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CHAPTER
13
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
Kinetics of Particles:
Energy and Momentum
Methods
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Work of a Force
Principle of Work & Energy
Applications of the Principle of
Work & Energy
Power and Efficiency
Sample Problem 13.1
Sample Problem 13.2
Sample Problem 13.3
Sample Problem 13.4
Sample Problem 13.5
Potential Energy
Conservative Forces
Conservation of Energy
Motion Under a Conservative
Central Force
Sample Problem 13.6
Sample Problem 13.7
Sample Problem 13.9
Principle of Impulse and Momentum
Impulsive Motion
Sample Problem 13.10
Sample Problem 13.11
Sample Problem 13.12
Impact
Direct Central Impact
Oblique Central Impact
Problems Involving Energy and
Momentum
Sample Problem 13.14
Sample Problem 13.15
Sample Problems 13.16
Sample Problem 13.17
13 - 2
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
• Previously, problems dealing with the motion of particles
 were

solved through the fundamental equation of motion, F  ma.
Current chapter introduces two additional methods of analysis.
• Method of work and energy: directly relates force, mass,
velocity and displacement.
• Method of impulse and momentum: directly relates force,
mass, velocity, and time.
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Vector Mechanics for Engineers: Dynamics
Work of a Force

• Differential vector dr is the particle displacement.
• Work of the force is
 
dU  F  dr
 F ds cos 
 Fx dx  Fy dy  Fz dz
• Work is a scalar quantity, i.e., it has magnitude and
sign but not direction.
• Dimensions of work are length  force. Units are
1 J  joule   1 N 1 m 
1ft  lb  1.356 J
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Vector Mechanics for Engineers: Dynamics
Work of a Force
• Work of a force during a finite displacement,
U12 
A2 

F

d
r

A1


s2
s2
s1
s1
 F cos ds   Ft ds
A2
 Fx dx  Fy dy  Fz dz 
A1
• Work is represented by the area under the
curve of Ft plotted against s.
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Vector Mechanics for Engineers: Dynamics
Work of a Force
• Work of a constant force in rectilinear motion,
U12  F cos  x
• Work of the force of gravity,
dU  Fx dx  Fy dy  Fz dz
 W dy
y2
U12    W dy
y1
 W  y 2  y1   W y
• Work of the weight is equal to product of
weight W and vertical displacement y.
• Work of the weight is positive when y < 0,
i.e., when the weight moves down.
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Vector Mechanics for Engineers: Dynamics
Work of a Force
• Magnitude of the force exerted by a spring is
proportional to deflection,
F  kx
k  spring constant N/m or lb/in.
• Work of the force exerted by spring,
dU   F dx  kx dx
x2
U12    kx dx  12 kx12  12 kx22
x1
• Work of the force exerted by spring is positive
when x2 < x1, i.e., when the spring is returning to
its undeformed position.
• Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against x,
U12   12  F1  F2  x
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Vector Mechanics for Engineers: Dynamics
Work of a Force
Work of a gravitational force (assume particle M
occupies fixed position O while particle m follows path
shown),
Mm
dU   Fdr  G 2 dr
r
r2
Mm
r1
r2
U12    G
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dr  G
Mm
Mm
G
r2
r1
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Vector Mechanics for Engineers: Dynamics
Work of a Force
Forces which do not do work (ds = 0 or cos   0:
• reaction at frictionless pin supporting rotating body,
• reaction at frictionless surface when body in contact
moves along surface,
• reaction at a roller moving along its track, and
• weight of a body when its center of gravity moves
horizontally.
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Vector Mechanics for Engineers: Dynamics
Particle Kinetic Energy: Principle of Work & Energy

• Consider a particle of mass m acted upon by force F
dv
Ft  mat  m
dt
dv ds
dv
m
 mv
ds dt
ds
F t ds  mv dv
• Integrating from A1 to A2 ,
s2
v2
s1
v1
 Ft ds  m  v dv  12 mv2  12 mv1
2
2
U12  T2  T1
T  12 mv 2  kinetic energy

• The work of the force F is equal to the change in
kinetic energy of the particle.
• Units of work and kinetic energy are the same:
2
m
 m


2
T  12 mv  kg    kg 2 m  N  m  J
s
 s 
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Vector Mechanics for Engineers: Dynamics
Applications of the Principle of Work and Energy
• Wish to determine velocity of pendulum bob
at A2. Consider work & kinetic energy.

