15 lecture ppt

VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Tenth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
California Polytechnic State University
CHAPTER
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
15
Kinematics of
Rigid Bodies
Kinematics of Rigid
Bodies

Vector Mechanics for Engineers: Dynamics
Tenth
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

Tenth
Edition
Applications
15 - 3
A battering ram is an example of curvilinear translation – the
ram stays horizontal as it swings through its motion.

Tenth
Edition
Applications
15 - 4
How can we determine the velocity of the tip of a turbine blade?

Tenth
Edition
Applications
15 - 5
Planetary gear systems are used to get high reduction ratios
with minimum weight and space. How can we design the
correct gear ratios?

Tenth
Edition
Applications
15 - 6
Biomedical engineers must determine the velocities and
accelerations of the leg in order to design prostheses.

Tenth
Edition
Introduction
15 - 7
• 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:

Tenth
Edition
Translation
15 - 8
• 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 (rB A being constant)
• 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.

Tenth
Edition
Rotation About a Fixed Axis. Velocity
15 - 9
• 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

Tenth
Edition
Concept Quiz
15 - 10
What is the direction of the velocity
of point A on the turbine blade?
A
a) →
b) ←
c) ↑
d) ↓
vA
=w ´r
x
y
vA
=wk̂ ´-Lˆ
i
vA
= -Lw ĵ

L

Tenth
Edition
Rotation About a Fixed Axis. Acceleration
15 - 11
• 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,

Tenth
Edition
Rotation About a Fixed Axis. Representative Slab
15 - 12
• 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












Tenth
Edition
Concept Quiz
15 - 13
What is the direction of the normal
acceleration of point A on the turbine
blade?
A
a) →
b) ←
c) ↑
d) ↓
an
= -w2
r
x
y
an
= -w2
(-Lˆ
i)
an
= Lw2ˆ
i

L

Tenth
Edition
Equations Defining the Rotation of a Rigid Body About a Fixed Axis
15 - 14
• 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

Tenth
Edition
Sample Problem 5.1
15 - 15
Cable C has a constant acceleration of
225 m/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.

Tenth
Edition
Sample Problem 5.1
15 - 16
SOLUTION:
• The tangential velocity and acceleration of D are equal to the
velocity and acceleration of C.
• 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



( )( ) ( )( )2
1 1
2 2
0 2 2
4rad s 2 s 3rad s 2 s
14 rad
t t
= + = +
= i
q w a
  revs
of
number
rad
2
rev
1
rad
14 








N rev
23
.
2

N
( )( )
( )( )
125 mm 10rad s
125 mm 14 rad
B
B
v r
y r
= =
= =
w
D q
1.25m s
1.75 m
B
B
v
y
=
D =

Tenth
Edition
Sample Problem 5.1
15 - 17
• Evaluate the initial tangential and normal acceleration
components of D.
aD
( )t
= aC = 25mm s2
®
aD
( )n
= rDw0
2
= 75mm
( ) 4rad s
( )
2
=1200mm s2
aD
( )t
= 225mm s2
® aD
( )n
=1200mm s2
¯
Magnitude and direction of the total acceleration,
( ) ( )
2 2
2 2
2
(225) (1200)
1220.9mm/s
D D D
t n
a a a
= +
= +
=
2
1.221m s
D
a =
( )
( )
tan
1200
225
D n
D t
a
a
=
=
f

 4
.
79


Tenth
Edition
Group Problem Solving
15 - 18
A series of small machine components
being moved by a conveyor belt pass over
a 150 mm-radius idler pulley. At the instant
shown, the velocity of point A is 375 mm/s
to the left and its acceleration is 225 mm/s2
to the right. Determine (a) the angular
velocity and angular acceleration of the
idler pulley, (b) the total acceleration of the
machine
component at B.
SOLUTION:
• Using the linear velocity and
accelerations, calculate the angular
velocity and acceleration.
• Using the angular velocity,
determine the normal acceleration.
• Determine the total acceleration
using the tangential and normal
acceleration components of B.

