Tenth Edition
CHAPTER
15
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
Kinematics of
Rigid Bodies
California Polytechnic State University
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Absolute and Relative Acceleration in
Introduction
Plane Motion
Translation
Sample Problem 15.6
Rotation About a Fixed Axis: Velocity
Sample Problem 15.7
Rotation About a Fixed Axis:
Sample Problem 15.8
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
15 - 2
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McGraw-Hill
Companies, Inc. All rights reserved.
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Applications
How can we determine the velocity of the tip of a turbine blade?
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15 - 3
Tenth
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Vector Mechanics for Engineers: Dynamics
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
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15 - 4
Tenth
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Vector Mechanics for Engineers: Dynamics
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.
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Rotational motion video
https://www.youtube.com/watch?v=fmXFWi-WfyU
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2-6
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
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
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Tenth
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Vector Mechanics for Engineers: Dynamics
Concept Quiz
What is the direction of the velocity
of point A on the turbine blade?

a) →
A
b) ←
y
L
c) ↑
x
d) ↓
vA    r
v A   kˆ   Liˆ
v A   L ˆj
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Vector Mechanics for Engineers: Dynamics
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
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Vector Mechanics for Engineers: Dynamics
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
a n  r 2
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15 - 10
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Quiz
What is the direction of the normal
acceleration of point A on the turbine
blade?
a) →
b) ←

A
y
L
c) ↑
d) ↓
x
an   r
2
an   ( Liˆ)
2ˆ
an  L i
2
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15 - 11
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Vector Mechanics for Engineers: Dynamics
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 
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 5.1
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.
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.
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• 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.
15 - 13
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 5.1
SOLUTION:
• 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 s 2 2 s   10 rad s


   0t  12 t 2  4 rad s 2 s   12 3 rad s 2 2 s 2
 14 rad
 1 rev 
N  14 rad 
  number of revs
 2 rad 
vB  r  5 in.10 rad s 
y B  r  5 in.14 rad 
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N  2.23 rev

vB  50 in. s 
y B  70 in.
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Vector Mechanics for Engineers: Dynamics
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.
s2 
aD n  48 in.
s2
s2 
Magnitude and direction of the total acceleration,
aD t2  aD 2n
aD 
 92  482
tan  

aD  48.8 in. s 2
aD n
aD t
48
9
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  79.4
15 - 15
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Using the linear velocity and
accelerations, calculate the angular
velocity and acceleration.
• Using the angular velocity,
determine the normal acceleration.
A series of small machine components
being moved by a conveyor belt pass over
a 6-in.-radius idler pulley. At the instant
shown, the velocity of point A is 15 in./s to
the left and its acceleration is 9 in./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.
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• Determine the total acceleration
using the tangential and normal
acceleration components of B.
15 - 16
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Find the angular velocity of the idler
pulley using the linear velocity at B.
v= 15 in/s
B
at= 9 in/s2
v  r
15 in./s  (6 in.)
  2.50 rad/s
Find the angular velocity of the idler
pulley using the linear velocity at B.
a  r
9 in./s  (6 in.)
2
  1.500 rad/s 2
Find the normal acceleration of point B.
an  r 2
 (6 in.)(2.5 rad/s) 2
a n  37.5 in./s 2
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What is the direction of
the normal acceleration
of point B?
Downwards, towards
the center
15 - 17
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
B
at= 9 in/s2
an= 37.5
in./s2
Find the total acceleration of the
machine component at point B.
at  9.0 in./s 2
an  37.5 in./s2
Calculate the magnitude
a  9.02  37.52  38.6 in./s 2
at= 9 in/s2
Calculate the angle from
the horizontal
 37.5 
o

76.5

 9.0 
  arctan 
Combine for a final answer
aB
an= 37.5
in/s2
a B  38.6 in./s 2
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76.5
15 - 18
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Vector Mechanics for Engineers: Dynamics
Example – General Plane Motion
The knee has linear velocity and acceleration from both
translation (the runner moving forward) as well as rotation
(the leg rotating about the hip).
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Tenth
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Vector Mechanics for Engineers: Dynamics
General Plane Motion Video
https://www.youtube.com/watch?v=sA_6U-aMK4s
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2 - 20
Tenth
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Vector Mechanics for Engineers: Dynamics
General Plane Motion
• General plane motion is neither a translation nor
a rotation.
• General plane motion can be considered as the
sum of a translation and rotation.
• Displacement of particles A and B to A2 and B2
can be divided into two parts:
- translation to A2 and B1
- rotation of B1 about A2 to B2
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Velocity in Plane Motion
• Any plane motion can be replaced by a translation of an
arbitrary reference point A and a simultaneous rotation
about A. 


vB  v A  vB A

vB
A
 
  k  rB
A
 


v B  v A   k  rB
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vB
A
 r
A
15 - 22
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Vector Mechanics for Engineers: Dynamics
Absolute and Relative Velocity in Plane Motion
• 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.
vB
 tan 
vA
v B  v A tan 
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vA
vA

 cos
v B A l

vA
l cos
15 - 23
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Vector Mechanics for Engineers: Dynamics
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.
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15 - 24
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.3
SOLUTION:
• Will determine the absolute velocity of
point D with



vD  vB  vD B

• The velocity v B is obtained from the
given crank rotation data.

