kinematics-of-rigid-body.ppt

VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ninth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
15
Kinematics of
Rigid Bodies

Vector Mechanics for Engineers: Dynamics
Ninth
Edition
Contents
15 - 2
Introduction
Translation
Rotation About a Fixed Axis: Velocity
Rotation About a Fixed Axis:
Acceleration
Rotation About a Fixed Axis:
Representative Slab
Equations Defining the Rotation of a
Rigid Body About a Fixed Axis
Sample Problem 5.1
General Plane Motion
Absolute and Relative Velocity in Plane
Motion
Sample Problem 15.2
Sample Problem 15.3
Instantaneous Center of Rotation in
Plane Motion
Sample Problem 15.4
Sample Problem 15.5
Absolute and Relative Acceleration in
Plane Motion
Analysis of Plane Motion in Terms of a
Parameter
Sample Problem 15.6
Sample Problem 15.7
Sample Problem 15.8
Rate of Change With Respect to a
Rotating Frame
Coriolis Acceleration
Sample Problem 15.9
Sample Problem 15.10
Motion About a Fixed Point
General Motion
Three Dimensional Motion. Coriolis
Acceleration
Frame of Reference in General Motion

Ninth
Edition
Introduction
15 - 3
• 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:
- general motion
- motion about a fixed point
- general plane motion
- rotation about a fixed axis
• curvilinear translation
• rectilinear translation
- translation:

Ninth
Edition
Translation
15 - 4
• 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,
A
B
A
B r
r
r





• Differentiating with respect to time,
A
B
A
A
B
A
B
v
v
r
r
r
r














All particles have the same velocity.
A
B
A
A
B
A
B
a
a
r
r
r
r


















• Differentiating with respect to time again,
All particles have the same acceleration.

Ninth
Edition
Rotation About a Fixed Axis. Velocity
15 - 5
• Consider rotation of rigid body about a
fixed axis AA’
• Velocity vector of the particle P is
tangent to the path with magnitude
dt
r
d
v



dt
ds
v 
   
  






sin
sin
lim
sin
0

r
t
r
dt
ds
v
r
BP
s
t












locity
angular ve
k
k
r
dt
r
d
v


















• The same result is obtained from

Ninth
Edition
Rotation About a Fixed Axis. Acceleration
15 - 6
• Differentiating to determine the acceleration,
 
v
r
dt
d
dt
r
d
r
dt
d
r
dt
d
dt
v
d
a




























•
k
k
k
celeration
angular ac
dt
d


















component
on
accelerati
radial
component
on
accelerati
l
tangentia










r
r
r
r
a

















• Acceleration of P is combination of two
vectors,

Ninth
Edition
Rotation About a Fixed Axis. Representative Slab
15 - 7
• Consider the motion of a representative slab in
a plane perpendicular to the axis of rotation.
• Velocity of any point P of the slab,



r
v
r
k
r
v










• Acceleration of any point P of the slab,
r
r
k
r
r
a









2













• Resolving the acceleration into tangential and
normal components,
2
2




r
a
r
a
r
a
r
k
a
n
n
t
t












Ninth
Edition
Equations Defining the Rotation of a Rigid Body About a Fixed Axis
15 - 8
• Motion of a rigid body rotating around a fixed axis is
often specified by the type of angular acceleration.










d
d
dt
d
dt
d
d
dt
dt
d





2
2
or
• Recall
• Uniform Rotation,  = 0:
t


 
 0
• Uniformly Accelerated Rotation,  = constant:
 
0
2
0
2
2
2
1
0
0
0
2 



















t
t
t

Ninth
Edition
Sample Problem 15.1
15 - 9
Cable C has a constant acceleration of
225 mm/s2 and an initial velocity of 300
mm/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:
• Due to the action of the cable, the
tangential velocity and acceleration of
D are equal to the velocity and
acceleration of C. Calculate the initial
angular velocity and acceleration.
• Apply the relations for uniformly
accelerated rotation to determine the
velocity and angular position of the
pulley after 2 s.
• Evaluate the initial tangential and
normal acceleration components of D.