Tenth
Edition
15 - 19
v= 375 mm/s at= 225 mm/s2
Find the angular velocity of the idler
pulley using the linear velocity at B.
375 mm/s (150 mm)
w
w
v r

 2.50 rad/s


2
225 mm/s (150 mm)
a
a
a r


2
1.500 rad/s


B
Find the angular acceleration of the idler
pulley using the linear velocity at B.
Find the normal acceleration of point B.
2
2
(150 mm)(2.5 rad/s)
w
n
a r


2
937.5 mm/s
an 
What is the direction of
the normal acceleration
of point B?
Downwards, towards
the center

Tenth
Edition
15 - 20
an= 937.5 mm/s2
Find the total acceleration of the
machine component at point B.
at= 225 mm/s2
2
937.5 mm/s
an 
2
964 mm/s
aB  76.5
at= 225 mm/s2
an= 937.5 mm/s2
B
a
2
225 mm/s
at 
B
2 2 2
(225) (937.5) 964 mm/s
a   
Calculate the magnitude
Calculate the angle from
the horizontal
Combine for a final answer
q = arctan 937.5
225
( )= 76.5

Tenth
Edition
Golf Robot
15 - 21
,
If the arm is shortened to ¾ of its
original length, what happens to
the tangential acceleration of the
club head?
A golf robot is used to test new
equipment. If the angular
velocity of the arm is doubled,
what happens to the normal
acceleration of the club head?
If the arm is shortened to ¾ of its
original length, what happens to
the tangential acceleration of the
club head?
If the speed of the club head is
constant, does the club head have
any linear accelertion ?
Not ours – maybe Tom Mase has pic?

Tenth
Edition
Example – General Plane Motion
15 - 22
The knee has linear velocity and acceleration from both
translation (the runner moving forward) as well as rotation
(the leg rotating about the hip).

Tenth
Edition
General Plane Motion
15 - 23
• 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

Tenth
Edition
Absolute and Relative Velocity in Plane Motion
15 - 24
• 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






 

Tenth
Edition
15 - 25
• 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




Tenth
Edition
15 - 26
• 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.

Tenth
Edition
Sample Problem 15.2
15 - 27
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










 

Tenth
Edition
Sample Problem 15.2
15 - 28
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


 


Tenth
Edition
Sample Problem 15.2
15 - 29
• 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




Tenth
Edition
Sample Problem 15.3
15 - 30
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:
• Will determine the absolute velocity of
point D with
B
D
B
D v
v
v





• The velocity is obtained from the
given crank rotation data.
B
v

• The directions of the absolute velocity
and the relative velocity are
determined from the problem geometry.
D
v

B
D
v

• The unknowns in the vector expression
are the velocity magnitudes
which may be determined from the
corresponding vector triangle.
B
D
D v
v and
• The angular velocity of the connecting
rod is calculated from .
B
D
v

Tenth
Edition
Sample Problem 15.3
15 - 31
SOLUTION:
• Will determine the absolute velocity of point D with
B
D
B
D v
v
v





• The velocity is obtained from the crank rotation data.
B
v

wAB = 2000 rev
min
( ) min
60s
æ
è
ç
ö
ø
÷
2p rad
rev
æ
è
ç
ö
ø
÷ = 209.4 rad s
vB = AB
( )wAB = 75mm
( ) 209.4 rad s
( )=15705 mm/s
The velocity direction is as shown.
• The direction of the absolute velocity is horizontal.
The direction of the relative velocity is
perpendicular to BD. Compute the angle between the
horizontal and the connecting rod from the law of sines.
D
v

B
D
v

sin40°
200mm
=
sinb
75mm
b =13.95°

Tenth
Edition
Sample Problem 15.3
15 - 32
• Determine the velocity magnitudes
from the vector triangle.
B
D
D v
v and
B
D
B
D v
v
v





vD
sin53.95°
=
vD B
sin50°
=
15705mm s
sin76.05°
13083mm s 13.08m s
12396mm s
D
D B
v
v
= =
=
12396mm s
200 mm
62.0 rad s
D B BD
D B
BD
v l
v
l
=
= =
=
w
w
13.08m s
P D
v v
= =
 k
BD


s
rad
0
.
62



Tenth
Edition
15 - 33
In the position shown, bar AB
has an angular velocity of 4 rad/s
clockwise. Determine the angular
velocity of bars BD and DE.
Which of the following is true?
a) The direction of vB is ↑
b) The direction of vD is →
c) Both a) and b) are correct

Tenth
Edition
15 - 34
In the position shown, bar AB has an
angular velocity of 4 rad/s clockwise.
Determine the angular velocity of bars
BD and DE.
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










 

Tenth
Edition
15 - 35
How should you proceed?
Determine vB with respect to A, then work
your way along the linkage to point E.
Determine the angular velocity of bars
BD and DE.
AB= 4 rad/s
(4 rad/s)
AB   k

Does it make sense that vB is in the +j direction?
x
y
/A
B A AB B
  
v v r

Write vB in terms of point A, calculate vB.

