HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9. MC1 · HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential...

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HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of point’s 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 2π 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” θ that an object with angular acceleration α rotates through at time t is 2 2 1 t α θ = (1) (assuming that objects starts with zero angular velocity at angle of zero) [This is analogous to 2 2 1 at x = for linear motion.] We know that t = 2T from the problem. All we need now is α. Have T initial final ω ω α - = The change in angular acceleration over the first revolution (2) 2 initial final average ω ω ω + = The definition of average angular velocity (3) We are told that 0 = initial ω . We are told that T revolution average 1 = ω . Inserting this into (3) gives T s revolution final 2 = ω . Inserting this into (2) gives 2 2 T s revolution = α Inserting this into (1) gives 2 2 2 1 ) 2 ( 2 T T s revolution = θ = 4 revolutions. [C]

Transcript of HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9. MC1 · HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential...

Page 1: HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9. MC1 · HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of point’s acceleration

HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of point’s 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 2π 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”

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

2

21 tαθ = (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 α. Have

T

initialfinal ωωα

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

2

initialfinal

average

ωωω

+= The definition of average angular velocity (3)

We are told that 0=initial

ω .

We are told that T

revolutionaverage

1=ω .

Inserting this into (3) gives T

srevolutionfinal

2=ω .

Inserting this into (2) gives 2

2

T

srevolution=α

Inserting this into (1) gives 2

22

1 )2(2

TT

srevolution

=θ = 4 revolutions. [C]

Page 2: HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9. MC1 · HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of point’s acceleration

9. MC10

From the definition of the moment of inertial I

2226)2()2( mddmdmI =+=

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

2229)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 β= where 5

2=β for the solid sphere, and

3

2=β 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)

2

2

12

2

1 ωIMvMgh +=

Page 3: HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9. MC1 · HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of point’s acceleration

For rotational motion have ωRv = . Also, 2MRI β=

So

( )2

2

212

21

+=

R

vMRMvMgh β

2

2

1 )1( Mvβ+=

Solve for v

β+

=1

2ghv

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

2=β ) is moving more slowly than the solid

sphere (5

2=β ) at the bottom of the ramp.

Answer is A. The solid sphere is faster.

CHAPTER 10

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Chapter 11 Multiple Choice

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Chapter 11 Problems

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