•
The
directions
of
the
absolute
velocity
v
The crank AB has a constant clockwise
D

and the relative velocity v D B are
angular velocity of 2000 rpm.
determined from the problem geometry.
For the crank position indicated,
• The unknowns in the vector expression
determine (a) the angular velocity of
are the velocity magnitudes v D and v D B
the connecting rod BD, and (b) the
which may be determined from the
velocity of the piston P.
corresponding vector triangle.
• The angular velocity of the connecting
rod is calculated from v D B .
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15 - 25
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.3
SOLUTION:
• Will determine the absolute velocity of point D with



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 velocity direction is as shown.

• The direction of the absolute velocity vD is horizontal.

The direction of the relative velocity vD B is
perpendicular to BD. Compute the angle between the
horizontal and the connecting rod from the law of sines.
sin 40 sin 

8 in.
3 in.
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  13.95
15 - 26
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.3
• Determine the velocity magnitudes vD and vD B
from the vector triangle.
vD B
vD
628.3 in. s


sin 53.95 sin 50 sin76.05
vD  523.4 in. s  43.6 ft s



vD  vB  vD B
vD
B
 495.9 in. s
vD
B
 l BD
vD
495.9 in. s
l
8 in.
 62.0 rad s
 BD 
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B
vP  vD  43.6 ft s


 BD

 62.0 rad s k
15 - 27
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
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
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2 - 28
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
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



vP  v A  vP
In the position shown, bar AB has an
angular velocity of 4 rad/s clockwise.
Determine the angular velocity of bars
BD and DE.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
A
 

 v A  k  rP
A
Evaluate the velocities of points B and D.
2 - 29
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
y
Determine the angular velocity of bars
BD and DE.
x
How should you proceed?
AB= 4 rad/s
Determine vB with respect to A, then work
your way along the linkage to point E.
Write vB in terms of point A, calculate vB.
vB  v A   AB  rB /A
 AB  (4 rad/s)k
rB/A  (7 in.)i
vB   AB  rB/A  (4k )  (7i )
vB  (28 in./s)j
Does it make sense that vB is in the +j direction?
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Vector Mechanics for Engineers: Dynamics
vD ,
Group Problem Solving
y
Determine vD with respect to B.
x
 BD  BD k rD/B  (8 in.)j
vD  vB   BD  rD/B  28 j  (BD k )  (8 j)
vD  28 j  8BD i
AB= 4 rad/s
Determine vD with respect to E, then
equate it to equation above.
 DE  DE k
rD /E  (11 in.)i  (3 in.)j
vD   DE  rD /E  (DE k )  (11i  3j)
vD  11DE j  3DE i
Equating components of the two expressions for vD
j:
28  11DE
i : 8BD  3DE
DE  2.5455 rad/s
3
8
BD   BD
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
 DE  2.55 rad/s
 BD  0.955 rad/s
2 - 31
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
• 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.
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15 - 32
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
• 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 are parallel, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
• 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 magnitudes are equal, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
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15 - 33
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
• The instantaneous center of rotation lies at the intersection of
the perpendiculars to the velocity vectors through A and B .
v
v
v
v B   BC   l sin   A
 A  A
l cos
AC l cos
 v A tan 
• 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.
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15 - 34
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Vector Mechanics for Engineers: Dynamics
Instantaneous Center of Rotation in Plane Motion
At the instant shown, what is the
approximate direction of the velocity
of point G, the center of bar AB?
a)
G
b)
c)
d)
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15 - 35
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.5
SOLUTION:
• Determine the velocity at B from the
given crank rotation data.
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.
• 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.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 15.5
SOLUTION:
• From Sample Problem 15.3,



vB  403.9i  481.3 j in. s 
vB  628.3 in. s
  13.95
• The instantaneous center of rotation is at the intersection
of the perpendiculars to the velocities through B and D.
 B  40    53.95
 D  90    76.05
BC
CD
8 in.


sin 76.05 sin 53.95 sin50
BC  10.14 in. CD  8.44 in.
• Determine the angular velocity about the center of
rotation based on the velocity at B.
vB  BC  BD
 BD 
vB 628.3 in. s

BC 10.14 in.
 BD  62.0 rad s
• Calculate the velocity at D based on its rotation about
the instantaneous center of rotation.
vD  CD  BD  8.44 in.62.0 rad s 
vP  vD  523 in. s  43.6 ft s
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
In the position shown, bar AB has an angular velocity of 4
rad/s clockwise. Determine the angular velocity of bars BD
and DE.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
vB  ( AB) AB  (0.25 m)(4 rad/s)  1 m/s
What is the velocity of B?
What direction is the velocity of B?
What direction is the velocity of D?
AB= 4 rad/s
vB

