ENGR 214
Chapter 15
Kinematics of Rigid Bodies
All figures taken from Vector Mechanics for Engineers: Dynamics, Beer
and Johnston, 2004
1
Introduction
• 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
2
Motion of the plate: is it translation or rotation?
Curvilinear translation
Rotation
3
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
All particles have the same velocity.
• Differentiating with respect to time again,
rB  rA  rB A  rA


aB  a A
All particles have the same acceleration.
4
Rotation About a Fixed Axis
When a body rotates about a fixed axis, any
point P in the body travels along a circular path.
The angular position of P is defined by .
The change in angular position, d, is called the
angular displacement, with units of either
radians or revolutions. They are related by
1 revolution = 2 radians
Angular velocity, , is obtained by taking the
time derivative of angular displacement:
 = d/dt (rad/s) +
Similarly, angular acceleration is
 = d2/dt2 = d/dt or  = (d/d) +
rad/s2
5
Rotation About a Fixed Axis: Velocity
• Consider rotation of rigid body about a
fixed axis AA’


• Velocity vector v  dr dt of the particle P is
tangent to the path with magnitude v  ds dt
s  BP   r sin  
v
ds

 lim r sin  
 r sin 
dt t 0
t
• The same result is obtained from

 dr  
v
 r
dt




   k   k  angular velocity
6
Rotation About a Fixed Axis: Acceleration
• Differentiating to determine the acceleration,

 dv d  
a
   r 
dt dt


d   dr

r  
dt
dt

d   

r  v
dt

d 
   angular acceleration
•
dt





  k   k   k
• Acceleration of P is combination of two
vectors,
     
a    r     r
 
  r  tangential acceleration component
  
    r  radial acceleration component
7
Rotation About a Fixed Axis: Representative Slab
• Consider the motion of a representative slab in
a plane perpendicular to the axis of rotation.
• Velocity of any point P of the slab,
 
  
v    r  k  r
v  r
• Acceleration of any point P of the slab,
     
a    r     r
 

  k  r   2r
• Resolving the acceleration into tangential and
normal components,
 

at  k  r
a t  r


an   2 r
an  r 2
8
Examples
9
Equations Defining the Rotation of a Rigid Body
About a Fixed Axis
• Motion of a rigid body rotating around a fixed axis is
often specified by the type of angular acceleration.
• Recall  
d
dt
or
dt 
d

d d 2
d

 2 
dt
d
dt
• Uniform Rotation,  = 0:
   0  t
• Uniformly Accelerated Rotation,  = constant:
  0  t
   0   0t  12  t 2
 2   02  2    0 
10
Sample Problem 5.1
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.
11
Sample Problem 5.1
• The tangential velocity and acceleration of D are equal to the
velocity and acceleration of C.






a

a
D t
C  9 in. s 
v   v   12 in. s 
D 0
C 0
vD 0  r0
vD 0
0 
r
aD t  r
aD t

12
  4 rad s
3
r

9
 3 rad s 2
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 s2 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
 2 rad 
vB  r  5 in. 10 rad s 
yB  r  5 in. 14 rad 
N  2.23 rev

vB  50 in. s 
yB  70 in.
12
Sample Problem 5.1
• 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
aD t
 9 in. s 2 
aD n  48 in.
s2
s2 
Magnitude and direction of the total acceleration,
aD t2  aD 2n
aD 
 92  482
tan  

aD  48.8in. s2
aD n
aD t
48
9
  79.4
13
General Plane Motion
A combination of translation & rotation
14
General Plane Motion
Pure translation, followed by rotation about A2 (to move B'1 to B' 2)
Motion of B w.r.t. A is pure rotation, i.e. B draws a circle centered at A
Any plane motion can be represented as a translation of an
arbitrary reference point A and a rotation about A.
15
Absolute and Relative Velocity
For any two points lying on the same rigid body:
Note:
vB A  r
vB  vA  vB A
r = distance from A to B
vB A   k  rB / A
 


