2012 F = MA exam #10 - (Rolling without slipping down a ramp)

[SignaturePF]

Homework Statement

Four objects are placed at rest at the top of an inclined plane and allowed to roll without slipping to the bottom
in the absence of rolling resistance and air resistance.
• Object A is a solid brass ball of diameter d.
• Object B is a solid brass ball of diameter 2d.
• Object C is a hollow brass sphere of diameter d.
• Object D is a solid aluminum ball of diameter d. (Aluminum is less dense than brass.)
The balls are placed so that their centers of mass all travel the same distance. In each case, the time of motion T
is measured. Which of the following statements is correct?

(A) TB > TC > TA = TD
(B) TA = TB = TC > TD
(C) TB > TA = TC = TD
(D) TC > TA = TB = TD← CORRECT
(E) TA = TB = TC = TD

Homework Equations

T = 2pir/ v
K(rotation and translation) = 1/2mv^2 + 1/2Iw^2

The Attempt at a Solution

No idea, for some reason I tried to find the total kinetic energy of each object, didn't really work though.

 

[haruspex]

It's all to with the proportion of PE that goes into rotational KE instead of into linear KE. The higher the proportion, the lower the linear speed, so the longer the time taken. Try to turn that into equations.

 

[SignaturePF]

So what you are saying is that in the beginning of the roll down the ramp, there is X potential energy that is converted to both linear and rotational kinetic energy. The more PE that goes to rotational KE takes away from linear KE and thus reduces linear speed. This means that it will take longer, so T will be higher. I'm not sure how to convert this to equations.
Here's an attempt using Cons. of Energy:
K_0 + U_0 = K + U
0 + mgh = 1/2Iw^2 + 1/2mv^2
I = mR^2 / 2
w = v / R
mgh = 1/4mv^2 + 1/2mv^2
mgh = 3/4 mv^2
gh = 3/4v^2
4/3gh = v^2
- I must be doing it wrong here because my calculations imply that linear velocity is independent of the radii.
Where did I go wrong?

 

[haruspex]

4/3gh = v^2
- I must be doing it wrong here because my calculations imply that linear velocity is independent of the radii.

No, that's the correct result. You assumed I = λmR² where λ = 1/2. That will lead to a fraction of energy going into linear KE that depends only on λ. Since the total is mgh, and the m's will cancel, you get a linear velocity √(2gh f(λ)).

 

[SignaturePF]

Ahh, I think I understand. So with the result I showed earlier, velocity is independent of entire mass and radius. Therefore, T_A = T_B = T_D. But since ball C will have the most mass concentrated furthest away, it has the highest rotational KE and thus it will have the slowest velocity and the highest period. These two lead us to:
T_C > T_A = T_B = T_D

 

[haruspex]

Ahh, I think I understand. So with the result I showed earlier, velocity is independent of entire mass and radius. Therefore, T_A = T_B = T_D. But since ball C will have the most mass concentrated furthest away, it has the highest rotational KE and thus it will have the slowest velocity and the highest period. These two lead us to:
T_C > T_A = T_B = T_D

Exactly.

 

[SignaturePF]

Thanks, I really appreciate it!