• Force P acts normal to path and does no
work.
T1  U12  T2
0  Wl 
1W 2
v2
2g
v2  2 gl
• Velocity found without determining
expression for acceleration and integrating.
• All quantities are scalars and can be added
directly.
• Forces which do no work are eliminated from
the problem.
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Vector Mechanics for Engineers: Dynamics
Applications of the Principle of Work and Energy
• Principle of work and energy cannot be
applied to directly determine the acceleration
of the pendulum bob.
• Calculating the tension in the cord requires
supplementing the method of work and energy
with an application of Newton’s second law.
• As the bob passes through A2 ,
 Fn  m an
v2  2 gl
W v22
P W 
g l
W 2 gl
P W 
 3W
g l
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Vector Mechanics for Engineers: Dynamics
Power and Efficiency
• Power  rate at which work is done.
 
dU F  dr


dt
dt
 
 F v
• Dimensions of power are work/time or force*velocity.
Units for power are
J
m
ft  lb
1 W (watt)  1  1 N 
or 1 hp  550
 746 W
s
s
s
•   efficiency
output work

input work
power output

power input
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
• Determine the distance required for the
work to equal the kinetic energy change.
An automobile weighing 4000 lb is
driven down a 5o incline at a speed of
60 mi/h when the brakes are applied
causing a constant total breaking force
of 1500 lb.
Determine the distance traveled by the
automobile as it comes to a stop.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
 mi  5280 ft  h 
v1   60 

  88 ft s
h  mi  3600 s 

T1  12 mv12  12 4000 32.2 882  481000ft  lb
v2  0
T2  0
• Determine the distance required for the work
to equal the kinetic energy change.
U12   1500lbx  4000lbsin 5x
 1151lbx
T1  U12  T2
481000ft  lb  1151lbx  0
x  418 ft
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.2
SOLUTION:
• Apply the principle of work and
energy separately to blocks A and B.
• When the two relations are combined,
the work of the cable forces cancel.
Solve for the velocity.
Two blocks are joined by an inextensible
cable as shown. If the system is released
from rest, determine the velocity of block
A after it has moved 2 m. Assume that the
coefficient of friction between block A
and the plane is mk = 0.25 and that the
pulley is weightless and frictionless.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.2
SOLUTION:
• Apply the principle of work and energy separately
to blocks A and B.


W A  200 kg  9.81m s 2  1962 N
FA  m k N A  m k W A  0.251962 N   490 N
T1  U12  T2 :
0  FC 2 m   FA 2 m   12 m Av 2
FC 2 m   490 N 2 m   12 200 kg v 2


WB  300 kg  9.81m s 2  2940 N
T1  U12  T2 :
0  Fc 2 m   WB 2 m   12 m B v 2
 Fc 2 m   2940 N 2 m   12 300 kg v 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.2
• When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.
FC 2 m   490 N 2 m   12 200 kg v 2
 Fc 2 m   2940 N 2 m   12 300 kg v 2
2940 N 2 m   490 N 2 m   12 200 kg  300 kg v 2
4900 J  12 500 kg v 2
v  4.43 m s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.3
SOLUTION:
• Apply the principle of work and energy
between the initial position and the
point at which the spring is fully
compressed and the velocity is zero.
The only unknown in the relation is the
friction coefficient.
A spring is used to stop a 60 kg package
which is sliding on a horizontal surface.
The spring has a constant k = 20 kN/m
and is held by cables so that it is initially • Apply the principle of work and energy
for the rebound of the package. The
compressed 120 mm. The package has a
only unknown in the relation is the
velocity of 2.5 m/s in the position shown
and the maximum deflection of the spring velocity at the final position.
is 40 mm.
Determine (a) the coefficient of kinetic
friction between the package and surface
and (b) the velocity of the package as it
passes again through the position shown.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
T1  12 mv12  12 60 kg 2.5 m s 2  187.5 J
U12  f
  m kW x

