978-0073380308 Chapter 8 Solution Manual Part 9

Computation. Substituting Eqs. (2)–(4) and the last three of Eqs. (5) in Eq. (1), we have

G2

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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 ftlb

, 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,

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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

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permission of McGraw-Hill, is prohibited.

Computation. Substituting Eqs. (2)–(6) into Eq. (1), we have

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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

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permission of McGraw-Hill, is prohibited.

In going from ¿to ¡, the disk rolls counterclockwise, so the angular velocity of the disk is

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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

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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

,