Tenth
Edition
15 - 36
Determine vD with respect to B.
AB= 4 rad/s
x
y
/
/
(200 mm)
700 ( ) ( 200 )
700 200
k r j
v v r j k j
v j i
BD BD D B
D B BD D B BD
D BD



  
      
 


/
/
(275 mm) (75 mm)
( ) ( 275 75 )
275 75
k r i j
v r k i j
v j i
DE DE D E
D DE D E DE
D DE DE


 
   
     
  


Determine vD with respect to E, then
equate it to equation above.
Equating components of the two expressions for vD
,
D
v
j: 700 275 2.5455 rad/s
DE DE
 
   
3
: 200 75
8
i BD DE BD DE
   
    0.955 rad/s
BD 

2.55 rad/s
DE 


Tenth
Edition
Instantaneous Center of Rotation in Plane Motion
15 - 37
• 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.

Tenth
Edition
15 - 38
• 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.

Tenth
Edition
15 - 39
• 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.

Tenth
Edition
15 - 40
At the instant shown, what is the
approximate direction of the velocity
of point G, the center of bar AB?
a)
b)
c)
d)
G

Tenth
Edition
Sample Problem 15.4
15 - 41
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.

Tenth
Edition
Sample Problem 15.4
15 - 42
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







Tenth
Edition
Sample Problem 15.5
15 - 43
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:
• Determine the velocity at B from the
given crank rotation data.
• The direction of the velocity vectors at B
and D are known. The instantaneous
center of rotation is at the intersection of
the perpendiculars to the velocities
through B and D.
• Determine the angular velocity about the
center of rotation based on the velocity
at B.
• Calculate the velocity at D based on its
rotation about the instantaneous center
of rotation.

Tenth
Edition
Sample Problem 15.5
15 - 44
SOLUTION:
• From Sample Problem 15.3,
vB = 10095i -12031j
( ) mm s
( ) vB =15705mm s
b =13.95°
• The instantaneous center of rotation is at the intersection
of the perpendiculars to the velocities through B and D.










05
.
76
90
95
.
53
40




D
B
BC
sin76.05°
=
CD
sin53.95°
=
200 mm
sin50°
253.4 mm 211.1 mm
BC CD
= =
• Determine the angular velocity about the center of
rotation based on the velocity at B.
( )
15705mm s
253.4 mm
B BD
B
BD
v BC
v
BC
w
w
=
= =
• Calculate the velocity at D based on its rotation about
the instantaneous center of rotation.
( ) ( )( )
211.1 mm 62.0rad s
D BD
v CD w
= =
13090mm s 1.309m s
P D
v v
= = =
s
rad
0
.
62

BD


Tenth
Edition
Instantaneous Center of Zero Velocity
15 - 45
What happens to the location of the instantaneous center of
velocity if the crankshaft angular velocity increases from
2000 rpm in the previous problem to 3000 rpm?
What happens to the location of the instantaneous center of
velocity if the angle  is 0?

Tenth
Edition
15 - 46
In the position shown, bar AB has an angular velocity of 4
rad/s clockwise. Determine the angular velocity of bars BD
and DE.

Tenth
Edition
15 - 47
vD
What direction is the velocity of B?
vB
What direction is the velocity of D?
AB= 4 rad/s
( ) (0.25 m)(4 rad/s) 1m/s
B AB
AB 
  
v
What is the velocity of B?
1 0.06 m
tan 21.8
0.15 m
 
  
Find  

Tenth
Edition
15 - 48
vD
vB
B
D

Locate instantaneous center C at intersection of lines drawn
perpendicular to vB and vD.
C
0.1m 0.1m
0.25 m
tan tan 21.8°
0.25 m 0.25 m
0.2693 m
cos cos21.8°
BC
DC


  
  
100 mm
1m/s (0.25 m) BD


4 rad/s
BD 

Find DE
0.25 m
( ) (4 rad/s)
cos
D BD
v DC 

 
1m/s 0.15 m
( ) ; ;
cos cos
D DE DE
v DE  
 
  6.67 rad/s
DE 

Find distances BC and DC
( )
B BD
v BC 

Calculate BD

Tenth
Edition
Absolute and Relative Acceleration in Plane Motion
15 - 49
As the bicycle accelerates, a point on the top of the wheel will
have acceleration due to the acceleration from the axle (the
overall linear acceleration of the bike), the tangential
acceleration of the wheel from the angular acceleration, and
the normal acceleration due to the angular velocity.