T2  0

  m k 60 kg  9.81m s 2 0.640 m   377 J m k
Pmin  kx0  20 kN m 0.120 m   2400 N
Pmax  k  x0  x   20 kN m 0.160 m   3200 N
U12 e   12 Pmin  Pmax x
  12 2400 N  3200 N 0.040 m   112.0 J
U12  U12  f  U12 e  377 J m k  112 J
T1  U12  T2 :
187.5 J - 377 J m k  112 J  0
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mk  0.20
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.3
• Apply the principle of work and energy for the rebound
of the package.
T 3 12 mv32  12 60kg v32
T2  0
U 23  U 23  f  U 23 e  377 J m k  112 J
 36.5 J
T2  U 23  T3 :
0  36.5 J  12 60 kg v32
v3  1.103 m s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to
determine velocity at point 2.
• Apply Newton’s second law to find
normal force by the track at point 2.
A 2000 lb car starts from rest at point 1
• Apply principle of work and energy to
and moves without friction down the
determine velocity at point 3.
track shown.
• Apply Newton’s second law to find
Determine:
minimum radius of curvature at point 3
such that a positive normal force is
a) the force exerted by the track on
exerted by the track.
the car at point 2, and
b) the minimum safe value of the
radius of curvature at point 3.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
T1  0
T2  12 mv22 
U12  W 40 ft 
T1  U12  T2 :
1W 2
v2
2g
0  W 40 ft  

v22  240 ft g  240 ft  32.2 ft s 2

1W 2
v2
2g
v2  50.8 ft s
• Apply Newton’s second law to find normal force by
the track at point 2.
   Fn  m an :
W v22 W 240 ft g
 W  N  m an 

g  2 g 20 ft
N  5W
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N  10000 lb
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.4
• Apply principle of work and energy to determine
velocity at point 3.
T1  U13  T3
0  W 25 ft  
v32  225 ft g  225 ft 32.2 ft s 
1W 2
v3
2g
v3  40.1ft s
• Apply Newton’s second law to find minimum radius of
curvature at point 3 such that a positive normal force is
exerted by the track.
   Fn  m an :
W  m an
W v32 W 225 ft g


g 3 g
3
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
3  50 ft
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.5
SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.
Power delivered by motor is
equal to FvD, vD = 8 ft/s.
The dumbwaiter D and its load have a
combined weight of 600 lb, while the
counterweight C weighs 800 lb.
• In the first case, bodies are in uniform
motion. Determine force exerted by
motor cable from conditions for static
equilibrium.
Determine the power delivered by the
electric motor M when the dumbwaiter
(a) is moving up at a constant speed of
8 ft/s and (b) has an instantaneous
velocity of 8 ft/s and an acceleration of
2.5 ft/s2, both directed upwards.
• In the second case, both bodies are
accelerating. Apply Newton’s
second law to each body to
determine the required motor cable
force.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.5
• In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
Free-body C:
   Fy  0 :
2T  800 lb  0
T  400 lb
Free-body D:
   Fy  0 : F  T  600 lb  0
F  600 lb  T  600 lb  400 lb  200 lb
Power  Fv D  200 lb8 ft s 
 1600ft  lb s
Power  1600ft  lb s 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
1 hp
 2.91 hp
550 ft  lb s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.5
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
a D  2.5 ft s 2 
aC   12 a D  1.25 ft s 2 
Free-body C:
   Fy  mC aC : 800  2T 
800
1.25
32.2
T  384.5 lb
Free-body D:
600
2.5
32.2
F  384.5  600  46.6
   Fy  mD a D : F  T  600 
F  262.1 lb
Power  Fv D  262.1 lb 8 ft s   2097 ft  lb s
Power  2097 ft  lb s 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
1 hp
 3.81 hp
550 ft  lb s
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Vector Mechanics for Engineers: Dynamics
Potential Energy

