Transcript Dt © 2013 The McGraw-Hill Companies, Inc. All rights reserved

Tenth Edition
CHAPTER
17
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
California State Polytechnic University
Plane Motion of Rigid
Bodies:
Energy and Momentum
Methods
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Principle of Work and Energy for a
Rigid Body
Work of Forces Acting on a Rigid Body
Kinetic Energy of a Rigid Body in Plane
Motion
Systems of Rigid Bodies
Conservation of Energy
Power
Sample Problem 17.1
Sample Problem 17.2
Sample Problem 17.3
Sample Problem 17.4
Sample Problem 17.5
Principle of Impulse and Momentum
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Systems of Rigid Bodies
Conservation of Angular Momentum
Sample Problem 17.6
Sample Problem 17.7
Sample Problem 17.8
Eccentric Impact
Sample Problem 17.9
Sample Problem 17.10
Sample Problem 17.11
17 - 2
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
To predict the launch from a catapult, you
must apply the principle of work-energy.
To determine the forces acting on the stopper pin when the
catapult reaches its final position, angular impulse momentum
equations are used.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 3
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
• Method of work and energy and the method of impulse and
momentum will be used to analyze the plane motion of rigid
bodies and systems of rigid bodies.
• Principle of work and energy is well suited to the solution of
problems involving displacements and velocities.
T1  U12  T2
• Principle of impulse and momentum is appropriate for
problems involving velocities and time.
t2 
t2 




H O 1    M O dt  H O 2
L1    Fdt  L2
t1
t1
• Problems involving eccentric impact are solved by supplementing
the principle of impulse and momentum with the application of
the coefficient of restitution.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 4
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
Approaches to Rigid Body Kinetics Problems
Forces and
Accelerations
Velocities and
Displacements
Velocities and
Time
Newton’s Second
Law (last chapter)
Work-Energy
ImpulseMomentum
 F  ma
M  H
t2
G
G
T1  U12  T2
G
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
mv1   F dt  mv2
t1
t2
I G1   M G dt  I G2
t1
2-5
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Principle of Work and Energy for a Rigid Body
• Work and kinetic energy are scalar quantities.
• Assume that the rigid body is made of a large
number of particles.
T1  U12  T2
T1 , T2  initial and final total kinetic energy of
particles forming body
U12  total work of internal and external forces
acting on particles of body.
• Internal forces between particles A
and B are equal and opposite.
• Therefore, the net work of internal
forces is zero.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 6
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Work of Forces Acting on a Rigid Body
• Work of a force during a displacement of its
point of application,
A2 
s
 2
U12   F  dr   F cos ds
A1
s1


• Consider the net work of two forces
F and  F

forming a couple of moment M during a
displacement of their points of application.
     
dU  F  dr1  F  dr1  F  dr2
 F ds2  Fr d
 M d
2
U12   M d
1
 M  2  1  if M is constant.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 7
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Work of Forces Acting on a Rigid Body
Do the pin forces at point
A do work?
YES
NO
Does the force P do work?
YES
NO
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 8
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Work of Forces Acting on a Rigid Body
Does the normal force N
do work on the disk?
YES
NO
Does the weight W do work?
YES
NO
If the disk rolls without slip, does
the friction force F do work?
YES
NO
dU  F dsC  F vc dt   0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2-9
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Kinetic Energy of a Rigid Body in Plane Motion
• Consider a rigid body of mass m in plane motion consisting of individual
particles i. The kinetic energy of the body can then be expressed as:
T  12 mv 2  12  Δmi vi2
 12 mv 2  12   ri2 Δmi   2
 12 mv 2  12 I  2
• Kinetic energy of a rigid body can be
separated into:
- the kinetic energy associated with the
motion of the mass center G and
- the kinetic energy associated with the
rotation of the body about G.
T  12 mv 2 
1
2
I 2
Translation + Rotation
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 10
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Kinetic Energy of a Rigid Body in Plane Motion
• Consider a rigid body rotating about a fixed axis through O.
T
1
2
 Δm v 
2
i i
1
2
2
2
1
Δ
m
r


r
Δ
m





 i i
i
2  i
2
 12 I O 2
• This is equivalent to using:
T  12 mv 2 
1
2
I 2
• Remember to only use
T  12 I O 2
when O is a fixed axis of rotation
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 11
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Quiz
The solid cylinder A and the pipe B
have the same diameter and mass.
If they are both released from rest
at the top of the hill, which will
reach the bottom the fastest?
a) A will reach the bottom first
b) B will reach the bottom first
c) They will reach the bottom
at the same time
A
B
Which will have the greatest
kinetic energy when it reaches the
bottom?
a) Cylinder A
b) Pipe B
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
c) Same kinetic energy
2 - 12
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Systems of Rigid Bodies
• For problems involving systems consisting of several rigid bodies, the
principle of work and energy can be applied to each body.
• We may also apply the principle of work and energy to the entire system,
T1  U12  T2
T1 ,T2 = arithmetic sum of the kinetic energies of
all bodies forming the system
U12 = work of all forces acting on the various
bodies, whether these forces are internal
or external to the system as a whole.
T
T
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 13
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Systems of Rigid Bodies
• For problems involving pin connected members, blocks and pulleys
connected by inextensible cords, and meshed gears,
- internal forces occur in pairs of equal and opposite forces
- points of application of each pair move through equal distances
- net work of the internal forces is zero
- work on the system reduces to the work of the external forces
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 14
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Conservation of Energy
• Expressing the work of conservative forces as a
change in potential energy, the principle of work
and energy becomes
T1  V1  T2  V2
• Consider the slender rod of mass m.
T1  0, V1  0
T2  12 mv22  12 I  22
 