Ninth
Edition
Sample Problem 5.1
15 - 10
SOLUTION:
• The tangential velocity and acceleration of D are equal to the
velocity and acceleration of C.
   
 
 
s
rad
4
75
300
s
mm
300
0
0
0
0
0
0







r
v
r
v
v
v
D
D
C
D



  
 
  2
s
rad
3
3
9
s
in.
9







r
a
r
a
a
a
t
D
t
D
C
t
D




• Apply the relations for uniformly accelerated rotation to
determine velocity and angular position of pulley after 2 s.
   s
rad
10
s
2
s
rad
3
s
rad
4 2
0 



 t



     
rad
14
s
2
s
rad
3
s
2
s
rad
4 2
2
2
1
2
2
1
0




 t
t 


  revs
of
number
rad
2
rev
1
rad
14 








N rev
23
.
2

N
  
  
rad
14
mm
125
s
rad
10
mm
125







r
y
r
v
B
B
m
75
.
1
s
m
25
.
1




B
B
y
v


Ninth
Edition
Sample Problem 5.1
15 - 11
• Evaluate the initial tangential and normal acceleration
components of D.
  


2
s
mm
225
C
t
D a
a


     2
2
2
0 s
mm
1200
s
rad
4
mm
5
7 

 
D
n
D r
a
    


 2
2
s
mm
1200
s
mm
225 n
D
t
D a
a


Magnitude and direction of the total acceleration,
   
2
2
2
2
1200
225 


 n
D
t
D
D a
a
a
2
s
mm
1220

D
a
 
 
225
1200
tan


t
D
n
D
a
a


 4
.
79


Ninth
Edition
General Plane Motion
15 - 12
• 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
1
B
1
B

Ninth
Edition
Absolute and Relative Velocity in Plane Motion
15 - 13
• Any plane motion can be replaced by a translation of an
arbitrary reference point A and a simultaneous rotation
about A.
A
B
A
B v
v
v






 r
v
r
k
v A
B
A
B
A
B 





A
B
A
B r
k
v
v






 

Ninth
Edition
15 - 14
• Assuming that the velocity vA of end A is known, wish to determine the
velocity vB of end B and the angular velocity  in terms of vA, l, and .
• The direction of vB and vB/A are known. Complete the velocity diagram.


tan
tan
A
B
A
B
v
v
v
v






cos
cos
l
v
l
v
v
v
A
A
A
B
A




Ninth
Edition
15 - 15
• 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.

Ninth
Edition
Sample Problem 15.2
15 - 16
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. Relate the translational
and angular displacements. Differentiate
to relate the translational and angular
velocities.
• The velocity for any point P on the gear
may be written as
Evaluate the velocities of points B and D.
A
P
A
A
P
A
P r
k
v
v
v
v










 

Ninth
Edition
Sample Problem 15.2
15 - 17
x
y
SOLUTION:
• The displacement of the gear center in one revolution is
equal to the outer circumference.
For xA > 0 (moves to right),  < 0 (rotates clockwise).



 1
2
2
r
x
r
x
A
A 



Differentiate to relate the translational and angular
velocities.
m
0.150
s
m
2
.
1
1
1






r
v
r
v
A
A


 k
k



s
rad
8


 


Ninth
Edition
Sample Problem 15.2
15 - 18
• For any point P on the gear, A
P
A
A
P
A
P r
k
v
v
v
v










 
Velocity of the upper rack is equal to
velocity of point B:
     
   i
i
j
k
i
r
k
v
v
v A
B
A
B
R










s
m
8
.
0
s
m
2
.
1
m
10
.
0
s
rad
8
s
m
2
.
1








 
 i
vR


s
m
2

Velocity of the point D:
     i
k
i
r
k
v
v A
D
A
D







m
150
.
0
s
rad
8
s
m
2
.
1 





 
   
s
m
697
.
1
s
m
2
.
1
s
m
2
.
1



D
D
v
j
i
v




Ninth
Edition
Instantaneous Center of Rotation in Plane Motion
15 - 19
• Plane motion of all particles in a slab can always be
replaced by the translation of an arbitrary point A and a
rotation about A with an angular velocity that is
independent of the choice of A.
• The same translational and rotational velocities at A are
obtained by allowing the slab to rotate with the same
angular velocity about the point C on a perpendicular to
the velocity at A.
• The velocity of all other particles in the slab are the same
as originally defined since the angular velocity and
translational velocity at A are equivalent.
• As far as the velocities are concerned, the slab seems to
rotate about the instantaneous center of rotation C.

