# HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9.· HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration

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HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of points acceleration vector that is perpendicular to the radial component. Unlike the radial acceleration, the tangential acceleration is zero for a uniformly rotating object (i.e one whose angular velocity is constant). 9. MC1

B. One turn is 2 radians = 2 x 3.14 = 6.28 6. 9. MC2 Angular motion is mathematically similar to linear motion. In particular have for the angular distance

that an object with angular acceleration rotates through at time t is

2

21 t = (1)

(assuming that objects starts with zero angular velocity at angle of zero) [This is analogous to 2

2

1 atx = for linear motion.]

We know that t = 2T from the problem. All we need now is . Have

T

initialfinal

= The change in angular acceleration over the first revolution (2)

2

initialfinal

average

+= The definition of average angular velocity (3)

We are told that 0=initial

.

We are told that T

revolutionaverage

1= .

Inserting this into (3) gives T

srevolutionfinal

2= .

Inserting this into (2) gives 2

2

T

srevolution=

Inserting this into (1) gives 222

1 )2(2

TT

srevolution

= = 4 revolutions. [C]

9. MC10

From the definition of the moment of inertial I

222 6)2()2( mddmdmI =+=

After the masses switch positions, the new moment of inertial I is

222 9)2)(2(' mdmddmI =+=

Solve first equation for d: m

Id

6= .

Insert this into the second equation:

Im

ImI

2

3

69'

2

=

= [C]

9. MC11

The general formula for moment of inertial I is

2

i

i

irmI = .

If the ri s double, I increases by a factor of 4. If the mi s double, I increase by a factor of 2. 4 x 2 = 8.

[C] 9. MC15

Let 2MRI = where 5

2= for the solid sphere, and

3

2= for the hollow sphere.

Initial potential energy completely converted to kinetic energy. Have energy conservation initial potential energy = (final kinetic energy of linear motion) + (final kinetic energy of rotational motion)

22

12

2

1 IMvMgh +=

For rotational motion have Rv = . Also, 2MRI =

So

( )2

2

212

21

+=

R

vMRMvMgh

22

1 )1( Mv+=

Solve for v

+

=1

2ghv

The larger is , the smaller is v. So the hollow sphere (3

2= ) is moving more slowly than the solid

sphere (5

2= ) at the bottom of the ramp.

Answer is A. The solid sphere is faster. CHAPTER 10

Chapter 11 Multiple Choice

Chapter 11 Problems

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