Tenth
Edition
15 - 50
• 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



Tenth
Edition
15 - 51
• 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 .

Tenth
Edition
15 - 52

 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






Tenth
Edition
Analysis of Plane Motion in Terms of a Parameter
15 - 53
• 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













Tenth
Edition
Concept Question
15 - 54
A) Energy will not be conserved when I kick this ball
B) In general, the linear acceleration of my knee is equal to
the linear acceleration of my foot
C) Throughout the kick, my foot will only have tangential
acceleration.
D) In general, the angular velocity of the upper leg (thigh)
will be the same as the angular velocity of the lower leg
You have made it to the kickball
championship game. As you try to kick
home the winning run, your mind
naturally drifts towards dynamics.
Which of your following thoughts is
TRUE, and causes you to shank the ball
horribly straight to the pitcher?

Tenth
Edition
Sample Problem 15.6
15 - 55
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.

Tenth
Edition
Sample Problem 15.6
15 - 56
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





Tenth
Edition
Sample Problem 15.6
15 - 57
   
         
     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.

Tenth
Edition
Sample Problem 15.6
15 - 58
         
     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




Tenth
Edition
Sample Problem 15.7
15 - 59
Crank AB 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.

Tenth
Edition
Sample Problem 15.7
15 - 60
• 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











( )( )
AB
2
2 2
75
100
2000rpm 209.4rad s constant
0
m 209.4rad s 3289m s
AB
B AB
a r
w
a
w
= = =
=
= = =
aB = 3289m s2
( ) -cos40°i -sin40°j
( )

Tenth
Edition
Sample Problem 15.7
15 - 61
• 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,  = 13.95o.
( ) ( ) ( )( )2
2 2
200
1000
m 62.0rad s 768.8m s
D B BD
n
a BD w
= = =
aD B
( )n
= 768.8m s2
( ) -cos13.95°i +sin13.95°j
( )
( ) ( ) ( )
200
1000
m 0.2
D B BD BD BD
t
a BD a a a
= = =
The direction of (aD/B)t is known but the sense is not known,
aD B
( )t
= 0.2aBD
( ) ±sin76.05°i ±cos76.05°j
( )
i
a
a D
D





Tenth
Edition
Sample Problem 15.7
15 - 62
   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:
-aD = -3289cos40°-768.8cos13.95°+0.2aBD sin13.95°
0 = -3289sin40°+768.8sin13.95°+0.2aBD cos13.95°
y components:
aBD = 9940rad s2
( )k
aD = - 2790m s2
( )i

Tenth
Edition
Sample Problem 15.8
15 - 63
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






Tenth
Edition
Sample Problem 15.8
15 - 64
SOLUTION:
• The angular velocities are determined by simultaneously
solving the component equations for
B
D
B
D v
v
v





vD =wDE ´rD =wDEk ´ -340i +340 j
( )
= -340wDEi -340wDE j
vB =wAB ´rB = 20k ´ 160i +280 j
( )
= -5600i +3200 j
vD B =wBD ´rD B =wBDk ´ 240i +60 j
( )
= -60wBDi + 240wBD j
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 

 


Tenth
Edition
Sample Problem 15.8
15 - 65
• The angular accelerations are determined by
simultaneously solving the component equations for
B
D
B
D a
a
a





aD =aDE ´rD -wDE
2
rD
=aDEk ´ -0.34i + 0.34 j
( )- 11.29
( )
2
-0.34i +0.34 j
( )
= -0.34aDEi - 0.34aDE j + 43.33i - 43.33j
aB =aAB ´rB -wAB
2
rB = 0- 20
( )
2
0.16i +0.28j
( )
= -64i +112 j
aD B =aBD ´rD/B -wBD
2
rD/B
=aB Dk ´ 0.24i + 0.06 j
( )- 29.33
( )
2
0.24i + 0.06 j
( )
= -0.06aB Di + 0.24aB D j -206.4i - 51.61j
x components: 0.34 0.06 313.7
DE BD
a a
- + = -
y components: 0.34 0.24 120.28
DE BD
a a
- - = -
   k
k DE
BD



 2
2
s
rad
809
s
rad
645 

 

(r raw expressed in m)

Tenth
Edition
15 - 66
Knowing that at the instant
shown bar AB has a constant
angular velocity of 4 rad/s
clockwise, determine the angular
acceleration of bars BD and DE.
Which of the following is true?
a) The direction of aD is
b) The angular acceleration of BD must also be constant
c) The direction of the linear acceleration of B is →

Tenth
Edition
15 - 67
Knowing that at the instant
shown bar AB has a constant
angular velocity of 4 rad/s
clockwise, determine the
angular acceleration of bars
BD and DE.
SOLUTION:
• The angular velocities were determined
in a previous problem by simultaneously
solving the component equations for
B
D
B
D v
v
v





• The angular accelerations are now
determined by simultaneously solving
the component equations for the relative
acceleration equation.

