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r1p4c3

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Before I start the problem, I would like to apologize for not using the formating. I am posting this from my high school's network and they have an absurd filtering algorithm that blocks css and other web scripts so I can't use the normal formatting since I can't remember the format tags for vBulletin.

This problem is from my AP Physics rotation test, and of my peers that I talked to, we can't agree what the answer is.

A ball of uniform density is placed at the top of the left side of a U shaped track. The left half of the track is friction less and the right half has friction. The top of the track is a height h above the bottom of the track. The ball has a mass of m and a radius of r.

What is the maximum height the ball will reach on the left side??

v = r x w

I = 2/5 m*r^2

GPE = mgh

KE = 1/2 m*v^2 + 1/2 I*w^2

Energy is conserved throughout the path of the ball. Since the left half of the track is friction less, so it won't roll but slide.

GPE = KE ==> mgh = 1/2 m*(v0)^2 ==> v0 = sqrt(2gh)

Energy before entering the friction part should be the same after entering it.

1/2 m*(v0)^2 = 1/2 m*v^2 + 1/2 I*w^2 ==> 1/2 m*(v0)^2 = 7/5 m*v^2 ==> v = (v0)*sqrt(5/7) = sqr (10gh/7)

Finally, energy is conserved while it rolls up the right side

1/2 m*v^2 + 1/2 I*w^2 = mgh

7/10 m*v^2 = mgh

7/10 v^2 = gh

h = 7/(5g) v^2 = 7/5g * 10gh/7 = h

The math says the final height will be the same as the initial height, but that answer seems too simple for this question.

Edit: I pressed submit too early. Is the math and assumptions I posted correct? ?

This problem is from my AP Physics rotation test, and of my peers that I talked to, we can't agree what the answer is.

## Homework Statement

A ball of uniform density is placed at the top of the left side of a U shaped track. The left half of the track is friction less and the right half has friction. The top of the track is a height h above the bottom of the track. The ball has a mass of m and a radius of r.

What is the maximum height the ball will reach on the left side??

## Homework Equations

v = r x w

I = 2/5 m*r^2

GPE = mgh

KE = 1/2 m*v^2 + 1/2 I*w^2

## The Attempt at a Solution

Energy is conserved throughout the path of the ball. Since the left half of the track is friction less, so it won't roll but slide.

GPE = KE ==> mgh = 1/2 m*(v0)^2 ==> v0 = sqrt(2gh)

Energy before entering the friction part should be the same after entering it.

1/2 m*(v0)^2 = 1/2 m*v^2 + 1/2 I*w^2 ==> 1/2 m*(v0)^2 = 7/5 m*v^2 ==> v = (v0)*sqrt(5/7) = sqr (10gh/7)

Finally, energy is conserved while it rolls up the right side

1/2 m*v^2 + 1/2 I*w^2 = mgh

7/10 m*v^2 = mgh

7/10 v^2 = gh

h = 7/(5g) v^2 = 7/5g * 10gh/7 = h

The math says the final height will be the same as the initial height, but that answer seems too simple for this question.

Edit: I pressed submit too early. Is the math and assumptions I posted correct? ?

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