Ratio of angular speed with conservation of energy

[RoboNerd]

Homework Statement

1. A ball rolls down an incline plane without slipping. What is the ratio of its angular velocity at h/3 to its angular velocity at 2h/3?
   1) 1:2
2. 2) 1:sqrt(2)
3. 3) 1:1
4. 4) sqrt(2):1
5. 5) 2:1

Homework Equations

Conservation of energy with provisions for rotational and translational motion.

The Attempt at a Solution

Hi everyone!

I got this funky problem today. Here's what I tried to do:

at h/3 I have potential energy resulting from lowering the ball a height of 2*h/3 that is converted to kinetic energy.

Thus, I did the following
m * g * (2h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * I * omega^2]
m * g * (2h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * (2 / 5) * m * r^2 * (v^2 / r^2)]

Rearrange to get an expression with v, and then substitute v = r * omega to get
(20 * h * g)/ (21 * r^2) = omega^2 for the h/3 height

OK. then I said that at height 2h/3 I have potential energy from descending h/3 converted to kinetic energy
m * g * (h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * I * omega^2]
m * g * (h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * (2 / 5) * m * r^2 * (v^2 / r^2)]

Rearrange to get an expression with v and then substitute v = r * omega to get
(10 * g * h) / (21 * r^2 ) = omega squared.

Thus, my ratio would have been for the two omega squared terms to be:
(20 * h * g)/ (21 * r^2) = omega^2 for the h/3 height : (10 * g * h) / (21 * r^2 ) = omega squared. for the 2h/3 height

Simplifying the ratio for the omega squared gives 20 : 10 -----> 2:1.
Taking the square root to compare just for the omegas gives sqrt(2) : 1.

Thus, I got 4. The answer according to the writers is 2.

Why is this the case?

Thanks a lot for the help in advance!

 

[haruspex]

It's not entirely clear what "at h/3" means. Is that h/3 from the bottom or from the top? Is there any further information that clarifies this?

 

[RoboNerd]

They did not provide any information. I believe that 'h' corresponds to the height at which the ball is initially launched on the ramp.

 

[haruspex]

They did not provide any information. I believe that 'h' corresponds to the height at which the ball is initially launched on the ramp.

Sure, but that still doesn't nail down the meaning of "at h/3". It could still be measured from top or from bottom. Measured from bottom, your answer is correct.

 

[RoboNerd]

I have no idea... these authors write really crummy questions, which does not help as I have a final on monday. :-(

 

[haruspex]

I have no idea... these authors write really crummy questions, which does not help as I have a final on monday. :-(

So you quoted the question exactly as given to you, word for word, and there is no diagram?

 

[RoboNerd]

Yes, sir!

 

[RoboNerd]

Yes, sir! I just assumed they had a ramp at an arbitrary angle, with it vertically ascending "h".

 

[haruspex]

Yes, sir! I just assumed they had a ramp at an arbitrary angle, with it vertically ascending "h".

Well, either they were thinking of h/3 etc. as descent, i.e. measured from the top, or they messed up and gave the wrong answer.

 

[RoboNerd]

OK, thanks so much for helping and confirming what I did. Really appreciated!