Ninth
Edition
15 - 20
• If the velocity at two points A and B are known, the
instantaneous center of rotation 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, the instantaneous center of rotation 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 parallel, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
• If the velocity magnitudes are equal, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.

Ninth
Edition
15 - 21
• The instantaneous center of rotation lies at the intersection of
the perpendiculars to the velocity vectors through A and B .


cos
l
v
AC
v A
A

    




tan
cos
sin
A
A
B
v
l
v
l
BC
v



• The velocities of all particles on the rod are as if they were
rotated about C.
• The particle at the center of rotation has zero velocity.
• The particle coinciding with the center of rotation changes
with time and the acceleration of the particle at the
instantaneous center of rotation is not zero.
• The acceleration of the particles in the slab cannot be
determined as if the slab were simply rotating about C.
• The trace of the locus of the center of rotation on the body
is the body centrode and in space is the space centrode.

Ninth
Edition
Sample Problem 15.4
15 - 22
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 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.
• Determine the angular velocity about C
based on the given velocity at A.
• Evaluate the velocities at B and D based on
their rotation about C.

Ninth
Edition
Sample Problem 15.4
15 - 23
SOLUTION:
• The 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.
• Determine the angular velocity about C based on the
given velocity at A.
s
rad
8
m
0.15
s
m
2
.
1




A
A
A
A
r
v
r
v 

• Evaluate the velocities at B and D based on their rotation
about C.
  
s
rad
8
m
25
.
0


 
B
B
R r
v
v
 i
vR


s
m
2

 
  
s
rad
8
m
2121
.
0
m
2121
.
0
2
m
15
.
0





D
D
D
r
v
r
  
s
m
2
.
1
2
.
1
s
m
697
.
1
j
i
v
v
D
D







Ninth
Edition
Absolute and Relative Acceleration in Plane Motion
15 - 24
• Absolute acceleration of a particle of the slab,
A
B
A
B a
a
a





• Relative acceleration associated with rotation about A includes
tangential and normal components,
A
B
a

 
  A
B
n
A
B
A
B
t
A
B
r
a
r
k
a





2





  
  2


r
a
r
a
n
A
B
t
A
B



Ninth
Edition
15 - 25
• Given
determine
,
and A
A v
a


.
and 


B
a
   t
A
B
n
A
B
A
A
B
A
B
a
a
a
a
a
a











• Vector result depends on sense of and the
relative magnitudes of  n
A
B
A a
a and
A
a

• Must also know angular velocity .

Ninth
Edition
15 - 26

 x components: 


 cos
sin
0 2
l
l
aA 



 y components: 


 sin
cos
2
l
l
aB 



• Solve for aB and .
• Write in terms of the two component equations,
A
B
A
B a
a
a






Ninth
Edition
Analysis of Plane Motion in Terms of a Parameter
15 - 27
• In some cases, it is advantageous to determine the
absolute velocity and acceleration of a mechanism
directly.

sin
l
xA  
cos
l
yB 




cos
cos
l
l
x
v A
A









sin
sin
l
l
y
v B
B















cos
sin
cos
sin
2
2
l
l
l
l
x
a A
A




















sin
cos
sin
cos
2
2
l
l
l
l
y
a B
B













Ninth
Edition
Sample Problem 15.6
15 - 28
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.
SOLUTION:
• The expression of the gear position as a
function of  is differentiated twice to
define the relationship between the
translational and angular accelerations.
• The acceleration of each point on the
gear is obtained by adding the
acceleration of the gear center and the
relative accelerations with respect to the
center. The latter includes normal and
tangential acceleration components.

Ninth
Edition
Sample Problem 15.6
15 - 29
SOLUTION:
• The expression of the gear position as a function of 
is differentiated twice to define the relationship
between the translational and angular accelerations.