• Work of the force of gravity W,
U12  W y1  W y2
• Work is independent of path followed; depends
only on the initial and final values of Wy.
V g  Wy
 potential energy of the body with respect
to force of gravity.
 1  Vg 2
U12  V g
• Choice of datum from which the elevation y is
measured is arbitrary.
• Units of work and potential energy are the same:
Vg  Wy  N  m  J
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Vector Mechanics for Engineers: Dynamics
Potential Energy
• Previous expression for potential energy of a body
with respect to gravity is only valid when the
weight of the body can be assumed constant.
• For a space vehicle, the variation of the force of
gravity with distance from the center of the earth
should be considered.
• Work of a gravitational force,
GMm GMm
U12 

r2
r1
• Potential energy Vg when the variation in the
force of gravity can not be neglected,
GMm
WR 2
Vg  

r
r
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Vector Mechanics for Engineers: Dynamics
Potential Energy
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
U12  12 kx12  12 kx22
• The potential energy of the body with respect
to the elastic force,
Ve  12 kx 2
U12  Ve 1  Ve 2
• Note that the preceding expression for Ve is
valid only if the deflection of the spring is
measured from its undeformed position.
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Vector Mechanics for Engineers: Dynamics
Conservative Forces
• Concept of potential energy can be applied if the
work of the force is independent of the path
followed by its point of application.
U12  V  x1, y1, z1   V  x2 , y2 , z2 
Such forces are described as conservative forces.
• For any conservative force applied on a closed path,
 
 F  dr  0
• Elementary work corresponding to displacement
between two neighboring points,
dU  V  x, y, z   V  x  dx, y  dy, z  dz 
 dV  x, y, z 
V
V 
 V
Fx dx  Fy dy  Fz dz  
dx 
dy 
dz 
y
z 
 x

 V V V 
F  


  gradV
 x y z 
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Vector Mechanics for Engineers: Dynamics
Conservation of Energy
• Work of a conservative force,
U12  V1  V2
• Concept of work and energy,
U12  T2  T1
• Follows that
T1  V1  T2  V2
E  T  V  constant
T1  0 V1  W
T1  V1  W
T2  12 mv22 
T2  V2  W
1W
2 g   W V2  0
2g
• When a particle moves under the action of
conservative forces, the total mechanical
energy is constant.
• Friction forces are not conservative. Total
mechanical energy of a system involving
friction decreases.
• Mechanical energy is dissipated by friction
into thermal energy. Total energy is constant.
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Vector Mechanics for Engineers: Dynamics
Motion Under a Conservative Central Force
• When a particle moves under a conservative central
force, both the principle of conservation of angular
momentum
r0 mv0 sin 0  rmv sin 
and the principle of conservation of energy
T0  V0  T  V
1 mv 2
0
2

GMm 1 2 GMm
 2 mv 
r0
r
may be applied.
• Given r, the equations may be solved for v and j.
• At minimum and maximum r, j  90o. Given the
launch conditions, the equations may be solved for
rmin, rmax, vmin, and vmax.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of
energy between positions 1 and 2.
• The elastic and gravitational potential
energies at 1 and 2 are evaluated from
the given information. The initial kinetic
energy is zero.
A 20 lb collar slides without friction
along a vertical rod as shown. The
spring attached to the collar has an
undeflected length of 4 in. and a
constant of 3 lb/in.
• Solve for the kinetic energy and velocity
at 2.
If the collar is released from rest at
position 1, determine its velocity after
it has moved 6 in. to position 2.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: Ve  12 kx12  12 3 lb in.8 in.  4 in.2  24 in.  lb
V1  Ve  Vg  24 in.  lb  0  2 ft  lb
T1  0
Position 2: Ve  12 kx22  12 3 lb in.10 in.  4 in.2  54 in.  lb
Vg  Wy  20 lb  6 in.  120 in.  lb
V2  Ve  Vg  54  120  66 in.  lb  5.5 ft  lb
T2  12 mv22 
1 20 2
v2  0.311v22
2 32.2
Conservation of Energy:
T1  V1  T2  V2
0  2 ft  lb  0.311v22  5.5 ft  lb
v2  4.91ft s 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.7
SOLUTION:
• Since the pellet must remain in contact
with the loop, the force exerted on the
pellet must be greater than or equal to
zero. Setting the force exerted by the
loop to zero, solve for the minimum
velocity at D.
The 0.5 lb pellet is pushed against the
spring and released from rest at A.
Neglecting friction, determine the
smallest deflection of the spring for
which the pellet will travel around the
loop and remain in contact with the
loop at all times.
• Apply the principle of conservation of
energy between points A and D. Solve
for the spring deflection required to
produce the required velocity and
kinetic energy at D.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.7
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
2
   Fn  man : W  man
mg  m vD
r
2
vD
 rg  2 ft 32.2 ft s   64.4 ft 2 s 2
• Apply the principle of conservation of energy between
points A and D.
V1  Ve  Vg  12 kx2  0  12 36 lb ft x 2  18 x 2
T1  0
V2  Ve  Vg  0  Wy  0.5 lb 4 ft   2 ft  lb
2
T2  12 mvD



