Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

Chapter Three
Kinematics of Rigid Bodies
3-1 Introduction to Dynamics
Kinematics of rigid bodies: relations between time and the positions,
velocities, and accelerations of the particles forming a rigid body.

Classification of rigid body motions:

- translation:
 rectilinear translation
 curvilinear translation
- rotation about a fixed axis
- general plane motion
- motion about a fixed point
- general motion

3-2 Translation
• Consider rigid body in translation:
- direction of any straight line inside the
body is constant,
- all particles forming the body move in
parallel lines.

• For any two particles in the body:

  
rB  rA  rB A

     
• Differentiating with respect to time: rB  rA  rB A  rA ....vB  v A
Then all particles have the same velocity.
• Differentiating with respect to time again:
r  r  r  r ...........a  a
B A B A A B A

And all particles have the same acceleration.

56 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

3-3 Rotation About a Fixed Axis: Velocity and Acceleration

Consider rotation of rigid body about a fixed axis

AA’. Velocity vector of the particle P is tangent to the
path with magnitude:

s  BP   r sin  


 lim r sin  
ds
v  r sin 
dt t  0 t

The same result is obtained from:


 dr  
v r
dt
  
   k  k  angular velocity

Differentiating to determine the acceleration:

   
 dv d   d   dr d   
a    r   r   r v
dt dt dt dt dt

d    
   angular acceleration   k   k  k
dt
     
Acceleration
 
of P is combination of two vectors: a    r      r
  r  tangentia l accelerati on component
  
    r  radial accelerati on component

Motion of a rigid body rotating around a fixed axis is often specified by the
type of angular acceleration.
d d d d 2 d
 or dt  .....   2 
dt  dt dt d

• Uniform Rotation, a = 0:   0  t
• Uniformly Accelerated Rotation, a = constant:
  0   t
   0  0t  12  t 2
 2  02  2    0 

57 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

3-4 Comparison Between Rotational and Linear Equations

The kinematics equations for rotational and translation motion:

Example1: Cable C 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 of
the pulley in 2 s, (b) the velocity and change in
position of the load B after 2 s, and (c) the acceleration
of the point D on the rim of the inner pulley at t = 0.

SOLUTION:

The tangential velocity and acceleration of D are equal

to the velocity and acceleration of C.
vD 0  vC 0  12 in. s  aD t  aC  9 in. s 
v D 0  r 0 aD t  r
v D 0 12 aD t 9
0    4 rad s    3 rad s 2
r 3 r 3

Apply the relations for uniformly accelerated rotation to

determine velocity and angular position of pulley after 2 s.
  0  t  4 rad s  3 rad s 2 2 s  10 rad s

  0t  12 t 2  4 rad s2 s  12 3 rad s 2 2 s2  14 rad

 1 rev 
N  14 rad    number of revs N  2.23 rev
 2 rad 

v B  r  5 in.10 rad s .........v B  50 in. s 
y B  r  5 in.14 rad ...........y B  70 in.

58 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

Evaluate the initial tangential and normal acceleration components of D.

aD t  aC  9 in. s 
aD n  rD02  3 in.4 rad s2  48 in s2

Magnitude and direction of the total acceleration:

aD  a D t2  a D 2n  9 2  48 2  48.8 in. s 2

aD n 48
tan    ……….   79.4
aD t 9

3-5 General Plane Motion

General plane motion is neither a translation nor a rotation. General
plane motion can be considered as the sum of a translation and rotation.

Displacement of particles A and B to A2 and B2 can be divided into two

parts:
- translation to A2 and
- rotation of about A2 to B2

59 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

3-6 Absolute and Relative Velocity in Plane Motion

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

reference point A and a simultaneous rotation about A.

  
vB  v A  vB
  A
v B A   k  rB A v B A  r
   
v B  v A   k  rB A

Assuming that the velocity vA of end A is known, wish to determine the

velocity vB of end B and the angular velocity ω in terms of vA, l, and θ.

The direction of vB and vB/A are known. Complete the velocity diagram.
vA v
vB
 tan   A  cos 
vA v B A l
vB  v A tan  
vA
l cos 

60 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

Selecting point B as the reference point and solving for the velocity vA of end A
and the angular velocity ω leads to an equivalent velocity triangle.

vA/B has the same magnitude but opposite sense of vB/A. The sense of the
relative velocity is dependent on the choice of reference point. Angular velocity
ω of the rod in its rotation about B is the same as its rotation about A. Angular
velocity is not dependent on the choice of reference point.

Example 2: The double gear rolls on the

stationary lower rack: the velocity of its center
is 1.2 m/s. Determine (a) the angular velocity
of the gear, and (b) the velocities of the upper
rack R and point D of the gear.

SOLUTION:

The displacement of the gear center in one revolution is equal to the outer
circumference. For xA > 0 (moves to right), θ < 0 (rotates clockwise).

xA 
 x A  r1
2 r 2

Differentiate to relate the translational and angular velocities.

v A  r1
  
vA 1.2 m s   k  8 rad s k
 
r1 0.150 m

61 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

For any point P on the gear:

Velocity of the upper rack is equal Velocity of the point D:

to velocity of point B:
        
v R  v B  v A  k  rB A vD  v A  k  rD A
     
 1.2 m s i  8 rad s k  0.10 m  j  1.2 m s i  8 rad s k   0.150 m i
    
 1.2 m s i  0.8 m s i  2 m s i  1.2 m s i  1.2 m s  j vD  1.697 m s

Example 3: 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, and (b) the velocity of the piston P.

SOLUTION:

The velocity vB is obtained from the crank rotation data:
 rev  min  2 rad 
 AB   2000     209.4 rad s
 min  60 s  rev 
vB   AB  AB  3 in.209.4 rad s 


The direction of the absolute velocity vD is horizontal. The direction of the

relative velocity vD B is perpendicular to BD. Compute the angle between the
horizontal and the connecting rod from the law of sines.

sin 40 sin 

   13.95
8 in. 3 in.

62 Asst. Prof. Dr. Sheelan M. Hama

Engineering Mechanics-Dynamics Kinematics of Rigid Bodies

Determine the velocity magnitudes from the vector

triangle:

vD vD B 628.3 in. s
 
sin 53.95 sin 50 sin76.05

v D  523.4 in. s  43.6 ft s

………. vP  vD  43.6 ft s
v D B  495.9 in. s

vD B 495.9 in. s
vD B  lBD ...........BD    62.0 rad s
 l 8 in.

BD  62.0 rad s k

63 Asst. Prof. Dr. Sheelan M. Hama