v B  v A   k  rB
A
Equation can be represented graphically by a velocity diagram
16
Absolute and Relative Velocity
Assuming that the velocity vA of end A is known, determine
the velocity vB of end B and the angular velocity .
Locus for vB
vA
vB
Locus
for vB/A

vB/A
vB  vA  vB A
The direction of vB and vB/A are known. Complete the velocity diagram.
tan  
vB
vA
 vB  v A tan 
vB A  cos
vB A  l   


l
l  cos
vB A
vA



 l cos
17
Absolute and Relative Velocity in Plane Motion
• Selecting point B as the reference point and solving for the velocity vA of end A
and the angular velocity  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  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.
18
Rolling Motion
Consider a circular disc that rolls without slipping on a flat surface
O1
O2
A2
r

From geometry:
A1
s
s  r
s = displacement of center
v  r  r
19
Sample Problem 15.2
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.
20
Sample Problem 15.2
vD/A
vB/A
vA  ArA
v A 1.2
A  
 8 rad / s
rA 0.15



For any point P on the gear: vP  vA  vP
 

A  vA  k  rP
For point B: vB  v A  vB A  1.2   8  0.1  2 m / s 
A
vR  vB  2 m / s 
For point D: vD  v A  vD A  1.2i   8  0.15 j  1.2i  1.2 j
21
Sample Problem 15.3
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.
22
Sample Problem 15.3



vD  vB  vD
B

• The velocity vB is obtained from the crank rotation data.
rev  min  2 rad 


  209.4 rad s
min  60 s  rev 

vB   AB  AB  3 in. 209.4 rad s 
 AB   2000

• The direction of the absolute velocity vD is horizontal.

The direction of the relative velocity vD B is
Locus for vD perpendicular to BD.
Locus for vD/B
23
Sample Problem 15.3
• Determine the velocity magnitudes vD and vD B
from the vector triangle drawn to scale.
vD  523.4 in. s  43.6 ft s



vD  vB  vD
vD
B
vD
B
 495.9 in. s
 l BD
B
vD
495.9 in. s
l
8 in.
 62.0 rad s
 BD 
B

24
Instantaneous Center of Rotation
For any body undergoing planar motion, there always exists a
point in the plane of motion at which the velocity is
instantaneously zero (if it were rigidly connected to the body).
This point is called the instantaneous center of
rotation, or C.
It may or may not lie on the body!
If the location of this point can be determined, the velocity
analysis can be simplified because the body appears to rotate
about this point at that instant.
25
Instantaneous Center of Rotation
To locate the C, we use the fact that the velocity of a point on a
body is always perpendicular to the position vector from C to that
point.
If the velocity at two points A and B are
known, C lies at the intersection of the
perpendiculars to the velocity vectors
through A and B .
If the velocity vectors at A and B
are perpendicular to the line AB,
C lies at the intersection of the
line AB with the line joining the
extremities of the velocity
vectors at A and B.
If the velocity vectors are equal & parallel, C is at infinity and the angular
velocity is zero (pure translation)
26
Instantaneous Center of Rotation
If the velocity vA of a point A on the body and
the angular velocity  of the body are
known, C is located along the line drawn
perpendicular to vA at A, at a distance
r = vA/ from A. Note that the C lies up and
to the right of A since vA must cause a
clockwise angular velocity  about C.
27
Velocity Analysis using Instantaneous Center
The velocity of any point on a body undergoing general plane
motion can be determined easily if the instantaneous center
is located.
Since the body seems to rotate about
the IC at any instant, the magnitude of
velocity of any arbitrary point is v =  r,
where r is the radial distance from the IC
to that point. The velocity’s line of action
is perpendicular to its associated radial
line. Note the velocity has a direction
which tends to move the point in a
manner consistent with the angular
rotation direction.
28
Velocity Analysis using Instantaneous Center
29
Instantaneous Center of Rotation
C lies at the intersection of the
perpendiculars to the velocity
vectors through A and B .
vA
vA


AC l cos
vA
v B  BC   l sin  
l cos
 v A tan 
The velocity of any point on the rod can be obtained.
Accelerations cannot be determined using C.
30
Sample Problem 15.4, using instantaneous center
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.
31
Sample Problem 15.4
Point C is in contact with the stationary lower rack
and, instantaneously, has zero velocity. It must be
the location of the instantaneous center of rotation.
v A 1.2
v A   rA   