1 0.5 lb
2 2
64
.
4
ft
s  0.5 ft  lb
2 32.2 ft s 2
T1  V1  T2  V2
0  18 x 2  0.5  2
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
x  0.3727 ft  4.47 in.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
SOLUTION:
• For motion under a conservative central
force, the principles of conservation of
energy and conservation of angular
momentum may be applied simultaneously.
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 36900 km/h from
an altitude of 500 km.
• Apply the principles to the points of
minimum and maximum altitude to
determine the maximum altitude.
• Apply the principles to the orbit insertion
point and the point of minimum altitude to
determine maximum allowable orbit
Determine (a) the maximum altitude
insertion angle error.
reached by the satellite, and (b) the
maximum allowable error in the
direction of launching if the satellite
is to come no closer than 200 km to
the surface of the earth
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
• Apply the principles of conservation of energy and
conservation of angular momentum to the points of minimum
and maximum altitude to determine the maximum altitude.
Conservation of energy:
TA  VA  TA  VA
1 mv 2
0
2

GMm 1 2 GMm
 2 mv1 
r0
r1
Conservation of angular momentum:
r
r0mv0  r1mv1
v1  v0 0
r1
Combining,
2

r0 2GM
1 v 2 1  r0   GM 1  r0 
1




2 0
2
r0  r1 
r1 r0v02
 r1 
r0  6370 km  500 km  6870 km
v0  36900 km h  10.25  106 m s


2
GM  gR 2  9.81m s 2 6.37  106 m  398  1012 m3 s 2
r1  60.4  106 m  60400 km
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
• Apply the principles to the orbit insertion point and the point
of minimum altitude to determine maximum allowable orbit
insertion angle error.
Conservation of energy:
GMm
1 mv 2  GMm  1 mv 2
T0  V0  TA  VA

0
max
2
2
r0
rmin
Conservation of angular momentum:
r
r0mv0 sin 0  rmin mvmax
vmax  v0 sin 0 0
rmin
Combining and solving for sin j0,
sin 0  0.9801
j0  90  11.5
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
allowable error  11.5
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Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• From Newton’s second law,
 d


F  mv 
mv  linear momentum
dt


Fdt  d mv 
t2



F
dt

m
v

m
v

2
1
t1
• Dimensions of the impulse of
a force are
force*time.
• Units for the impulse of a
force are


N  s  kg  m s  s  kg  m s
2
t2


 Fdt  Imp12  impulse of the force F
t1


mv1  Imp12  mv2
• The final momentum of the particle can be
obtained by adding vectorially its initial
momentum and the impulse of the force during
the time interval.
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Vector Mechanics for Engineers: Dynamics
Impulsive Motion
• Force acting on a particle during a very short
time interval that is large enough to cause a
significant change in momentum is called an
impulsive force.
• When impulsive forces act on a particle,



mv1   F t  mv2
• When a baseball is struck by a bat, contact
occurs over a short time interval but force is
large enough to change sense of ball motion.
• Nonimpulsive
forces are forces for which

Ft is small and therefore, may be
neglected.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.10
SOLUTION:
• Apply the principle of impulse and
momentum. The impulse is equal to the
product of the constant forces and the
time interval.
An automobile weighing 4000 lb is
driven down a 5o incline at a speed of
60 mi/h when the brakes are applied,
causing a constant total braking force of
1500 lb.
Determine the time required for the
automobile to come to a stop.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.10
SOLUTION:
• Apply the principle of impulse and
momentum.


mv1   Imp12  mv2
Taking components parallel to the
incline,
mv1  W sin 5t  Ft  0
 4000 