 12 m 12 l
2
 12


2
1
ml
2
1 ml  

12
2 3
2
2
V2   12 Wl sin    12 mgl sin 
T1  V1  T2  V2
• mass m
• released with zero velocity
• determine  at 
1 ml 2 2 1
0
  mgl sin 
2 3
2
 3g
   sin  
 l

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 15
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Power
• Power = rate at which work is done


• For a body acted upon by force F and moving with velocity v ,
dU  
Power 
 F v
dt


• For a rigid body rotating with an
angular
velocity
and acted

upon by a couple of moment M parallel to the axis of rotation,
Power 
dU M d

 M
dt
dt
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 16
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.1
SOLUTION:
• Consider the system of the
flywheel and block. The work
done by the internal forces exerted
by the cable cancels.
• Note that the velocity of the block
and the angular velocity of the
drum and flywheel are related by
v  r
For the drum and flywheel, I  10.5 lb  ft  s 2 .
The bearing friction is equivalent to a
couple of 60 lb  ft. At the instant shown,
the block is moving downward at 6 ft/s.
• Apply the principle of work and
kinetic energy to develop an
expression for the final velocity.
Determine the velocity of the block after it
has moved 4 ft downward.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 17
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.1
SOLUTION:
• Consider the system of the flywheel and block. The work
done by the internal forces exerted by the cable cancels.
• Note that the velocity of the block and the angular velocity of
the drum and flywheel are related by
v
6 ft s
v
v
v  r
1  1 
 4.80 rad s
2  2  2
r 1.25 ft
r 1.25
• Apply the principle of work and kinetic energy to develop an
expression for the final velocity.
T1  12 mv12  12 I12

1 240 lb
2 1
2





6
ft
s

10
.
5
lb

ft

s
4
.
80
rad
s
2 32.2 ft s 2
2
 255 ft  lb
T2  12 mv22  12 I  22
2
1 240 2 1
 v 

v2  10.5 2   7.09v22
2 32.2
2
 1.25 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 18
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.1
T1  12 mv12  12 I12  255 ft  lb
T2  12 mv22  12 I22  7.09v22
• Note that the block displacement and pulley
rotation are related by
s
4 ft
2  2 
 3.20 rad
r 1.25 ft
Then,
U12  W s2  s1   M  2  1 
 240 lb 4 ft   60 lb  ft 3.20 rad 
 768 ft  lb
• Principle of work and energy:
T1  U12  T2
255 ft  lb  768 ft  lb  7.09 v22
v2  12.01ft s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
v2  12.01ft s
17 - 19
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.2
SOLUTION:
• Consider a system consisting of the two
gears. Noting that the gear rotational
speeds are related, evaluate the final
kinetic energy of the system.
• Apply the principle of work and energy.
Calculate the number of revolutions
m A  10 kg k A  200 mm
required for the work of the applied
mB  3 kg k B  80 mm
moment to equal the final kinetic energy
of the system.
The system is at rest when a moment
• Apply the principle of work and energy to
of M  6 N  m is applied to gear B.
a system consisting of gear A. With the
Neglecting friction, a) determine the
final kinetic energy and number of
number of revolutions of gear B before revolutions known, calculate the moment
its angular velocity reaches 600 rpm,
and tangential force required for the
and b) tangential force exerted by gear
indicated work.
B on gear A.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 20
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.2
SOLUTION:
• Consider a system consisting of the two gears. Noting
that the gear rotational speeds are related, evaluate the
final kinetic energy of the system.
B 
600 rpm 2 rad rev  62.8 rad s
60 s min
r
0.100
 A   B B  62.8
 25.1rad s
rA
0.250
I A  m Ak A2  10kg 0.200m 2  0.400 kg  m 2
I B  mB k B2  3kg 0.080m 2  0.0192 kg  m 2
T2  12 I A A2  12 I B B2
 12 0.400 25.1 2  12 0.019262.82
 163.9 J
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 21
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.2
• Apply the principle of work and energy. Calculate
the number of revolutions required for the work.
T1  U12  T2
0  6 B J  163.9J
 B  27.32 rad
B 
27.32
 4.35 rev
2
• Apply the principle of work and energy to a system
consisting of gear A. Calculate the moment and
tangential force required for the indicated work.
r
0.100
 A   B B  27.32
 10.93 rad
rA
0.250
T2  12 I A A2  12 0.40025.1 2  126.0 J
T1  U1 2  T2
0  M A 10.93 rad   126.0J
M A  rA F  11.52 N  m
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
F
11.52
 46.2 N
0.250
17 - 22
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.3
SOLUTION:
• The work done by the weight of the
bodies is the same. From the principle
of work and energy, it follows that each
body will have the same kinetic energy
after the change of elevation.
A sphere, cylinder, and hoop, each
having the same mass and radius, are
released from rest on an incline.
Determine the velocity of each body
after it has rolled through a distance
corresponding to a change of elevation h.
• Because each of the bodies has a
different centroidal moment of inertia,
the distribution of the total kinetic
energy between the linear and rotational
components will be different as well.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 23
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.3
SOLUTION:
• The work done by the weight of the bodies is the
same. From the principle of work and energy, it
follows that each body will have the same kinetic
energy after the change of elevation.
v
With  
r
v 
T2  12 mv  12 I   12 mv  12 I  
r
I 

