Rolling Ball Height In Terms of Ho

[tater08]

Homework Statement

A basketball (which can be closely modeled as a hollow spherical shell) rolls down a mountainside into a valley and then up the opposite side, starting from rest at a height H0 above the bottom. In the figure, the rough part of the terrain prevents slipping while the smooth part has no friction.

How high, in terms of H0, will it go up the other side?

Homework Equations

THe KE of a thin hollow sphere is (2/3 mr^2)v^2 but i do not need it here

The Attempt at a Solution

i guess i just need a push into the right direction. I assuming that the terms of the ball reaching the height on the other side will have to do something of its circumference.

 

[Doc Al]

What's different about the ball's trip down the rough side compared to its trip up the smooth side?

Hint: What's the translational KE of the ball when it reaches the bottom?

 

[tater08]

the translational KE of the ball when it reaches the bottom is 0.5 mv^2 but the trip down and the trip up will have a difference due to gravity

 

[Doc Al]

the translational KE of the ball when it reaches the bottom is 0.5 mv^2

Sure, but what is it in term of the original height?

but the trip down and the trip up will have a difference due to gravity

Gravity applies in the same way for both sides. What's different is that one side is rough, the other smooth. How does that affect the motion?

 

[tater08]

well on the rough decline, the rotational energy is being turned into the translational energy which can be converted to potential gravitational energy while up the smooth incline, the only energy would be the rotational.

 

[Doc Al]

well on the rough decline, the rotational energy is being turned into the translational energy which can be converted to potential gravitational energy while up the smooth incline, the only energy would be the rotational.

Not exactly. On the rough decline, the gravitational PE is being transformed into both translational and rotational KE. What happens to that energy on the way up the smooth side?

The key difference between the rough and smooth sides is that only a rough side (with friction exerting a torque) can change the rotational energy of the ball. What does that imply?

 

[tater08]

If there is no friction, the disk going up the frictionless ramp will continue rotating at the same angular velocity the entire time it's on the ramp. Therefore when it gets to its maximum height on the ramp, the disk will still have a significant amount of its energy stored in the rotation of the disk, leaving less energy to be transferred into gravitational energy. For the frictional ramp, the friction is causing a torque on the disk. While this torque doesn't cause the speed of the disk to change, it does affect the angular velocity of the disk. This decrease in angular velocity (and decrease in rotational kinetic energy) causes the disk to roll further up the ramp.

mgh = 1/2*mv^2 + 1/2*(2/5*mR^2)*(v/R)^2 but because there is no translational energy going up the hill mgh = 7/10*mv^2 making h=(7/10 mv^2) /g

 

[Doc Al]

If there is no friction, the disk going up the frictionless ramp will continue rotating at the same angular velocity the entire time it's on the ramp. Therefore when it gets to its maximum height on the ramp, the disk will still have a significant amount of its energy stored in the rotation of the disk, leaving less energy to be transferred into gravitational energy.

Exactly.

For the frictional ramp, the friction is causing a torque on the disk. While this torque doesn't cause the speed of the disk to change, it does affect the angular velocity of the disk.

The friction does reduce the translational speed of the disk compared to what it would have been with no friction. (Energy is conserved. If some energy goes into rotation, that leaves less for translation.)

This decrease in angular velocity (and decrease in rotational kinetic energy) causes the disk to roll further up the ramp.

mgh = 1/2*mv^2 + 1/2*(2/5*mR^2)*(v/R)^2 but because there is no translational energy going up the hill mgh = 7/10*mv^2 making h=(7/10 mv^2) /g

Two problems here:
(1) You need to find the final height in terms of the original height, not in terms of v. Hint: Since the only energy contributing to the final height is the translational energy the ball had at the bottom of the hill, compute that translational energy.
(2) The ball is a hollow sphere, not a solid ball. So correct your moment of inertia.

 

[tater08]

The translational energy at the bottom of the hill would be 1/2 mv^2 and the moment of inertia for the hollow sphere is (2mr^2)/3 =I the height at the bottom of the hill is 0. I just can't figure out the final height in terms of the original height with the given information.

 

[tater08]

i do not know how to come about to solving this part

(1) You need to find the final height in terms of the original height, not in terms of v. Hint: Since the only energy contributing to the final height is the translational energy the ball had at the bottom of the hill, compute that translational energy.

 

[Doc Al]

The translational energy at the bottom of the hill would be 1/2 mv^2 and the moment of inertia for the hollow sphere is (2mr^2)/3 =I the height at the bottom of the hill is 0. I just can't figure out the final height in terms of the original height with the given information.

Hint: Solve for 1/2mv^2 at the bottom of the ramp in terms of the initial height. That's the only portion of the energy that will contribute to getting the ball up the ramp.