88 ft s   4000sin 5t  1500t  0
 32.2 
t  9.49 s
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.11
SOLUTION:
• Apply the principle of impulse and
momentum in terms of horizontal and
vertical component equations.
A 4 oz baseball is pitched with a
velocity of 80 ft/s. After the ball is hit
by the bat, it has a velocity of 120 ft/s
in the direction shown. If the bat and
ball are in contact for 0.015 s,
determine the average impulsive force
exerted on the ball during the impact.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.11
SOLUTION:
• Apply the principle of impulse and momentum in
terms of horizontal and vertical component equations.


mv1  Imp12  mv2
x component equation:
 mv1  Fx t  mv2 cos 40
4 16
80  Fx 0.15  4 16 120 cos 40
32.2
32.2
Fx  89 lb

y component equation:
y
0  Fy t  mv2 sin 40
x
4 16
120 cos 40
32.2
Fy  39.9 lb



F  89 lb i  39.9 lb  j , F  97.5 lb
Fy 0.15 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
SOLUTION:
A 10 kg package drops from a chute
into a 24 kg cart with a velocity of 3
m/s. Knowing that the cart is initially at
rest and can roll freely, determine (a)
the final velocity of the cart, (b) the
impulse exerted by the cart on the
package, and (c) the fraction of the
initial energy lost in the impact.
• Apply the principle of impulse and
momentum to the package-cart system
to determine the final velocity.
• Apply the same principle to the package
alone to determine the impulse exerted
on it from the change in its momentum.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 47
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
SOLUTION:
• Apply the principle of impulse and momentum to the package-cart
system to determine the final velocity.
y
x




m pv1   Imp12  m p  mc v2
x components:


m p v1 cos 30  0  m p  mc v2
10 kg 3 m/s cos 30  10 kg  25 kg v2
v2  0.742 m/s
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 48
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
• Apply the same principle to the package alone to determine the impulse
exerted on it from the change in its momentum.
y
x


m pv1   Imp12  m pv2
x components:
m p v1 cos 30  Fx t  m p v2
10 kg 3 m/s cos 30  Fx t  10 kg v2
y components:
Fx t  18.56 N  s
 m p v1 sin 30  Fy t  0
 10 kg 3 m/s sin 30  Fy t  0



 Imp12  Ft   18.56 N  si  15 N  s j
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
Fy t  15 N  s
Ft  23.9 N  s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.12
To determine the fraction of energy lost,
T1  12 m p v12  12 10 kg 3 m s 2  45 J


T1  12 m p  mc v22  12 10 kg  25 kg 0.742 m s 2  9.63 J
T1  T2 45 J  9.63 J

 0.786
T1
45 J
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Vector Mechanics for Engineers: Dynamics
Impact
• Impact: Collision between two bodies which
occurs during a small time interval and during
which the bodies exert large forces on each other.
• Line of Impact: Common normal to the surfaces
in contact during impact.
Direct Central Impact
• Central Impact: Impact for which the mass
centers of the two bodies lie on the line of impact;
otherwise, it is an eccentric impact..
• Direct Impact: Impact for which the velocities of
the two bodies are directed along the line of
impact.
• Oblique Impact: Impact for which one or both of
the bodies move along a line other than the line of
impact.
Oblique Central Impact
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Vector Mechanics for Engineers: Dynamics
Direct Central Impact
• Bodies moving in the same straight line,
vA > vB .
• Upon impact the bodies undergo a
period of deformation, at the end of which,
they are in contact and moving at a
common velocity.
• A period of restitution follows during
which the bodies either regain their
original shape or remain permanently
deformed.
• Wish to determine the final velocities of the
two bodies. The total momentum of the
two body system is preserved,
m Av A  mB v B  mB vB  mB vB
• A second relation between the final
velocities is required.
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Vector Mechanics for Engineers: Dynamics
Direct Central Impact
e  coefficient of restitution
• Period of deformation: m Av A   Pdt  m Au