 12  m  2 v 2

r 
2
2
2
2
T1  U1 2  T2
I 

0  Wh  12  m  2 v 2

r 
2Wh
2 gh
v2 

m  I r 2 1  I mr 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 24
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.3
• Because each of the bodies has a different
centroidal moment of inertia, the distribution of the
total kinetic energy between the linear and
rotational components will be different as well.
2 gh
v2 
1  I mr 2
I  52 mr 2
v  0.845 2 gh
Cylinder : I  12 mr 2
v  0.816 2 gh
Sphere :
Hoop :
I  mr 2
v  0.707 2 gh
NOTE:
• For a frictionless block sliding through the same
distance,   0, v  2 gh
• The velocity of the body is independent of its mass
and radius.
• The velocity of the body does depend on
k2
I

mr 2
r2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 25
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.4
SOLUTION:
• The weight and spring forces are
conservative. The principle of work and
energy can be expressed as
T1  V1  T2  V2
• Evaluate the initial and final potential
energy.
A 30-lb slender rod pivots about the
point O. The other end is pressed
against a spring (k = 1800 lb/in) until
the spring is compressed one inch and
the rod is in a horizontal position.
• Express the final kinetic energy in terms
of the final angular velocity of the rod.
• Based on the free-body-diagram
equation, solve for the reactions at the
pivot.
If the rod is released from this position,
determine its angular velocity and the
reaction at the pivot as the rod passes
through a vertical position.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 26
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.4
SOLUTION:
• The weight and spring forces are conservative. The
principle of work and energy can be expressed as
T1  V1  T2  V2
• Evaluate the initial and final potential energy.
V1  Vg  Ve  0  12 kx12  12 1800 lb in.1in.2
 900 in  lb  75 ft  lb
V2  Vg  Ve  Wh  0  30 lb 1.5 ft 
1 ml 2
I  12
1  30 lb 
2


 
5
ft
12  32.2 ft s 2 
 1.941lb  ft  s 2
 45 ft  lb
• Express the final kinetic energy in terms of the angular
velocity of the rod.
T2  12 mv22  12 I  22  12 mr 2 2  12 I  22

1 30
1.5 2 2  12 1.941 22  2.019 22
2 32.2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 27
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.4
From the principle of work and energy,
T1  V1  T2  V2
2  3.86 rad s
0  75 ft  lb  2.019 22  45 ft  lb
• Based on the free-body-diagram equation, solve for the
reactions at the pivot.

2
2
2
an  22.3 ft s 2
an  r  2  1.5 ft 3.86 rad s   22.3 ft s

at  r
at  r
 M O   M O eff
 Fx   Fx eff
 Fy   Fy eff
0  I  mr  r
 0
Rx  mr  
Rx  0
R y  30 lb  man

2


22
.
3
ft
s
2
32.2 ft s
30 lb
R y  9.22 lb
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

R  9.22
17 - 28
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.5
SOLUTION:
• Consider a system consisting of the two
rods. With the conservative weight force,
T1  V1  T2  V2
• Evaluate the initial and final potential
energy.
• Express the final kinetic energy of the
Each of the two slender rods has a
system in terms of the angular velocities of
mass of 6 kg. The system is released the rods.
from rest with b = 60o.
• Solve the energy equation for the angular
Determine a) the angular velocity of
velocity, then evaluate the velocity of the
o
rod AB when b = 20 , and b) the
point D.
velocity of the point D at the same
instant.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 29
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.5
SOLUTION:
• Consider a system consisting of the two rods. With
the conservative weight force,
T1  V1  T2  V2
• Evaluate the initial and final potential energy.
V1  2Wy1  258.86 N 0.325 m 
 38.26 J
V2  2Wy2  258.86 N 0.1283 m 
 15.10 J

