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Using the (full precision) values of
IG
and
IQ
in Eqs. (3), and recalling that
mSD4:2 kg
,
mAB D7kg
,
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.
Dynamics 2e 1759
Problem 8.62
Revisit Example 8.5 on p. 598 and replace the two springs with a
system of two counterweights
P
(only one counterweight is shown)
each of weight
WP
. Recalling that the door’s weight is
WD
800 lb
and that the total height of the door is
HD30 ft
, if
the door is released from rest in the fully open position and friction
is negligible, determine the minimum value of
WP
so that
A
will strike
the left end of the horizontal guide with a speed no greater than 0:5 ft=s.
Solution
We modify the modeling assumptions in Example 8.5 simply by saying that the counterweights are modeled
Balance Principles.
Applying the work-energy principle as a statement of conservation of energy, we have
T1CV1DT2CV2;(1)
Keeping in mind that
W
is the weight of the entire door and that the door consists of two identical parts, for
the masses and the mass moments of inertia, we have
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1761
Problem 8.63
The double pulley
D
has mass of
15 kg
, center of mass
G
coinciding with its geomet-
ric 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
R
by a cord of negligible
mass that unwinds without slip from the inner and outer spools of the double pulley
D
. The crate
C
, which has a mass of
20 kg
, is released from rest, and the inner and
outer parts of the double pulley rotate together as a single unit.
Neglecting the mass of the pulley
P
, determine the speed of the crate
C
and
the angular velocity of the pulley
D
after the crate has dropped a distance
hD2
m.
Solution
Kinematic Equations.
Denoting by
E
and
F
the points on
D
from which the cord separates, because
G
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1763
Problem 8.64
The double pulley
D
has mass of
15 kg
, center of mass
G
coinciding with its geomet-
ric 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
R
by a cord of negligible
mass that unwinds without slip from the inner and outer spools of the double pulley
D
. The crate
C
, which has a mass of
20 kg
, is released from rest, and the inner and
outer parts of the double pulley rotate together as a single unit.
Assuming that the pulley
P
has a mass of
1:5 kg
and a radius of gyration
kAD3:5 cm
, determine the speed of the crate
C
and the angular velocity of the
pulley Dafter the crate has dropped a distance hD2m.
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permission of McGraw-Hill, is prohibited.
Force Laws.
For the potential energy, observing that
A
and
C
are at a constant distance from one another,
we have
V1D.mCCmP/gyC1 and V2D.mCCmP/g.yC2 h/: (4)
Kinematic Equations.
Denoting by
E
and
F
the points on
D
from which the cord separates, because
G
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1765
Problem 8.65
Torsional springs provide a simple propulsion mechanism for toy cars. When the rear wheels are rotated as
if the car were moving backward, they cause a torsional spring (with one end attached to the axle and the
other to the body of the car) to wind up and store energy. Therefore, a simple way to charge the spring is
to place the car onto a surface and to pull it backward, making sure that the wheels roll without slipping.
Note that the torsional spring can only be wound by pulling the car backward; that is, the forward motion
of the car unwinds the spring.
Let the weight of the car (body and wheels) be
WD5oz
, the weight of each
of the wheels be
WwD0:15 oz
, and the radius of the wheels be
rD0:25 in:
,
where the wheels roll without slip and can be treated as uniform disks. Neglecting
friction internal to the car and letting the car’s torsional spring be linear with constant
ktD0:0002 ftlb=rad
, determine the maximum speed achieved by the car if it is released from rest
after pulling it back a distance LD0:75 ft from a position in which the spring is unwound.
Solution
Force Laws.
Letting
✓1
describe the initial wounding angle, i.e., the angle by which the wheels rotate
when the car is initially pulled backwards by the distance
L
, and recalling that the car’s torsional spring is
completely unwound at position two, we have
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1767
Problem 8.66
Torsional springs provide a simple propulsion mechanism for toy cars. When the rear wheels are rotated as
if the car were moving backward, they cause a torsional spring (with one end attached to the axle and the
other to the body of the car) to wind up and store energy. Therefore, a simple way to charge the spring is
to place the car onto a surface and to pull it backward, making sure that the wheels roll without slipping.
Note that the torsional spring can only be wound by pulling the car backward; that is, the forward motion
of the car unwinds the spring.
Let the weight of the car (body and wheels) be
WD5oz
, the weight of each of the wheels be
WwD0:15 oz
, and the radius of the wheels be
rD0:25 in:
, where the wheels roll without slip and can
be treated as uniform disks. In addition, let the torque
M
provided by the nonlinear torsional spring be
given by
MDˇ✓3
, where
ˇD0:5⇥106ftlb=rad3
,
✓
is the angular displacement of the rear axle, and
the minus sign in front of
ˇ
indicates that
M
acts opposite to the direction of
✓
. Neglecting any friction
internal to the car, determine the maximum speed achieved by the car if it is released from rest after pulling
it back a distance LD0:75 ft from a position in which the spring is unwound.
Solution