Tenth
Edition
15 - 68
From our previous problem, we used the relative
velocity equations to find that:
AB= 4 rad/s 0.955 rad/s
BD 

2.55 rad/s
DE 

0
AB 

We can now apply the relative acceleration
equation with
2
/A /A
B A AB B AB B

   
a a r r

2 2 2
/ (4) ( 0.175 ) 2.8 m/s
a r i i
B AB B A

     
Analyze
Bar AB
Analyze Bar BD
2 2
/ / 2.8 ( 0.2 ) (0.95455) ( 0.2 )
a a r r i k j j
D B BD D B BD D B BD
 
         

(2.8 0.2 ) 0.18223
a i j
D BD

  

Tenth
Edition
15 - 69
AB= 4 rad/s
Analyze Bar DE
2
/ /
2
( 0.275 0.075 ) (2.5455) ( 0.275 0.075 )
0.275 0.075 1.7819 0.486
a r
k i j i j
j i i j
D DE D E DE D E
DE
DE DE
r


 
  
      
    

( 0.075 1.7819) (0.275 0.486)
a i j
D DE DE
 
    
Equate like components of aD
j: 0.18223 (0.275 0.486)
DE

  
2
2.43 rad/s
DE
  
i: 2.8 0.2 [ (0.075)( 2.43) 1.7819]
BD

    
2
4.18 rad/s
BD
  
From previous page, we had: (2.8 0.2 ) 0.18223
a i j
D BD

  

Tenth
Edition
Concept Question
15 - 70
If the clockwise angular velocity of crankshaft AB is
constant, which of the following statement is true?
a) The angular velocity of BD is constant
b) The linear acceleration of point B is zero
c) The angular velocity of BD is counterclockwise
d) The linear acceleration of point B is tangent to the path

Tenth
Edition
Applications
15 - 71
Rotating coordinate systems are often used to analyze mechanisms
(such as amusement park rides) as well as weather patterns.

Tenth
Edition
Rate of Change With Respect to a Rotating Frame
15 - 72
• 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











Tenth
Edition
15 - 73
• 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
vP = v ¢
P +vP F

Tenth
Edition
15 - 74
 
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

Tenth
Edition
15 - 75
• 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.
aP = aA +aP F +ac
• 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

Tenth
Edition
15 - 76
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,

Tenth
Edition
Concept Question
15 - 77
v

a) +x
b) -x
c) +y
d) -y
e) Acceleration = 0
You are walking with a
constant velocity with
respect to the platform,
which rotates with a constant
angular velocity w. At the
instant shown, in which
direction(s) will you
experience an acceleration
(choose all that apply)?
x
y
     Oxy
Oxy
P r
r
r
r
a 























 2

Tenth
Edition
Sample Problem 15.9
15 - 78
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.

Tenth
Edition
Sample Problem 15.9
15 - 79
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



r







 6
.
17
30
4
.
42
90

The interior angle of the vector triangle is

Tenth
Edition
Sample Problem 15.9
15 - 80
• 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

Tenth
Edition
15 - 81
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.

Tenth
Edition
15 - 82
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 

• 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

Tenth
Edition
15 - 83
• 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




Tenth
Edition
15 - 84
The sleeve BC is welded to an arm that rotates about stationary point A
with a constant angular velocity  = (3 rad/s) j. In the position shown
rod DF is being moved to the left at a constant speed u = 400 mm/s
relative to the sleeve. Determine the acceleration of Point D.
SOLUTION:
• The absolute acceleration of point D may
be expressed as
aD
= aD'
+aD BC
+ac
• Determine the acceleration of the virtual
point D’.
• Calculate the Coriolis acceleration.
• Add the different components to get the
overall acceleration of point D.