W  mg  6 kg  9.81m s 2
 58.86 N

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 30
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.5
• Express the final kinetic energy of the system in terms
of the angular velocities of the rods.
vAB   0.375m 


Since vB is perpendicular to AB and vD is horizontal,
the instantaneous center of rotation for rod BD is C.
CD  20.75 msin 20  0.513 m
BC  0.75 m
and applying the law of cosines to CDE, EC = 0.522 m
Consider the velocity of point B

vB   AB   BC  AB
 BD  

vBD  0.522 m
For the final kinetic energy,
1 ml 2  1 6 kg 0.75 m 2  0.281kg  m 2
I AB  I BD  12
12
1 mv 2  1 I  2  1 mv 2  1 I  2
T2  12
AB 2 AB AB 12
BD 2 BD BD
1 6 0.375 2  1 0.281 2  1 6 0.522 2  1 0.281 2
 12
2
12
2
 1.520 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 31
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.5
• Solve the energy equation for the angular velocity,
then evaluate the velocity of the point D.
T1  V1  T2  V2
0  38.26 J  1.520 2  15.10 J
  3.90 rad s

 AB  3.90 rad s
vD  CD 
 0.513 m 3.90 rad s 
 2.00 m s

vD  2.00 m s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 32
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
SOLUTION:
• Because the problem deals with
positions and velocities, you should
apply the principle of work energy.
• Draw out the system at position 1 and
position 2 and define your datum
A slender 4-kg rod can rotate in a vertical
plane about a pivot at B. A spring of
constant k = 400 N/m and of unstretched
length 150 mm is attached to the rod as
shown. Knowing that the rod is released
from rest in the position shown, determine
its angular velocity after it has rotated
through
90o.
©
2013 The McGraw-Hill
Companies, Inc. All rights reserved.
• Use the work-energy equation
to determine the angular
velocity at position 2
2 - 33
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Draw your diagrams, set your datum and
apply the work energy equation
T1  V1  U1 2  T2  V2
Are any of the terms zero?
T1  V1  U1 2  T2  V2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 34
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
Determine the spring energy at position 1
Unstretched
Length
x1  CD  (150 mm )  370  150  220 mm  0.22 m
Ve 
1 2 1
kx1  (400 N/m)(0.22 m) 2  9.68 J
2
2
Determine the potential energy due to
gravity at position 1
Vg1  Wh  mgh  (4 kg)(9.81 m/s2 )(0.22 m)  7.063 J
Determine the spring energy at position 2
x2  230 mm  150 mm  80 mm  0.08 m
Ve 2 
1 2 1
kx2  (400 N/m)(0.08 m)2  1.28 J
2
2
Determine the potential energy due to
gravity at position 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vg 2  0
2 - 35
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
Determine an expression for T2
T2 
1
1
mv22  I 22
2
2
Can you relate v2 and 2?
v2  r2  (0.18 m)2
Find I and substitute in to T2
1
1
mL2  (4 kg)(0.6 m)2  0.12 kg  m2
12
12
1
1
1
1
T2  mv22  I 22  (4 kg)(0.182 )2  (0.12)22  0.124822
2
2
2
2
I 
Substitute into T1 + V1 = T2 + V2
9.68  7.063  0.1248 22  1.28 J
22
 10.713
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2  3.273 rad/s
2 - 36
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
For the previous problem, how would
you determine the reaction forces at B
when the bar is horizontal?
a) Apply linear-momentum to solve for BxDt and ByDt
b) Use work-energy to determine the work done by the
moment at C
c) Use sum of forces and sum of moments equations when
the bar is horizontal
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 37
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Angular Impulse Momentum
When two rigid bodies collide, we typically use principles
of angular impulse momentum. We often also use linear
impulse momentum (like we did for particles).
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 38
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
Approaches to Rigid Body Kinetics Problems
Forces and
Accelerations
Velocities and
Displacements
Velocities and
Time
Newton’s Second
Law (last chapter)
Work-Energy
ImpulseMomentum
 F  ma
M  H
t2
G
G
T1  U12  T2
G
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
mv1   F dt  mv2
t1
t2
I G1   M G dt  I G2
t1
2 - 39
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• Method of impulse and momentum:
- well suited to the solution of problems involving time and velocity
- the only practicable method for problems involving impulsive
motion and impact.
Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 40
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• The momenta of the particles of a system may be
reduced to a vector attached to the mass center
equal to their sum,
and a couple equal to the sum of their
moments about the mass center,
• For the plane motion of a rigid slab or of a rigid
body symmetrical with respect to the reference
plane,
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.