Find 
  tan 1
0.06 m
 21.8
0.15 m
vD
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Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Locate instantaneous center C at intersection of lines drawn
perpendicular to vB and vD.
Find distances BC and DC
C
B

vB
100 mm
D
BC 
0.1 m
0.1 m

 0.25 m
tan tan 21.8°
DC 
0.25 m
0.25 m

 0.2693 m
cos
cos21.8°
Calculate BD
vB  ( BC )BD
1 m/s  (0.25 m)BD
 BD  4 rad/s
vD
Find DE
vD  ( DC )BD 
0.25 m
(4 rad/s)
cos
vD  ( DE )DE ;
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
1 m/s 0.15 m

DE ;
cos
cos
 DE  6.67 rad/s
2 - 40
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion
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.
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2 - 41
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion
• Absolute acceleration of a particle of the slab,



aB  a A  aB A

• Relative acceleration a B A associated with rotation about A includes
tangential and normal components,
 

a B A t   k  rB A a B A t  r
aB A    2 rB A a B A n  r 2
n
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15 - 42
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion


• Given a A and v A ,


determine a B and  .



aB  a A  aB A



 a A  a B A   a B
n

• 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 .
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

A t
15 - 43
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Absolute and Relative Acceleration in Plane Motion



• Write a B  a A  a B
A

 x components:

in terms of the two component equations,
0  a A  l 2 sin   l cos
y components:  a B  l 2 cos  l sin 
• Solve for aB and .
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15 - 44
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
SOLUTION:
• The angular acceleration of the
connecting rod BD and the acceleration
of point D will be determined from






aD  aB  aD B  aB  aD B  aD B

t 
n
• The acceleration of B is determined from
the given rotation speed of AB.
Crank AG of the engine system has a
constant clockwise angular velocity of
2000 rpm.
• The directions of the accelerations



are
a D , a D B , and a D B
t
n
determined from the geometry.
For the crank position shown,
determine the angular acceleration of
the connecting rod BD and the
acceleration of point D.
• Component equations for acceleration
of point D are solved simultaneously for
acceleration of D and angular
acceleration of the connecting rod.

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.



15 - 45
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
SOLUTION:
• The angular acceleration of the connecting rod BD and
the acceleration of point D will be determined from






aD  aB  aD B  aB  aD B  aD B

t 
n
• The acceleration of B is determined from the given rotation
speed of AB.
 AB  2000 rpm  209.4 rad s  constant
 AB  0
2
aB  r AB


123 ft 209.4 rad s2  10,962 ft s2




2
aB  10,962 ft s  cos 40i  sin 40 j 
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15 - 46
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7

• The directions of the accelerations aD , aD
determined from the geometry.


aD   aD i



,
and
a
B t
D B n are
From Sample Problem 15.3, BD = 62.0 rad/s,  = 13.95o.
2
8 ft 62.0 rad s 2  2563 ft s 2
aD B n  BD  BD
 12
aD B n  2563 ft s2  cos13.95i  sin13.95 j 
aD B t  BD  BD  128 ft  BD  0.667 BD
The direction of (aD/B)t is known but the sense is not known,



aD B  0.667 BD  sin 76.05i  cos 76.05 j 

t
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15 - 47
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 15.7
• Component equations for acceleration of point D are solved
simultaneously.






aD  aB  aD B  aB  aD B  aD B

t 
n
x components:
 aD  10,962 cos 40  2563 cos13.95  0.667 BD sin 13.95
y components:
0  10,962 sin 40  2563 sin 13.95  0.667 BD cos13.95



2 
aD  9290 ft s  i

 BD  9940 rad s 2 k
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15 - 48
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
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 →
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2 - 49
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• The angular velocities were determined
in a previous problem by simultaneously
solving the component equations for



vD  vB  vD B
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.
• The angular accelerations are now
determined by simultaneously solving
the component equations for the relative
acceleration equation.
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2 - 50
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
From our previous problem, we used the relative
velocity equations to find that:
AB= 4 rad/s
 DE  2.55 rad/s
 BD  0.955 rad/s
We can now apply the relative acceleration
equation with  AB  0
2
a B  a A   AB  rB /A   AB
rB /A
Analyze
Bar AB
2
a B   AB
rB/A  (4) 2 (7i )  112 in./s 2 i
Analyze Bar BD
2
a D  a B   BD  rD /B  BD
rD /B  112i   BD k  (8 j)  (0.95455) 2 (8 j)
a D  (112  8  BD )i  7.289 j
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2 - 51
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Analyze Bar DE
2
a D   DE  rD /E  DE
rD /E
  DE k  (11i  3j)  (2.5455) 2 (11i  3 j)
AB= 4 rad/s
 11 DE j  3 DE i  71.275i  19.439 j
a D  (3 DE  71.275)i  (11 DE  19.439) j
From previous page, we had:
a D  (112  8  BD )i  7.289 j
Equate like components of aD
j:
i:
7.289  (11 DE  19.439)
112  8 BD  [(3)(2.4298)  71.275]
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 DE  2.4298 rad/s 2
 BD  4.1795 rad/s 2
2 - 52