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Computation. Substituting Eqs. (2)–(4) and the last three of Eqs. (5) in Eq. (1), we have
G2
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 1739
Problem 8.52
In a contraption built by a fraternity, a person sits at the center of a
swinging platform with weight
WpD800 lb
and length
LD12 ft
suspended by two identical arms each of length
HD10 ft
and weight
WaD200 lb
. The platform, which is at rest when
✓D0
, is put
in motion by a motor that pumps the ride by exerting a constant
moment
M
in the direction shown, during each upswing, whenever
0✓✓p, while exerting zero moment otherwise.
Neglecting the mass of the person, neglecting friction, letting
MD900 ftlb
, and letting
✓pD25ı
, find the minimum number of
swings necessary to achieve
✓> 90ı
and the ensuing speed achieved
by the person at the lowest point in the swing. Model the arms
AB
and CD as uniform thin bars.
Solution
Referring to the
FBD
shown, we model the arms and the platform
Kinematic Equations.
The system starts from rest and, to determine the minimum number of cycles,
¡
is
to be achieved with the system coming to a temporary stop. Hence,
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 1741
Problem 8.53
In a contraption built by a fraternity, a person sits at the center of a
swinging platform with weight
WpD800 lb
and length
LD12 ft
suspended by two identical arms each of length
HD10 ft
and weight
WaD200 lb
. The platform, which is at rest when
✓D0
, is put
in motion by a motor that pumps the ride by exerting a constant
moment
M
in the direction shown, during each upswing, whenever
0✓✓p, while exerting zero moment otherwise.
Neglecting the mass of the person, neglecting friction, and letting
✓pD20ı
, determine the value of
M
required to achieve a maximum
value of
✓
equal to
90ı
in 6 full swings. Model the arms
AB
and
CD
as uniform thin bars.
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.
Computation. Substituting Eqs. (2)–(6) into Eq. (1), 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.
Dynamics 2e 1743
Problem 8.54
The disk
D
, which has weight
W
, mass center
G
coinciding with the
disk’s geometric center, and radius of gyration
kG
, is at rest on an incline
when the constant moment
M
is applied to it. The disk is attached at
its center to a wall by a linear elastic spring of constant
k
. The spring
is unstretched when the system is at rest. Assuming that the disk rolls
without slipping and that it has not yet come to a stop, determine an
expression for the angular velocity of the disk after its center
G
has
moved a distance
d
down the incline. After doing so, using
kD5lb=ft
,
RD1:5 ft
,
WD10 lb
, and
✓D30ı
, determine the value of the
moment Mfor the disk to stop after rolling dsD5ft down the incline.
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.
In going from ¿to ¡, the disk rolls counterclockwise, so the angular velocity of the disk is
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 1745
Problem 8.55
In Example 8.2 on p. 592, we ignored the rotational inertia of the counterweight. Let’s re-
visit that example and remove that simplifying assumption. Assume that the arm
AD
is
still a uniform thin bar of length
LD15:7 ft
and weight
45 lb
. The hinge
O
is still
dD2:58 ft
from the right end of the arm, and the
160 lb
counterweight
C
is still
ıD1:4 ft
from
the hinge. Now model the counterweight as a uniform block of height
hD14 in:
and width
wD9in:
With this new assumption, solve for the angular velocity of the arm as it reaches the horizontal position
after being nudged from the vertical position. Determine the percent change in angular velocity compared
with that found in Example 8.2.
Solution
Balance Principles.
Applying the work-energy principle as a
statement of conservation of energy, we have
Kinematic Equations. The system is released from rest, so
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 1747
Problem 8.56
For the barrier gate shown, assume that the arm consists of a section of aluminum tubing from
A
to
B
of
length
lD11:6 ft
and weight
20 lb
and a steel support section from
B
to
D
of weight
40 lb
. The overall
length of the arm is
LD15:7 ft
. In addition, the
120 lb
counterweight
C
is placed a distance
ı
from the
hinge at
O;
and the hinge is
dD2:58 ft
from the right end of section BD. Model the two sections
AB
and
BD as uniform thin bars, and model the counterweight as a uniform block of height
hD14 in:
and width
wD9in:
Using these new assumptions, determine the distance
ı
so that the angular velocity of the arm is
0:25 rad=s as it reaches the horizontal position after being nudged from the vertical position.
Solution
We model the arm as a thin rigid body of mass
mAB DWAB =g
,