L   vi Δmi  mv

 
H G   ri  vi Δmi

H G  I
17 - 41
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• For plane motion problems, draw out an impulse-momentum diagram,
(similar to a free-body diagram)
• This leads to three equations of motion:
- summing and equating momenta and impulses in the x and y
directions
- summing and equating the moments of the momenta and impulses
with respect to any given point (often choose G)
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 42
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Impulse Momentum Diagrams
A sphere S hits a stationary bar
AB and sticks to it. Draw the
impulse-momentum diagram for
the ball and bar separately;
time 1 is immediately before
the impact and time 2 is
immediately after the impact.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 43
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Impulse Momentum Diagrams
Momentum of the
ball before impact
Impulse on
ball
Momentum of the
ball after impact
FimpDt
Momentum of the
bar before impact
Impulse on
bar
Momentum of the
bar after impact
FimpDt
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 44
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• Fixed axis rotation:
- The angular momentum about O
I O  I   mv r
 I   mr  r


 I  mr 2 
- Equating the moments of the momenta and
impulses about O,
t2
I O1    M O dt  I O 2
t1
The pin forces at point O now contribute no moment to the equation
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 45
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Systems of Rigid Bodies
• Motion of several rigid bodies can be analyzed by applying
the principle of impulse and momentum to each body
separately.
• For problems involving no more than three unknowns, it may
be convenient to apply the principle of impulse and
momentum to the system as a whole.
• For each moving part of the system, the diagrams of momenta
should include a momentum vector and/or a momentum couple.
• Internal forces occur in equal and opposite pairs of vectors and
generate impulses that cancel out.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 46
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Practice
From the previous problem, notice that the impulse acting on the
sphere is equal and opposite to the impulse acting on the bar. We can
take advantage of this by drawing the impulse-momentum diagram of
the entire system, as shown on the next slide.
FimpDt
FimpDt
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 47
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Practice – Diagram for combined system
FimpDt
FimpDt
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 48
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Conservation of Angular Momentum
The moments acting through the skater’s center of gravity are
negligible, so his angular momentum remains constant. He can
adjust his spin rate by changing his moment of inertia.
t2
I G1    M G dt  I G2
t1
IG1

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
IG2
2 - 49
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Conservation of Angular Momentum
• When no external force acts on a rigid body or a system of rigid
bodies, the system of momenta at t1 is equipollent to the system
at t2. The total linear momentum and angular momentum about
any point are conserved,


H 0 1  H 0 2
L1  L2
• When the sum of the angular impulses pass through O, the
linear momentum may not be conserved, yet the angular
momentum about O is conserved,
H 0 1  H 0 2
• Two additional equations may be written by summing x and
y components of momenta and may be used to determine
two unknown linear impulses, such as the impulses of the
reaction components at a fixed point.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 50
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
For the problem we looked at previously, is the
angular momentum about G conserved?
YES
NO
For the problem we looked at previously, is the
angular momentum about point A conserved?
YES
NO
For the problem we looked at previously, is the
linear momentum of the system conserved?
YES
NO
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 51
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.6
SOLUTION:
• Considering each gear separately, apply
the method of impulse and momentum.
• Solve the angular momentum equations
for the two gears simultaneously for the
unknown time and tangential force.
m A  10 kg k A  200 mm
mB  3 kg k B  80 mm
The system is at rest when a moment
of M  6 N  m is applied to gear B.
Neglecting friction, a) determine the
time required for gear B to reach an
angular velocity of 600 rpm, and b) the
tangential force exerted by gear B on
gear A.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 52
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.6
SOLUTION:
• Considering each gear separately, apply the method of impulse
and momentum.
moments about A:
0  FtrA   I A  A 2
Ft 0.250 m   0.400 kg  m 25.1rad s 
Ft  40.2 N  s
moments about B:
0  Mt  FtrB  I B  B 2
6 N  m t  Ft 0.100 m 
 0.0192 kg  m 2 62.8 rad s 
• Solve the angular momentum equations for the two gears simultaneously
for the unknown time and tangential force.
t  0.871 s
F  46.2 N
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 53
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.7
SOLUTION:
• Apply principle of impulse and momentum
to find variation of linear and angular
velocities with time.
Uniform sphere of mass m and
radius r is projected along a rough
horizontal surface with a linear
velocity v1 and no angular velocity.
The coefficient of kinetic friction is
k .
Determine a) the time t2 at which
the sphere will start rolling without
sliding and b) the linear and angular
velocities of the sphere at time t2.
• Relate the linear and angular velocities
when the sphere stops sliding by noting
that the velocity of the point of contact is
zero at that instant.
• Substitute for the linear and angular
velocities and solve for the time at which
sliding stops.
• Evaluate the linear and angular velocities
at that instant.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 54
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.7
SOLUTION:
• Apply principle of impulse and momentum
to find variation of linear and angular
velocities with time.
• Relate linear and angular velocities when
sphere stops sliding by noting that velocity
Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2 of point of contact is zero at that instant.
y components:
Nt Wt  0
N  W  mg
x components:
mv1  Ft  mv2
mv1   k mgt  mv2
v2  r 2
v2  v1  k gt
moments about G:
Ftr  I 2
k mg tr  52 mr 2 2
• Substitute for the linear and angular
velocities and solve for the time at which
sliding stops.
 5 k g 
v1   k gt  r 
t
2
r


t
2 
5 k g
t
2 r
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 v1
7 k g
17 - 55
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.7
• Evaluate the linear and angular velocities
at that instant.
 2 v1 

v2  v1   k g 
7

g

k 
5
v2  v1
7
Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2
y components:
N  W  mg
x components:
v2  v1  k gt
moments about G:
2 
5 k g
t
2 r
2 
5  k g  2 v1 