1
1
1
r
r
v
r
x
A
A







s
rad
8
m
0.150
s
m
2
.
1
1






r
vA


 1
1 r
r
aA 


 

m
150
.
0
s
m
3 2
1




r
aA

 k
k


 2
s
rad
20





Ninth
Edition
Sample Problem 15.6
15 - 30
   
         
     j
i
i
j
j
k
i
r
r
k
a
a
a
a
a
a
a
A
B
A
B
A
n
A
B
t
A
B
A
A
B
A
B

















2
2
2
2
2
2
2
s
m
40
.
6
s
m
2
s
m
3
m
100
.
0
s
rad
8
m
100
.
0
s
rad
20
s
m
3



















    2
2
2
s
m
12
.
8
s
m
40
.
6
m
5 

 B
B a
j
i
s
a



• The acceleration of each point
is obtained by adding the
acceleration of the gear center
and the relative accelerations
with respect to the center.
The latter includes normal and
tangential acceleration
components.

Ninth
Edition
Sample Problem 15.6
15 - 31
         
     j
i
i
j
j
k
i
r
r
k
a
a
a
a A
C
A
C
A
A
C
A
C














2
2
2
2
2
2
2
s
m
60
.
9
s
m
3
s
m
3
m
150
.
0
s
rad
8
m
150
.
0
s
rad
20
s
m
3














 

 j
ac

 2
s
m
60
.
9

         
     i
j
i
i
i
k
i
r
r
k
a
a
a
a A
D
A
D
A
A
D
A
D














2
2
2
2
2
2
2
s
m
60
.
9
s
m
3
s
m
3
m
150
.
0
s
rad
8
m
150
.
0
s
rad
20
s
m
3














 

    2
2
2
s
m
95
.
12
s
m
3
m
6
.
12 

 D
D a
j
i
s
a




Ninth
Edition
Sample Problem 15.7
15 - 32
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.
SOLUTION:
• The angular acceleration of the
connecting rod BD and the acceleration
of point D will be determined from
   n
B
D
t
B
D
B
B
D
B
D a
a
a
a
a
a











• The acceleration of B is determined from
the given rotation speed of AB.
• The directions of the accelerations
are
determined from the geometry.
   n
B
D
t
B
D
D a
a
a



and
,
,
• Component equations for acceleration
of point D are solved simultaneously for
acceleration of D and angular
acceleration of the connecting rod.

Ninth
Edition
Sample Problem 15.7
15 - 33
• The acceleration of B is determined from the given rotation
speed of AB.
SOLUTION:
• The angular acceleration of the connecting rod BD and
the acceleration of point D will be determined from
   n
B
D
t
B
D
B
B
D
B
D a
a
a
a
a
a











   2
2
1000
7
2
AB
s
m
3289
s
rad
4
.
209
m
0
constant
s
rad
209.4
rpm
2000







AB
B
AB
r
a 


  
j
i
aB







 40
sin
40
cos
s
m
3289 2

Ninth
Edition
Sample Problem 15.7
15 - 34
• The directions of the accelerations are
determined from the geometry.
   n
B
D
t
B
D
D a
a
a



and
,
,
From Sample Problem 15.3, BD = 62.0 rad/s, b = 13.95o.
       2
2
1000
200
2
s
m
8
.
768
s
rad
0
.
62
m 

 BD
n
B
D BD
a 
    
j
i
a n
B
D







 95
.
13
sin
95
.
13
cos
s
m
8
.
768 2
      BD
BD
BD
t
B
D BD
a 

 2
.
0
m
1000
200



The direction of (aD/B)t is known but the sense is not known,
    
j
i
a BD
t
B
D







 05
.
76
cos
05
.
76
sin
2
.
0 
i
a
a D
D





Ninth
Edition
Sample Problem 15.7
15 - 35
   n
B
D
t
B
D
B
B
D
B
D a
a
a
a
a
a











• Component equations for acceleration of point D are solved
simultaneously.
x components:







 95
.
13
sin
2
.
0
95
.
13
cos
8
.
768
40
cos
3289 BD
D
a 






 95
.
13
cos
2
.
0
95
.
13
sin
8
.
768
40
sin
3289
0 BD

y components:
 
 i
a
k
D
BD




2
2
s
m
2790
s
rad
9940





Ninth
Edition
Sample Problem 15.8
15 - 36
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.
SOLUTION:
• The angular velocities are determined by
simultaneously solving the component
equations for
B
D
B
D v
v
v