Tenth
Edition
15 - 85
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
Given: u= 400 mm/s,  = (3 rad/s) j.
Find: aD
Write overall expression for aD
Do any of the terms go to zero?
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
Determine the normal acceleration term of the virtual point D’
a ¢
D
= W´ W´r
( )
= (3 rad/s)j´ (3 rad/s)j´[-(100 mm)j+(300 mm)k]
{ }
= -(2700 mm/s2
)k
where r is from A to D

Tenth
Edition
15 - 86
Determine the Coriolis acceleration of point D
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
/
2
2
2(3 rad/s) (400 mm/s)
(2400 mm/s )
a v
j k
i
C D F
 
 


/
2 2
(2700 mm/s ) 0 (2400 mm/s )
a a a a
k i
D D D F C

  
   
Add the different components to obtain
the total acceleration of point D
2 2
(2400 mm/s ) (2700 mm/s )
a i k
D  

Tenth
Edition
15 - 87
In the previous problem, u and 
were both constant.
What would happen if u was
increasing?
a) The x-component of aD would increase
b) The y-component of aD would increase
c) The z-component of aD would increase
d) The acceleration of aD would stay the same
What would happen if  was increasing?
a) The x-component of aD would increase
b) The y-component of aD would increase
c) The z-component of aD would increase
d) The acceleration of aD would stay the same


Tenth
Edition
Motion About a Fixed Point
15 - 88
• 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 .

 


Tenth
Edition
General Motion
15 - 89
• 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.

Tenth
Edition
Concept Question
15 - 90
The figure depicts a model of a
coaster wheel. If both 1 and
2 are constant, what is true
about the angular acceleration
of the wheel?
a) It is zero.
b) It is in the +x direction
c) It is in the +z direction
d) It is in the -x direction
e) It is in the -z direction

Tenth
Edition
15 - 91
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







 


Tenth
Edition
15 - 92
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 




Tenth
Edition
15 - 93
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 




Tenth
Edition
Three-Dimensional Motion. Coriolis Acceleration
15 - 94
• 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























Tenth
Edition
Frame of Reference in General Motion
15 - 95
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

Tenth
Edition
15 - 96
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

Tenth
Edition
15 - 97
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 
 



Tenth
Edition
15 - 98
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

 

Tenth
Edition
15 - 99
The crane shown rotates at the constant rate 1= 0.25 rad/s; simultaneously, the
telescoping boom is being lowered at the constant rate 2= 0.40 rad/s. Knowing
that at the instant shown the length of the boom is 6 m and is increasing at
the constant rate u= 0.5 m/s determine the acceleration of Point B.
SOLUTION:
• Define a moving reference frame Axyz or
F attached to the arm at A.
aB
= aB'
+aB F
+ac
• The angular velocity and angular
acceleration of the disk are
w = W+wB F
a = w
( )F
+W´w

Tenth
Edition
15 - 100
Given: 1= 0.25 rad/s, 2= -0.40 rad/s. L= 6 m, u= 0.5 m/s
Find: aB.
Equation of overall acceleration of B
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
Do any of the terms go to zero?
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
Let the unextending portion of the boom AB be a rotating
frame of reference. What are W and W ?
2 1
(0.40 rad/s) (0.25 rad/s) .
 
 
 
i j
i j
 W =w1
j´w2
i
= -w1
w2
k
= -(0.10 rad/s2
)k.

Tenth
Edition
15 - 101
Find
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
W ´ r
Find W´ W´r
( )
 
2 2 2
(0.40 0.25 ) (0.40 0.25 ) (3 3 3 )
(0.3 m/s ) (0.48 m/s ) (1.1561 m/s )
r i j i j j k
i j k
B
 
       
 
  
 
/
(6 m)(sin30 cos30 )
(3 m) (3 3 m)
r r
j k
j k
B A B

   
 
Determine the position vector rB/A

Tenth
Edition
15 - 102
aD
= W´r +W´ W´r
( )+2W´ r
( )Oxy
+ r
( )Oxy
Determine the Coriolis acceleration – first
define the relative velocity term
/ (sin30 cos30 )
(0.5 m/s)sin30 (0.5 m/s)cos30
v j k
j k
B F u
   
   
/
2 2 2
2 (2)(0.40 0.25 ) (0.5sin30 0.5cos30 )
(0.2165 m/s ) (0.3464 m/s ) (0.2 m/s )
Ω v i j j k
i j k
B F
      
  
Calculate the Coriolis acceleration
Add the terms together
2 2 2
(0.817 m/s ) (0.826 m/s ) (0.956 m/s )
a i j k
B   

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15 lecture ppt