2 r  7  k g 
2 
5 v1
7r
v2  r 2
 5 k g 
v1   k gt  r 
t
2
r


t
2 v1
7 k g
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 56
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.8
SOLUTION:
• Observing that none of the external
forces produce a moment about the y
axis, the angular momentum is
conserved.
Two solid spheres (radius = 3 in.,
W = 2 lb) are mounted on a spinning
horizontal rod ( I R  0.25 lb  ft  s 2 ,
 = 6 rad/sec) as shown. The balls are
held together by a string which is
suddenly cut. Determine a) angular
velocity of the rod after the balls have
moved to A’ and B’, and b) the energy
lost due to the plastic impact of the
spheres and stops.
• Equate the initial and final angular
momenta. Solve for the final angular
velocity.
• The energy lost due to the plastic impact
is equal to the change in kinetic energy
of the system.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 57
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.8
Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2
SOLUTION:
• Observing that none of the
external forces produce a
moment about the y axis, the
angular momentum is
conserved.
• Equate the initial and final
angular momenta. Solve for
the final angular velocity.
2ms r11 r1  I S 1   I R1  2ms r2 2 r2  I S  2   I R 2
 2  1
ms r12  I S  I R
ms r22  I S  I R
I R  0.25 lb  ft  s 2
1  6 rad s
 2  2
 ft   0.00155 lb  ft  s 2
IS 

2
 32.2 ft s  12 
2
2
2  2  5 
2  2  25 
mS r1  
   0.0108 mS r2  
   0.2696
 32.2  12 
 32.2  12 
2 ma 2
5

2
5
2 lb
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2  2.08 rad s
17 - 58
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.8
• The energy lost due to the
plastic impact is equal to the
change in kinetic energy of the
system.
1  6 rad s
2  2.08 rad s
I R  0.25 lb  ft  s 2
I S  0.00155 lb  ft  s 2
mS r12  0.0108 lb  ft  s 2
mS r22  0.2696 lb  ft  s 2




T  2 12 mS v 2  12 I S  2  12 I R 2  12 2mS r 2  2 I S  I R  2
T1  12 0.27562  4.95 ft  lb
T2  12 0.7922.082  1.71ft  lb
ΔT  T2  T1  1.71  4.95
DT  3.24 ft  lb
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 59
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
SOLUTION:
• Consider the projectile and bar as a
single system. Apply the principle of
impulse and momentum.
• The moments about C of the momenta
and impulses provide a relation between
the final angular velocity of the rod and
velocity of the projectile.
• Use the principle of work-energy to
determine the angle through which the
bar swings.
A projectile weighing 0.08 lb is fired with a horizontal
velocity of 500 ft/s into the lower end of a slender 15-lb bar
of length L= 30 in. Knowing that h= 12 in. and that the bar is
initially at rest, determine the angular velocity of the bar
when it reaches the horizontal position.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 60
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
Draw the impulse momentum diagram
Apply the angular impulse momentum
equation about point C
m0 v0 ( L  h)  m0 vB ( L  h)  I C 
Given: Wo= 0.08 lb, vo= 500 ft/s
WAB = 15-lb L= 30 in. h= 12 in.
Find: AB when = 90o
Or you could use the relationship:
L

m0 v0 ( L  h)  m0 vB ( L  h)  mv0   h   I 
2

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 61
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
m0 v0 ( L  h)  m0 vB ( L  h)  I C 
Relate vB and  (after the impact)
vB  ( L  h)
Substitute into equation (1) and solve for 
m0 v0 ( L  h)  m0 ( L  h)2   IC

Find IC
L  30 in.  2.5 ft m 
m0 v0 ( L  h)
m0 ( L  h) 2  I C
15
 0.46584 lb  s 2 /ft
32.2
1
1
mL2  md 2  (0.46584)(2.5)2  (0.46584)(0.25)2
12
12
2
I C  0.27174 lb  s  ft
IC 
Substitute and solve

