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## Homework Statement

- 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) 1:sqrt(2)
- 3) 1:1
- 4) sqrt(2):1
- 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!