978-0073380308 Chapter 8 Solution Manual Part 18

subject Type Homework Help
subject Pages 9
subject Words 3978
subject Authors Francesco Costanzo, Gary Gray, Michael Plesha

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1828 Solutions Manual
!DriO|
, respectively. Since the cord is inextensible and the cord does not slip relative
to the pulleys, we must have
EvBDEvQD!DroO|
and
EvEDEvHD!DriO|
. We now
observe that
EvADvCy O|
because pulley
P
moves with the same velocity as that of
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Dynamics 2e 1829
Problem 8.102
The double pulley
D
has mass of
mDD15 kg
, center of mass
G
coinciding with
its geometric center, radius of gyration
kGD10 cm
, outer radius
roD15 cm
, and
inner radius
riD7:5 cm
. It is connected to the pulley
P
with radius
RD.rori/=2
by a cord of negligible mass that unwinds from the inner and outer spools of the
double pulley
D
. The crate
C
, which has a mass
mCD20 kg
, is released from
rest. The cord does not slip relative to the pulleys, and the inner and outer pulleys
rotate as a single unit.
Assuming that the pulley
P
has a mass of
1:5 kg
and a radius of gyration
kAD3:5 cm, use the impulse-momentum principles to determine the speed of the
crate 4s after release.
Solution
We model the crate as a particle subject to its own weight
mCg
impulse-momentum principle in the vertical direction. Since pul-
ley
D
is in a fixed axis rotation about
G
and we are not interested
in the reaction forces
Gx
and
Gy
, in this case we will only apply the angular impulse momentum principle in
the
´
direction using
G
as moment center. In the case of pulley
P
, which does not move in the horizontal
direction, we will apply the linear impulse-momentum principle in the vertical direction as well as the angular
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where
vCy
is the
y
component of the velocity of
C
,
IG
is the mass moment of inertia of
D
about
G
,
E!DD!DO
k
is the angular velocity of
D
,
vAy
is the
y
component of the velocity of
A
,
IA
is the mass moment
of inertia of Prelative to A, and E!PD!PO
kis the angular velocity of P. For IGand IA, we have
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
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Dynamics 2e 1831
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pf5
Problem 8.103
Some pipe sections of radius
r
and mass
m
are being unloaded and placed in a row against a wall. The
first of these pipe sections,
A
, is made to roll without slipping into a corner with an angular velocity
!0
as
shown. Upon touching the wall,
A
does not rebound, but slips against the ground and against the wall.
Modeling
A
as a uniform thin ring with center at
G
and letting
g
and
w
be the coefficients of kinetic
friction of the contacts between
A
and the ground and between
A
and the wall, respectively, determine an
expression for the angular velocity of
A
as a function of time from the moment
A
touches the wall until it
stops. Hint: Using the methods learned in Chapter 7, we can show that the friction forces at the ground
and at the wall are constant.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
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Dynamics 2e 1833
Computation.
Solving Eqs. (4) for
Ng
and
Nw
, and substituting the result into Eqs. (3) we have a system
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permission of McGraw-Hill, is prohibited.
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Problem 8.104
A
14 lb
bowling ball is thrown onto a lane with a backspin
!0D9rad=s
and
forward velocity
v0D18 mph
. Point
G
is both the geometric center and the
mass center of the ball. After a few seconds, the ball starts rolling without slip.
Let
rD4:25 in:
, and let the radius of gyration of the ball be
kGD2:6 in
. If
the coefficient of kinetic friction between the ball and the floor is
kD0:1
,
determine the speed v
fthat the ball will achieve when it starts rolling without
slip. In addition, determine the time
tr
the ball takes to achieve
v
f:
Hint: Using
the methods of Chapter 7, we can show that the force between the ball and the
floor is constant.
Solution
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
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Dynamics 2e 1835
Problem 8.105
A crate
A
with weight
WAD250 lb
is hanging from a rope wound around a uniform
drum
D
of radius
rD1:2 ft
, weight
WDD125 lb
, and center
C
. The system is
initially at rest when the restraining system holding the drum stationary fails, thus
causing the drum to rotate, the rope to unwind, and, consequently, the crate to fall.
Assuming that the rope does not stretch or slip relative to the drum and neglecting
the inertia of the rope, determine the speed of the crate
1:5
s after the system starts to
move.
Solution
0
0
2mDr.vAy /2:(4)
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pf9
Problem 8.106
A toy helicopter consists of a rotor
A
, a body
B
, and a small ballast
C
. The axis of rotation of the rotor goes through
G
, which is the
center of mass of the body
B
and ballast
C
. While holding the
body (and ballast) fixed, the rotor is spun as shown with a given
angular velocity
!0
. If there is no friction between the helicopter’s
body and the rotor’s shaft, will the body of the helicopter start
spinning once the toy is released?
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
page-pfa
Dynamics 2e 1837
Problem 8.107
A toy helicopter consists of a rotor
A
with diameter
dD10 in:
and weight
WrD0:09103oz
, a thin
body
B
of length
`D12 in:
and weight
WBD0:144103oz
, and a small ballast
C
placed at the
front end of the body and with weight
WCD0:0723103oz
. The ballast’s weight is such that the axis
of the rotation of the rotor goes through
G
, which is the center of mass of the body
B
and ballast
C
.
While holding the body (and ballast) fixed, the rotor is spun as shown with
!0D150 rpm
. Neglecting
aerodynamic effects, the weights of the rotor’s shaft and the body’s tail, and assuming there is friction
between the helicopter’s body and the rotor’s shaft, determine the angular velocity of the body once the toy
is released and the angular velocity of the rotor decreases to
120 rpm
. Model the body as a uniform thin
rod and the ballast as a particle. Assume that the rotor and the body remain horizontal after release.
Solution
Note: we model the rotor as uniform thin bar of length d.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.

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