 Rdt  u  vA
 Pdt v A  u
0  e 1
• Period of restitution:
m Au   Rdt  m AvA
• A similar analysis of particle B yields
vB  u
e
u  vB
• Combining the relations leads to the desired
second relation between the final velocities.
vB  vA  ev A  v B 
• Perfectly plastic impact, e = 0: vB  vA  v
m Av A  mB v B  m A  mB v
• Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.
vB  vA  v A  v B
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Vector Mechanics for Engineers: Dynamics
Oblique Central Impact
• Final velocities are
unknown in magnitude
and direction. Four
equations are required.
• No tangential impulse component;
tangential component of momentum
for each particle is conserved.
• Normal component of total
momentum of the two particles is
conserved.
• Normal components of relative
velocities before and after impact
are related by the coefficient of
restitution.
v A t  vA t
v B t  vB t
m A v A n  mB v B n  m A vA n  mB vB n
vB n  vA n  ev A n  v B n 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Oblique Central Impact
• Block constrained to move along horizontal
surface.


• Impulses from internal forces F and  F
along the n axis and from external force Fext
exerted by horizontal surface and directed
along the vertical to the surface.
• Final velocity of ball unknown in direction and
magnitude and unknown final block velocity
magnitude. Three equations required.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Oblique Central Impact
• Tangential momentum of ball is
conserved.
v B t  vB t
• Total horizontal momentum of block
and ball is conserved.
m A v A   mB v B  x  m A vA   mB vB  x
• Normal component of relative
velocities of block and ball are related
by coefficient of restitution.
vB n  vA n  ev A n  v B n 
• Note: Validity of last expression does not follow from previous relation for
the coefficient of restitution. A similar but separate derivation is required.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Problems Involving Energy and Momentum
• Three methods for the analysis of kinetics problems:
- Direct application of Newton’s second law
- Method of work and energy
- Method of impulse and momentum
• Select the method best suited for the problem or part of a problem
under consideration.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.14
SOLUTION:
• Resolve ball velocity into components
normal and tangential to wall.
• Impulse exerted by the wall is normal
to the wall. Component of ball
momentum tangential to wall is
conserved.
A ball is thrown against a frictionless,
vertical wall. Immediately before the
ball strikes the wall, its velocity has a
magnitude v and forms angle of 30o
with the horizontal. Knowing that
e = 0.90, determine the magnitude and
direction of the velocity of the ball as
it rebounds from the wall.
• Assume that the wall has infinite mass
so that wall velocity before and after
impact is zero. Apply coefficient of
restitution relation to find change in
normal relative velocity between wall
and ball, i.e., the normal ball velocity.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.14
SOLUTION:
• Resolve ball velocity into components parallel and
perpendicular to wall.
vn  v cos 30  0.866v
vt  v sin 30  0.500v
• Component of ball momentum tangential to wall is conserved.
vt  vt  0.500v
t
n
• Apply coefficient of restitution relation with zero wall
velocity.
0  vn  evn  0
vn  0.90.866v   0.779v



v   0.779v n  0.500v t
 0.779 
v  0.926v tan 1
  32.7
 0.500 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.15
SOLUTION:
• Resolve the ball velocities into components
normal and tangential to the contact plane.
• Tangential component of momentum for
each ball is conserved.
The magnitude and direction of the
velocities of two identical
frictionless balls before they strike
each other are as shown. Assuming
e = 0.9, determine the magnitude
and direction of the velocity of each
ball after the impact.
• Total normal component of the momentum
of the two ball system is conserved.
• The normal relative velocities of the
balls are related by the coefficient of
restitution.
• Solve the last two equations simultaneously
for the normal velocities of the balls after
the impact.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.15
SOLUTION:
• Resolve the ball velocities into components normal and
tangential to the contact plane.
v A n  v A cos 30  26.0 ft
s
vB n  vB cos 60  20.0 ft
s
v A t  v A sin 30  15.0 ft
s
vB t  vB sin 60  34.6 ft
s
• Tangential component of momentum for each ball is
conserved.
vA t  v A t  15.0 ft
s
vB t  vB t  34.6 ft
s
• Total normal component of the momentum of the two
ball system is conserved.
mA v A n  mB vB n  mA vA n  mB vB n
m26.0  m 20.0  mvA n  mvB n
vA n  vB n  6.0
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.15
• The normal relative velocities of the balls are related by the
coefficient of restitution.
vA n  vB n  ev A n  vB n 
 0.9026.0   20.0  41.4
• Solve the last two equations simultaneously for the normal
velocities of the balls after the impact.
vA n  17.7 ft s
vB n  23.7 ft s