m0 v0 ( L  h)
m0 ( L  h)2  I C

32.2 (1000)(2.5  1)
2
0.08lbs
32.2 (2.5  1)  0.27174
0.08lbs
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2  6.7189 rad/s
2 - 62
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
Draw position 1 and 2, set your datum and
apply the conservation of energy equation
T2  V2  T3  V3
Find T2
Find V2
T2 
B
2
DATUM
1
1
I C 22  (0.27174)(6.71892 )
2
2
1
T2  6.1337 lb ft
V2  m AB gy AB 2  mO gyO 2  WAB y AB 2  WO yO 2
V2  15( L2  h)  0.08( L  h)  15(0.25)  0.08(1.5)  3.87 lbs ft
Solve for 3
1
T3  I C 32  T2  V2
2
1
(0.27174)32  6.1337  3.87
2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
3  4.08 rad/s
2 - 63
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
For the previous problem, how would
you determine the reaction forces at C
when the bar is horizontal?
a) Apply linear-momentum to solve for CxDt and CyDt
b) Use work-energy to determine the work done by the
moment at C
c) Use sum of forces and sum of moments equations when
the bar is horizontal
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 64
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
For the previous problem, what would
happen if the coefficient of restitution
between the projectile and bar was 1.0
instead of zero?
a)
b)
c)
d)
The angular velocity after impact would be bigger
The angular velocity after impact would be smaller
The angular velocity after impact would be the same
Not enough information to tell
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 65
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Eccentric Impact
u A n  uB n
Period of deformation

Impulse   Rdt
Period of restitution

Impulse   Pdt
• Principle of impulse and momentum is supplemented by

Rdt

e  coefficient of restitution  
 Pdt
vB n  vA n

v A n  v B n
These velocities are for the
points of impact
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 66
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Questions
The cars collide, hitting at point
P as shown. Which of the
following can you use to help
analyze the collision?
a) The linear momentum of car A is
conserved.
b) The linear momentum of the
combined two cars is conserved
c) The total kinetic energy before the
impact equals the total kinetic
energy after the impact
d) The angular momentum about the
CG of car B is conserved
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
A
P
P
A
2 - 67
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.9
SOLUTION:
• Consider a system consisting of the
bullet and panel. Apply the principle of
impulse and momentum.
• The final angular velocity is found
from the moments of the momenta and
impulses about A.
A 0.05-lb bullet is fired into the side of a
20-lb square panel which is initially at
rest.
• The reaction at A is found from the
horizontal and vertical momenta and
impulses.
Determine a) the angular velocity of the
panel immediately after the bullet
becomes embedded and b) the impulsive
reaction at A, assuming that the bullet
becomes embedded in 0.0006 s.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 68
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.9
SOLUTION:
• Consider a system consisting
of the bullet and panel. Apply
the principle of impulse and
momentum.
• The final angular velocity is
found from the moments of
the momenta and impulses
about A.
moments about A:
mB v B
v2 
1412 ft   0 m P v2 129 ft   I P2
 
9 ft 
2
12
IP 
 
1 m b2
6 P

 0.05 
 20  9


1500 14

 12 2
12
32
.
2
32
.
2




 2  4.67 rad s
v2 
129 2  3.50 ft s
2
1  20  18 
2
 
   0.2329 lb  ft  s
6  32.2  12 
129   0.23292
2  4.67 rad s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 69
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.9
• The reactions at A are found
from the horizontal and
vertical momenta and
impulses.
 2  4.67 rad s
v2 
129 2  3.50 ft s
x components:
mB vB  Ax D t  m p v2
 0.05 
 20 




1500

A
0
.
0006




3.50 
x
32
.
2
32
.
2




Ax  259 lb
Ax  259 lb
y components:
0  Ay D t  0
Ay  0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 70
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.10
SOLUTION:
• Consider the sphere and rod as a single
system. Apply the principle of impulse
and momentum.
A 2-kg sphere with an initial velocity
of 5 m/s strikes the lower end of an 8kg rod AB. The rod is hinged at A and
initially at rest. The coefficient of
restitution between the rod and sphere
is 0.8.
Determine the angular velocity of the
rod and the velocity of the sphere
immediately after impact.
• The moments about A of the momenta
and impulses provide a relation between
the final angular velocity of the rod and
velocity of the sphere.
• The definition of the coefficient of
restitution provides a second
relationship between the final angular
velocity of the rod and velocity of the
sphere.
• Solve the two relations simultaneously
for the angular velocity of the rod and
velocity of the sphere.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 71
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.10
SOLUTION:
• Consider the sphere and rod as a
single system. Apply the
principle of impulse and
momentum.
moments about A:
ms vs 1.2 m  ms vs 1.2 m  mRvR 0.6 m  I 
• The moments about A of the
momenta and impulses provide a
relation between the final
angular velocity of the rod and
velocity of the rod.
vR  r    0.6 m  
1 mL2  1 8 kg 1.2 m 2  0.96 kg  m 2
I  12
12
2 kg 5 m s 1.2 m   2 kg vs 1.2 m   8 kg 0.6 m  0.6 m 


