t 2
w A  20 
t 6 s 

2
t
 2t
2
aA 
dw A
dt
w A  20 rad / s
wA

dw
20
6
A

 t  2dt
t 2
6

w A  86.83 rad / s (at t  6 s)
2
w A b   w B 2b  
w B  43.415 rad / s
N B  414.59 rev / min
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.
PROBLEM
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 s .
PROBLEM
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.
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



v A  vB  v A / B
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 Dt. This movement may be visualized
as occurring in two parts. First, the body translates to the parallel

position A´´B´ with the displacement DrB . 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 DrA / B of A with respect to B.

DrA / B
With B as the reference point, the total displacement of A is



DrA  DrB  DrA / B

Where DrA / B has the magnitude rDq as Dq approaches zero.
Dividing the time interval
relative velocity equation
Dt and passing to the limit, we obtain the



v A  vB  v A / B
The distance r between A
and B remains constant.
The magnitude of the relative velocity is thus seen to be

 DrA / B 
dq
 rDq 


v A / B  lim
 lim 
r

 Dt 0  Dt 
Dt 0 
D
t
dt


which, with w  q becomes
v A / B  rw


Using r to represent the vector rA / B , we may write the
relative velocity as the vector
 

vA/ B  w  r
Therefore, the relative velocity equation becomes
 


v A  vB  w  r
Here,

w
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.
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 v B , plus
 

vA/ B  w  r
the rotational portion
, which has the magnitude

vA/B=rw, where w  q , the absolute angular velocity of AB . The
relative linear velocity is always perpendicular to the line joining the
two points A and B.
Relative Acceleration
Equation of relative velocity is



v A  vB  v A / B
By differentiating the equation with respect to time, we obtain the
relative acceleration equation, which is
or



v A  vB  v A / B



a A  aB  a A / B
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.
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 v A / B and a tangential component perpendicular to AB due to the

v
change in magnitude of A / B . Thus, we may write,




a A  aB  a A / B n  a A / B t
Where the magnitudes of the relative acceleration components are
a A / B n  v A / B 2 / r  rw 2
a A / B t  v A / B  ra
In vector notation the acceleration components are
  

a A / B n  w  w  r 
 

a A / B t  a  r
The relative acceleration equation, thus, becomes
    


a A  aB  w  w  r   a  r
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.
PROBLEMS
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.
vo=45 cm/s
O
37o
A
y
4.5 cm
6 cm
B
1
y  x2
4
4 cm
x
x=2 cm
10 cm
ao=90 cm/s2