v A  17.7t  15.0n
n
 15.0 
vA  23.2 ft s tan 1
  40.3
 17.7 



vB  23.7t  34.6n
t
 34.6 
vB  41.9 ft s tan 1
  55.6
23
.
7


© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.16
SOLUTION:
• Determine orientation of impact line of
action.
• The momentum component of ball A
tangential to the contact plane is
conserved.
• The total horizontal momentum of the
two ball system is conserved.
Ball B is hanging from an inextensible
• The relative velocities along the line of
cord. An identical ball A is released
action before and after the impact are
from rest when it is just touching the
related by the coefficient of restitution.
cord and acquires a velocity v0 before
striking ball B. Assuming perfectly
• Solve the last two expressions for the
elastic impact (e = 1) and no friction,
velocity of ball A along the line of action
determine the velocity of each ball
and the velocity of ball B which is
immediately after impact.
horizontal.
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13 - 63
Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.16
r
 0.5
2r
  30
sin  
SOLUTION:
• Determine orientation of impact line of action.
• The momentum component of ball A
tangential to the contact plane is
conserved.



mv A  Ft  mv A
mv0 sin 30  0  mvA t
vA t  0.5v0
• The total horizontal (x component)
momentum of the two ball system is
conserved.




mv A  Tt  mv A  mvB
0  mvA t cos 30  mvA n sin 30  mvB
0  0.5v0 cos 30  vA n sin 30  vB
0.5vA n  vB  0.433v0
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.16
• The relative velocities along the line of action before
and after the impact are related by the coefficient of
restitution.
vB n  vA n  ev A n  vB n 
vB sin 30  vA n  v0 cos 30  0
0.5vB  vA n  0.866v0
• Solve the last two expressions for the velocity of ball
A along the line of action and the velocity of ball B
which is horizontal.
vA n  0.520v0
vB  0.693v0



v A  0.5v0t  0.520v0n
vA  0.721v0
  tan 1
0.52 
  46.1
 0.5 
  46.1  30  16.1
vB  0.693v0 
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
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Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.17
SOLUTION:
• Apply the principle of conservation of
energy to determine the velocity of the
block at the instant of impact.
• Since the impact is perfectly plastic, the
block and pan move together at the same
velocity after impact. Determine that
velocity from the requirement that the
total momentum of the block and pan is
conserved.
A 30 kg block is dropped from a height
of 2 m onto the the 10 kg pan of a
• Apply the principle of conservation of
spring scale. Assuming the impact to be
energy to determine the maximum
perfectly plastic, determine the
deflection of the spring.
maximum deflection of the pan. The
constant of the spring is k = 20 kN/m.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 66
Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.17
SOLUTION:
• Apply principle of conservation of energy to
determine velocity of the block at instant of impact.
T1  0
V1  WA y  309.812  588 J
T2  12 m A v A 22  12 30 v A 22
V2  0
T1  V1  T2  V2
0  588 J  12 30 v A 22  0
v A 2  6.26 m s
• Determine velocity after impact from requirement that
total momentum of the block and pan is conserved.
mA v A 2  mB vB 2  mA  mB v3
306.26  0  30  10v3
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
v3  4.70 m s
13 - 67
Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.17
• Apply the principle of conservation of energy to
determine the maximum deflection of the spring.
T3  12 m A  mB v32  12 30  10 4.7 2  442 J
V3  Vg  Ve
0
1 kx 2
2 3

1
2
20  10 4.91 10 
3
3 2
 0.241 J
T4  0
Initial spring deflection due to
pan weight:
x3 
WB 109.81
3


4
.
91

10
m
3
k
20  10
V4  Vg  Ve  WA  WB  h   12 kx42


 392x4  4.91  103   12 20  103 x42
 392 x4  x3   12 20  103 x42
T3  V3  T4  V4

 

442  0.241  0  392 x4  4.91  103  12 20  103 x42
x4  0.230 m
h  x4  x3  0.230 m  4.91  103 m
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
h  0.225 m
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