• The angular accelerations are determined
by simultaneously solving the component
equations for
B
D
B
D a
a
a






Ninth
Edition
Sample Problem 15.8
15 - 37
SOLUTION:
• The angular velocities are determined by simultaneously
solving the component equations for
B
D
B
D v
v
v





 
j
i
j
i
k
r
v
DE
DE
DE
D
DE
D












17
17
340
340









 
j
i
j
i
k
r
v B
AB
B








160
280
280
160
20







 
 
j
i
j
i
k
r
v
BD
BD
BD
B
D
BD
B
D












12
3
60
240








BD
DE 
 3
280
17 



x components:
BD
DE 
 12
160
17 



y components:
   k
k DE
BD




s
rad
29
.
11
s
rad
33
.
29 

 


Ninth
Edition
Sample Problem 15.8
15 - 38
• The angular accelerations are determined by
simultaneously solving the component equations for
B
D
B
D a
a
a





     
j
i
j
i
j
i
j
i
k
r
r
a
DE
DE
DE
D
DE
D
DE
D













33
.
43
33
.
43
34
.
0
34
.
0
34
.
0
34
.
0
29
.
11
34
.
0
34
.
0
2
2




















   
j
i
j
i
r
r
a B
AB
B
AB
B








112
64
28
.
0
16
20
0
2
2








 

     
j
i
j
i
j
i
j
i
k
r
r
a
D
B
D
B
D
B
D
B
BD
D
B
BD
B
D













61
.
51
4
.
206
24
.
0
06
.
0
06
.
0
24
.
0
33
.
29
06
.
0
24
2
2

















x components: 7
.
313
06
.
0
34
.
0 


 BD
DE 

y components: 28
.
120
24
.
0
34
.
0 


 BD
DE 

   k
k DE
BD



 2
2
s
rad
809
s
rad
645 

 


Ninth
Edition
Rate of Change With Respect to a Rotating Frame
15 - 39
• Frame OXYZ is fixed.
• Frame Oxyz rotates about
fixed axis OA with angular
velocity 

• Vector function varies
in direction and magnitude.
 
t
Q

  k
Q
j
Q
i
Q
Q z
y
x
Oxyz











• With respect to the fixed OXYZ frame,
  k
Q
j
Q
i
Q
k
Q
j
Q
i
Q
Q z
y
x
z
y
x
OXYZ




















• rate of change
with respect to rotating frame.
  


 Oxyz
z
y
x Q
k
Q
j
Q
i
Q 







• If were fixed within Oxyz then is
equivalent to velocity of a point in a rigid body
attached to Oxyz and
 OXYZ
Q


Q
k
Q
j
Q
i
Q z
y
x













Q

• With respect to the rotating Oxyz frame,
k
Q
j
Q
i
Q
Q z
y
x







• With respect to the fixed OXYZ frame,
    Q
Q
Q Oxyz
OXYZ











Ninth
Edition
15 - 40
• Frame OXY is fixed and frame Oxy rotates with angular
velocity .


• Position vector for the particle P is the same in both
frames but the rate of change depends on the choice of
frame.
P
r

• The absolute velocity of the particle P is
   Oxy
OXY
P r
r
r
v 










• Imagine a rigid slab attached to the rotating frame Oxy
or F for short. Let P’ be a point on the slab which
corresponds instantaneously to position of particle P.
  
