Post on 06-Jun-2018
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]
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 +=
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
Chapter 11 Multiple Choice
Chapter 11 Problems