978-0073380308 Chapter 8 Solution Manual Part 18

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

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.rori/=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

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

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Dynamics 2e 1831

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permission of McGraw-Hill, is prohibited.

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.

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permission of McGraw-Hill, is prohibited.

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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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

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permission of McGraw-Hill, is prohibited.

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

2mDr.vAy /2:(4)

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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?

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Dynamics 2e 1837

Problem 8.107

A toy helicopter consists of a rotor

A

with diameter

dD10 in:

and weight

WrD0:09⇥103oz

, a thin

body

B

of length

`D12 in:

and weight

WBD0:144⇥103oz

, and a small ballast

C

placed at the

front end of the body and with weight

WCD0:0723⇥103oz

. 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.

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