 Oxy
P r
v 


F velocity of P along its path on the slab

'
P
v

absolute velocity of point P’ on the slab
• Absolute velocity for the particle P may be written as
F
P
P
P v
v
v




 

Ninth
Edition
15 - 41
 
F
P
P
Oxy
P
v
v
r
r
v













• Absolute acceleration for the particle P is
   
 
Oxy
OXY
P r
dt
d
r
r
a 















     Oxy
Oxy
P r
r
r
r
a 























 2
   
 
     Oxy
Oxy
Oxy
Oxy
OXY
r
r
r
dt
d
r
r
r






















but,
 
 Oxy
P
P
r
a
r
r
a





















F
• Utilizing the conceptual point P’ on the slab,
• Absolute acceleration for the particle P becomes
 
  2
2
2

















F
F
F
P
Oxy
c
c
P
P
Oxy
P
P
P
v
r
a
a
a
a
r
a
a
a















Coriolis acceleration

Ninth
Edition
15 - 42
• Consider a collar P which is made to slide at constant
relative velocity u along rod OB. The rod is rotating at
a constant angular velocity . The point A on the rod
corresponds to the instantaneous position of P.
c
P
A
P a
a
a
a






 F
• Absolute acceleration of the collar is
  0

 Oxy
P r
a 



F
u
a
v
a c
P
c 
2
2 


 F



• The absolute acceleration consists of the radial and
tangential vectors shown
  2

r
a
r
r
a A
A 















where

Ninth
Edition
15 - 43
u
v
v
t
t
u
v
v
t
A
A















,
at
,
at
• Change in velocity over t is represented by the
sum of three vectors
T
T
T
T
R
R
v 










  2

r
a
r
r
a A
A 















recall,
• is due to change in direction of the velocity of
point A on the rod,
A
A
t
t
a
r
r
t
v
t
T
T








2
0
0
lim
lim 




 

T
T 

• result from combined effects of
relative motion of P and rotation of the rod
T
T
R
R 


 and
u
u
u
t
r
t
u
t
T
T
t
R
R
t
t










 

2
lim
lim
0
0


















 






u
a
v
a c
P
c 
2
2 


 F



recall,

Ninth
Edition
Sample Problem 15.9
15 - 44
Disk D of the Geneva mechanism rotates
with constant counterclockwise angular
velocity D = 10 rad/s.
At the instant when  = 150o, determine
(a) the angular velocity of disk S, and (b)
the velocity of pin P relative to disk S.
SOLUTION:
• The absolute velocity of the point P
may be written as
s
P
P
P v
v
v




 
• Magnitude and direction of velocity
of pin P are calculated from the
radius and angular velocity of disk D.
P
v

• Direction of velocity of point P’ on
S coinciding with P is perpendicular to
radius OP.
P
v 

• Direction of velocity of P with
respect to S is parallel to the slot.
s
P
v

• Solve the vector triangle for the
angular velocity of S and relative
velocity of P.

Ninth
Edition
Sample Problem 15.9
15 - 45
SOLUTION:
• The absolute velocity of the point P may be written as
s
P
P
P v
v
v




 
• Magnitude and direction of absolute velocity of pin P are
calculated from radius and angular velocity of disk D.
   s
mm
500
s
rad
10
mm
50 

 D
P R
v 
• Direction of velocity of P with respect to S is parallel to slot.
From the law of cosines,
mm
1
.
37
551
.
0
30
cos
2 2
2
2
2





 r
R
Rl
l
R
r
From the law of cosines,





 4
.
42
742
.
0
30
sin
sin
30
sin
R
sin
b
b
b
r







 6
.
17
30
4
.
42
90

The interior angle of the vector triangle is

Ninth
Edition
Sample Problem 15.9
15 - 46
• Direction of velocity of point P’ on S coinciding with P is
perpendicular to radius OP. From the velocity triangle,
 
mm
1
.
37
s
mm
2
.
151
s
mm
2
.
151
6
.
17
sin
s
mm
500
sin







s
s
P
P
r
v
v



 k
s


s
rad
08
.
4



  

 6
.
17
cos
s
m
500
cos
P
s
P v
v
  
j
i
v s
P







 4
.
42
sin
4
.
42
cos
s
m
477
s
mm
500

P
v

Ninth
Edition
15 - 47
In the Geneva mechanism, disk D
rotates with a constant counter-
clockwise angular velocity of 10
rad/s. At the instant when j = 150o,
determine angular acceleration of
disk S.
SOLUTION:
• The absolute acceleration of the pin P may
be expressed as
c
s
P
P
P a
a
a
a






 
• The instantaneous angular velocity of Disk
S is determined as in Sample Problem 15.9.
• The only unknown involved in the
acceleration equation is the instantaneous
angular acceleration of Disk S.
• Resolve each acceleration term into the
component parallel to the slot. Solve for
the angular acceleration of Disk S.

