978-0073380308 Chapter 8 Solution Manual Part 11

Using the (full precision) values of

IG

and

IQ

in Eqs. (3), and recalling that

mSD4:2 kg

,

mAB D7kg

,

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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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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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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 ftlb=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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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

MDˇ✓3

, where

ˇD0:5⇥106ftlb=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