Transcript chapt17_lecture_updated

Seventh Edition
CHAPTER
17
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Plane Motion of Rigid Bodies:
Energy and Momentum Methods
Lecture Notes:
J. Walt Oler
Texas Tech University
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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
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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
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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.
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Vector Mechanics for Engineers: Dynamics
Principle of Work and Energy for a Rigid Body
• Method of work and energy is well adapted to
problems involving velocities and displacements.
Main advantage is that the 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.
• In general, small displacements of the particles A
and B are not equal but the components of the
displacements along AB are equal.
• Therefore, the net work of internal forces is zero.
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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.
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Vector Mechanics for Engineers: Dynamics
Work of Forces Acting on a Rigid Body
Forces acting on rigid bodies which do no work:
• Forces applied to fixed points:
- reactions at a frictionless pin when the supported body
rotates about the pin.
• Forces acting in a direction perpendicular to the displacement
of their point of application:
- reaction at a frictionless surface to a body moving along
the surface
- weight of a body when its center of gravity moves
horizontally
• Friction force at the point of contact of a body rolling without
sliding on a fixed surface.
dU  F dsC  F vc dt   0
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Vector Mechanics for Engineers: Dynamics
Kinetic Energy of a Rigid Body in Plane Motion
• Consider a rigid body of mass m in plane motion.
T  12 mv 2  12  Δmi vi 2
 12 mv 2  12
 ri2Δmi  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.
• Consider a rigid body rotating about a fixed axis
through O.


T  12  Δmi vi2  12  Δmi ri 2  12  ri2 Δmi  2
 12 I O 2
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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.
• 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
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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

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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
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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.
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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 
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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
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v2  12.01ft s
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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.
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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
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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
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F
11.52
 46.2 N
0.250
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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.
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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
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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
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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.
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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
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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
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
R  9.22
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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.
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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

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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  0.375 m


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
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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
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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
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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,



L   vi Δmi  mv
and a couple equal to the sum of their moments about the mass center,

 
H G   ri  vi Δmi
• For the plane motion of a rigid slab or of a rigid body symmetrical with
respect to the reference plane,

H G  I
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Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• Principle of impulse and momentum for the plane motion of a rigid slab
or of a rigid body symmetrical with respect to the reference plane
expressed as a free-body-diagram equation,
• 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
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Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
• Noncentroidal 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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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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
do not generate nonzero net impulses.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 35
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
2 v1
7 k g
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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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 37
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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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 38
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
2  2.08 rad s
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Seventh
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
T  3.24 ft  lb
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17 - 40
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 41
Seventh
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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 42
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 43
Seventh
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  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  t  0
Ay  0
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 44
Seventh
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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 45
Seventh
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 
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 46
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 47
Seventh
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.
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 48
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
17 - 49
Seventh
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
© 2003 The McGraw-Hill Companies, Inc. All rights reserved.
0.285 g a
v1  0.712 ga
17 - 50