Ninth
Edition
15 - 48
SOLUTION:
• Absolute acceleration of the pin P may be expressed as
c
s
P
P
P a
a
a
a






 
• From Sample Problem 15.9.
 
  
j
i
v
k
s
P
S














4
.
42
sin
4
.
42
cos
s
mm
477
s
rad
08
.
4
4
.
42 
b
• Considering each term in the acceleration equation,
  
  
j
i
a
R
a
P
D
P










30
sin
30
cos
s
mm
5000
s
mm
5000
s
rad
10
mm
500
2
2
2
2

   
    
    
     
j
i
a
j
i
r
a
j
i
r
a
a
a
a
S
t
P
S
t
P
S
n
P
t
P
n
P
P



































4
.
42
cos
4
.
42
sin
mm
1
.
37
4
.
42
cos
4
.
42
sin
4
.
42
sin
4
.
42
cos
2



note: S may be positive or negative

Ninth
Edition
15 - 49
• The relative acceleration must be parallel to
the slot.
s
P
a

s
P
v

• The direction of the Coriolis acceleration is obtained
by rotating the direction of the relative velocity
by 90o in the sense of S.
  
   
  
j
i
j
i
j
i
v
a s
P
S
c







4
.
42
cos
4
.
42
sin
s
mm
3890
4
.
42
cos
4
.
42
sin
s
mm
477
s
rad
08
.
4
2
4
.
42
cos
4
.
42
sin
2
2











 
• Equating components of the acceleration terms
perpendicular to the slot,
s
rad
233
0
7
.
17
cos
5000
3890
1
.
37






S
S


 k
S


s
rad
233




Ninth
Edition
Motion About a Fixed Point
15 - 50
• The most general displacement of a rigid body with a
fixed point O is equivalent to a rotation of the body
about an axis through O.
• With the instantaneous axis of rotation and angular
velocity the velocity of a particle P of the body is
,


r
dt
r
d
v






 
and the acceleration of the particle P is
  .
dt
d
r
r
a



















• Angular velocities have magnitude and direction and
obey parallelogram law of addition. They are vectors.
• As the vector moves within the body and in space,
it generates a body cone and space cone which are
tangent along the instantaneous axis of rotation.


• The angular acceleration represents the velocity of
the tip of .

 


Ninth
Edition
General Motion
15 - 51
• For particles A and B of a rigid body,
A
B
A
B v
v
v





• Particle A is fixed within the body and motion of
the body relative to AX’Y’Z’ is the motion of a
body with a fixed point
A
B
A
B r
v
v






 
• Similarly, the acceleration of the particle P is
 
A
B
A
B
A
A
B
A
B
r
r
a
a
a
a




















• Most general motion of a rigid body is equivalent to:
- a translation in which all particles have the same
velocity and acceleration of a reference particle A, and
- of a motion in which particle A is assumed fixed.

Ninth
Edition
15 - 52
The crane rotates with a constant
angular velocity 1 = 0.30 rad/s and the
boom is being raised with a constant
angular velocity 2 = 0.50 rad/s. The
length of the boom is l = 12 m.
Determine:
• angular velocity of the boom,
• angular acceleration of the boom,
• velocity of the boom tip, and
• acceleration of the boom tip.
• Angular acceleration of the boom,
 
2
1
2
2
2
2
1





























 Oxyz
• Velocity of boom tip,
r
v




 
• Acceleration of boom tip,
  v
r
r
r
a


















 




SOLUTION:
With
• Angular velocity of the boom,
2
1 







 
j
i
j
i
r
k
j









6
39
.
10
30
sin
30
cos
12
50
.
0
30
.
0 2
1







 


Ninth
Edition
15 - 53
j
i
r
k
j







6
39
.
10
50
.
0
30
.
0 2
1



 