 8rad s
rA 0.15
vR  vB  rB  8 0.25


vR  2 m s i
rD   0.15
 2  0.2121 m
vD   rD   8  0.2121
vD  1.697 m s



vD  1.2i  1.2 j m s 
32
Sample Problem 15.5 using instantaneous center
Crank-slider mechanism
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.
33
Sample Problem 15.5
C is at the intersection of the perpendiculars to the velocities through B and D.
vB   AB  AB   BD  BC 
BD
vB

BC
vD  BD  CD 
34
Absolute and Relative Acceleration
Absolute acceleration of point B:

Relative acceleration aB



aB  a A  aB A
A includes tangential and normal components:
 aB A   r
t
2
a

r

 B A
n
35
Absolute and Relative Acceleration




• Given a A and v A , determine a B and  .



aB  a A  aB A



 a A  a B A   a B A 
n
t

• Vector result depends on sense of a A and the
relative magnitudes of a A and a B A 
n
• Must also know angular velocity .
36
Absolute and Relative Acceleration
Draw acceleration diagram to scale:
37
Sample Problem 15.6
The center of the double gear has a velocity and acceleration
to the right of 1.2 m/s and 3 m/s2, respectively. The lower rack
is stationary.
Determine (a) the angular acceleration of the gear, and (b) the
acceleration of points B, C, and D.
38
Sample Problem 15.6
vA 1.2

 8 rad / s
r 0.15
aA
3
aA   r   

 20 rad / s 2
r 0.15
v A  r   
aB  a A  aB A
aC  a A  aC
A
 3i   2 r ( j )   ri
 3i   2 r ( j )   r ( i )
 3i  (8) 2 (0.1) j  (20)(0.1)i
 3i  (8) 2 (0.15) j  (20)(0.15)i
 5i  6.4 j
 9.6 j
aD  a A  aD A
 3i   2 r (i )   rj
 3i  (8) 2 (0.15)i  (20)(0.15) j
 12.6i  3 j
39
Sample Problem 15.7
Crank AG of the engine system has a constant clockwise
angular velocity of 2000 rpm.
For the crank position shown, determine the angular
acceleration of the connecting rod BD and the acceleration
of point D.
40
Sample Problem 15.7

t 






aD  aB  aD B  aB  aD B  aD B
n
 AB  2000 rpm  209.4rad s  constant
 AB  0
aB  r
2
AB

3
12
ft   209.4rad s   10,962ft s 2
2
From Sample Problem 15.3, BD = 62.0 rad/s, b = 13.95o.
41
Sample Problem 15.7
Draw acceleration diagram:

t 






aD  aB  aD B  aB  aD B  aD B
2
8 ft 62.0 rad s 2  2563 ft
aD B n  BD BD
 12
aD B t  BD  BD  128 ft  BD  0.667 BD
n
s2
Drawn to scale
42
Sample Problem 15.8
Four-bar mechanism
In the position shown, crank AB has a constant angular velocity
1 = 20 rad/s counterclockwise. Determine the angular
velocities and angular accelerations of the connecting rod BD
and crank DE.
43
Sample Problem 15.8
vB
vD
Velocities



vD  vB  vD
B
vB  1 ( AB)
Velocity diagram
vD/B
vB
vD/B
vD
vD B  BD ( BD)
Shown here not to scale
vD  DE (DE)

 BD

 29.33 rad s k

 DE

 11.29 rad s k
44
Sample
a Problem 15.8
aD/E t
D/B t
Accelerations



aD  aB  aD B
aD/B n
aD/E n
aB
Acceleration diagram
aD/E n
aB   ( AB)
2
a


 D B  BD (BD)
2
AB
aB
n
a   
a   
D E n
2
DE
( DE )
DE t
DE
( DE )



 BD   645 rad s k

2
aD/B n
aD/E t

 DE



 809 rad s k
2
Shown here not to scale
45