 0.96 kg  m 2  
12  2.4 vs  3.84 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 72
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.10
• The definition of the coefficient
of restitution provides a second
relationship between the final
angular velocity of the rod and
velocity of the sphere.
• Solve the two relations
simultaneously for the angular
velocity of the rod and velocity
of the sphere.
Moments about A:
12  2.4 vs  3.84 
Relative velocities:
vB  vs  evB  vs 
1.2 m    vs  0.85 m s 
Solving,
   3.21rad/s
vs  0.143 m s
   3.21rad/s
vs  0.143 m s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 73
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.11
SOLUTION:
• Apply the principle of impulse and
momentum to relate the velocity of the
package on conveyor belt A before the
impact at B to the angular velocity about
B after impact.
A square package of mass m moves
down conveyor belt A with constant
velocity. At the end of the conveyor,
the corner of the package strikes a rigid
support at B. The impact is perfectly
plastic.
Derive an expression for the minimum
velocity of conveyor belt A for which
the package will rotate about B and
reach conveyor belt C.
• Apply the principle of conservation of
energy to determine the minimum initial
angular velocity such that the mass
center of the package will reach a
position directly above B.
• Relate the required angular velocity to
the velocity of conveyor belt A.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 74
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.11
SOLUTION:
• Apply the principle of impulse and momentum to relate the velocity of the package on
conveyor belt A before the impact at B to angular velocity about B after impact.
Moments about B:
mv1 12 a   0  mv2  22 a   I2
v2 
mv1 12 a   0  m 22 a2  22 a   16 ma 2 2
 a
2
2
2
I  16 m a 2
v1  43 a 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 75
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 17.11
• Apply the principle of conservation of energy to determine
the minimum initial angular velocity such that the mass
center of the package will reach a position directly above B.
T2  V2  T3  V3
T2  12 mv22  12 I  22
h2  GB sin 45  15

 a sin 60  0.612a
2
2


1 m 2 a
2
2
2
   ma 
2
1 1
2 6
2
2
2
 13 ma 2 22
V2  Wh2
T3  0
(solving for the minimum 2)
V3  Wh3
1 ma 2 2
2
3
 22 
h3 
2
a
2
 0.707a
 Wh2  0  Wh3
3W
ma
3g


0.707a  0.612a  
h

h

2
2 3
2
a
v1  43 a 2  43 a 0.285 g a
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
0.285 g a
v1  0.712 ga
17 - 76
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
SOLUTION:
• Consider the sphere and panel as a
single system. Apply the principle of
impulse and momentum.
• The moments about A of the momenta
and impulses provide a relation between
the angular velocity of the panel and
velocity of the sphere.
• Use the principle of work-energy to
determine the angle through which the
panel swings.
An 8-kg wooden panel P is suspended from a pin support at A and is
initially at rest. A 2-kg metal sphere S is released from rest at B and falls
into a hemispherical cup C attached to the panel at the same level as the
mass center G. Assuming that the impact is perfectly plastic, determine
the angular velocity of the panel immediately after the impact.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 77
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
Draw the impulse momentum diagram
Apply the angular impulse momentum
equation about point A
Given: mS= 2 kg, mP = 8 kg, mS (vC )1 (0.2 m)  0  mS (vC )2 ( AC)  I2  mP v2 (0.25 m)
hS= 0.250 m, e= 0.
HA of sphere
Find: Angle  through which
HA of panel
before impact
after impact
the panel and sphere swing
HA of sphere
after the impact
after impact
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 78
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
mS (vC )1 (0.2 m)  0  mS (vC )2 ( AC)  I2  mP v2 (0.25 m)
Determine velocity of sphere at impact (vS)1
You can apply work-energy or kinematics
(vS )1  2 gy
 2(9.81 m/s 2 )(0.5 m)
 3.1321 m/s
Determine velocity of sphere after impact in terms of 2
( v S )2  AC 2
AC   (0.2)2  (0.25)2  0.32016 m
( v S ) 2  0.320162
(perpendicular to AC.)
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 79
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Team Problem Solving
mS (vC )1 (0.2 m)  0  mS (vC )2 ( AC)  I2  mP v2 (0.25 m)
Determine mass moment of inertia for panel
I 
1
1
mP (0.5 m) 2  (8)(0.5) 2  0.3333 kg  m 2
6
6
Substitute into H equation and solve for 2
mS (vC  )1 (0.2 m)  0  mS (vC  ) 2 ( AC )  I 2  mP v2 (0.25 m)
(2 kg)(3.1321 m/s)(0.2 m)  (2 kg)(0.320162 )(0.32016 m)  0.33332  (8 kg)(0.25 m) 2 2
1.25284  (0.2050  0.3333  0.500)2
2  1.2066 rad/s
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 80
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
For the previous problem, what would
you do if you wanted to determine how
high up the panel swung after the
impact?
a) Apply linear-momentum to solve for mvG
b) Use work-energy and set Tfinal equal to zero
c) Use sum of forces and sum of moments equations
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 81
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
For the previous problem, what if the
ball was dropped closer to point A (e.g.,
at x= 100 mm instead of 200 mm)?
a)
b)
c)
d)
The angular velocity after impact would be bigger
The angular velocity after impact would be smaller
The angular velocity after impact would be the same
Not enough information to tell
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 82