SOLUTION:
• Angular velocity of the boom,
2
1 







   k
j



s
rad
50
.
0
s
rad
30
.
0 


• Angular acceleration of the boom,
 
   k
j
Oxyz















s
rad
50
.
0
s
rad
30
.
0
2
1
2
2
2
2
1



















 i

 2
s
rad
15
.
0


• Velocity of boom tip,
0
6
39
.
10
5
.
0
3
.
0
0
k
j
i
r
v








 
     k
j
i
v




s
m
12
.
3
s
m
20
.
5
s
m
54
.
3 




Ninth
Edition
15 - 54
j
i
r
k
j







6
39
.
10
50
.
0
30
.
0 2
1



 

• Acceleration of boom tip,
 
k
j
i
i
k
k
j
i
k
j
i
a
v
r
r
r
a






















90
.
0
50
.
1
60
.
2
94
.
0
90
.
0
12
.
3
20
.
5
3
50
.
0
30
.
0
0
0
6
39
.
10
0
0
15
.
0

















 




     k
j
i
a



 2
2
2
s
m
80
.
1
s
m
50
.
1
s
m
54
.
3 




Ninth
Edition
Three-Dimensional Motion. Coriolis Acceleration
15 - 55
• With respect to the fixed frame OXYZ and rotating
frame Oxyz,
    Q
Q
Q Oxyz
OXYZ










• Consider motion of particle P relative to a rotating
frame Oxyz or F for short. The absolute velocity can
be expressed as
 
F
P
P
Oxyz
P
v
v
r
r
v














• The absolute acceleration can be expressed as
     
  on
accelerati
Coriolis
2
2
2























F
F
P
Oxyz
c
c
P
p
Oxyz
Oxyz
P
v
r
a
a
a
a
r
r
r
r
a























Ninth
Edition
Frame of Reference in General Motion
15 - 56
Consider:
- fixed frame OXYZ,
- translating frame AX’Y’Z’, and
- translating and rotating frame Axyz
or F.
• With respect to OXYZ and AX’Y’Z’,
A
P
A
P
A
P
A
P
A
P
A
P
a
a
a
v
v
v
r
r
r















• The velocity and acceleration of P relative to
AX’Y’Z’can be found in terms of the velocity
and acceleration of P relative to Axyz.
 
F
P
P
Axyz
A
P
A
P
A
P
v
v
r
r
v
v
















 
   
c
P
P
Axyz
A
P
Axyz
A
P
A
P
A
P
A
P
a
a
a
r
r
r
r
a
a

































 F
2

Ninth
Edition
15 - 57
For the disk mounted on the arm, the
indicated angular rotation rates are
constant.
Determine:
• the velocity of the point P,
• the acceleration of P, and
• angular velocity and angular
acceleration of the disk.
SOLUTION:
• Define a fixed reference frame OXYZ at O
and a moving reference frame Axyz or F
attached to the arm at A.
• With P’ of the moving reference frame
coinciding with P, the velocity of the point
P is found from
F
P
P
P v
v
v




 
• The acceleration of P is found from
c
P
P
P a
a
a
a






  F
• The angular velocity and angular
acceleration of the disk are
  



















F
F
D

Ninth
Edition
15 - 58
SOLUTION:
• Define a fixed reference frame OXYZ at O and a
moving reference frame Axyz or F attached to the
arm at A.
j
j
R
i
L
r





1

 


k
j
R
r
D
A
P




2

 

F
• With P’ of the moving reference frame coinciding
with P, the velocity of the point P is found from
 
i
R
j
R
k
r
v
k
L
j
R
i
L
j
r
v
v
v
v
A
P
D
P
P
P
P
P
















2
2
1
1























F
F
F
k
L
i
R
vP



1
2 
 



Ninth
Edition
15 - 59
• The acceleration of P is found from
c
P
P
P a
a
a
a






  F
    i
L
k
L
j
r
aP






 2
1
1
1 



 








 
  j
R
i
R
k
r
a A
P
D
D
P







2
2
2
2 











 F
F
F
  k
R
i
R
j
v
a P
c






2
1
2
1 2
2
2










 F
k
R
j
R
i
L
aP




2
1
2
2
2
1 2 


 



• Angular velocity and acceleration of the disk,
F
D







 k
j



2
1 

 

 
 
k
j
j








2
1
1 











 F
i


2
1

 

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