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 cant remember the format tags for vBulletin.
This problem is from my AP Physics rotation test, and of my peers that I talked to, we cant agree what the answer is.
1. The problem statement, all variables and given/known data
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??
2. Relevant equations
v = r x w
I = 2/5 m*r^2
GPE = mgh
KE = 1/2 m*v^2 + 1/2 I*w^2
3. 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 wont 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.
This problem is from my AP Physics rotation test, and of my peers that I talked to, we cant agree what the answer is.
1. The problem statement, all variables and given/known data
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??
2. Relevant equations
v = r x w
I = 2/5 m*r^2
GPE = mgh
KE = 1/2 m*v^2 + 1/2 I*w^2
3. 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 wont 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.
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