A hollow spherical shell, our model for a ball, rolls down into a valley from a
ID: 1521097 • Letter: A
Question
A hollow spherical shell, our model for a ball, rolls down into a valley from a height H_0. The figure below illustrates that one side of the valley is rough whereas the other side is smooth. The difference has to do with whether or not the ball can slip, 0n the rough side the friction docs not allow the ball to slip, while 0n the smooth side there is no friction. In terms of H_0, how high does the ball move up the smooth side? Explain why the ball doesn't return to a height H_0. Has it lost any of it's original potential energy?Explanation / Answer
When the ball starts from rest at the top of the rough slope, it has gravitational potential energy of mgH and zero kinetic energy. When it reaches the bottom, all of its GPE has been converted to kinetic energy (both translational and rotational, mv²/2 and I²/2).
From the law of conservation of mechanical energy:
mgH = mv²/2 + I²/2
2mgH = mv² + (2mr²/5)(v/r)²
10gH = 5v² + 2v²
10gH = 7v²
v = [10gH/7]
With this speed, the ball will begin to travel up the smooth slope and it will come to a stop on the slope when all its KE has been converted to GPE, so:
0.5mv² = mgh
0.5m([10gH/7])² = mgh
10gH/7 = 2gh
h = 5H/7
b)The ball doesnt return to a height Ho because of the frictional forces present in the rough surface which result in loss of mechanical energy in the form of heat.Thus some of the energy is lost and the ball doesnt move upto its original height.
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