# Homework Help: Ball rolling over a step

1. Oct 27, 2015

### andyfeynman

1. The problem statement, all variables and given/known data
A uniform solid sphere of radius R rolls without slipping at velocity V on a level surface. It collides with a step of height h. Assume that after the collision, the sphere maintains contact with the step at point A with no slipping.

Find the minimum value of V for the sphere to be able to rise up to the top of the step, in terms of h, R and g. (Hint: If h = R/5, V = (14gR)1/2/6.)

2. Relevant equations
L = Iω

3. The attempt at a solution
Let the initial and final angular velocities be ω0 and ωf respectively. Let M be the mass of the sphere. Let I be the moment of inertia about point A.
Angular momentum about point A before collision = Iω0 + MV(R-h)
Angular momentum about point A after collision = Iωf
Conservation of angular momentum: Iω0 + MV(R-h) = Iωf
Conservation of energy after collision: (1/2)Iωf2 = Mgh

Then I used the above equations to solve for V in terms of h, R and g, which is different from the answer. What went wrong?

Last edited: Oct 27, 2015
2. Oct 27, 2015

### BvU

Hard to say. You don't post your steps. In fact your attempt at solution is mainly listing off equations. Your last one is weird: If h = 0, you don't expect $\omega$ to be zero.

3. Oct 27, 2015

### J Hann

You can't assume conservation of energy thru the collision. because energy is lost in collision with the step.
After the collision then energy is conserved.
The key is that initial angular momentum is around the center of mass of the sphere.
After the collision with the step the angular momentum is about the point of contact with the step (no slipping).
Appropriate moments of inertia must be used to calculate speed just after collision with the step.

4. Oct 27, 2015

### AstroK

I am working on problems similar to this one too. Please allow me to borrow your thread for a quick question -- for this particular problem, I have tried to use conservation of angular momentum then energy to solve, and I got everything right in my final expression except I don't get the 1/3 factor in it for some unknown reason. I have no idea where that 1/3 come from.

5. Oct 27, 2015

### haruspex

That's not quite right.
Angular momentum can be assessed about any point you choose. Maybe you mean the instantaneous centre of rotation is about the point of contact with the step after the collision, (but before collision it is about the point of contact with the ground, not its mass centre).
For applying conservation of angular momentum, you must pick either the (moving) mass centre as the axis, or any fixed point. It is often helpful to choose a fixed point through which one or more unknown impulses act. There is an obvious choice here.

I'm not completely sure what this is saying. In practice, it is not possible for the sphere to remain in contact with the step while it rises onto it. That would require (post collision) an acceleration towards the point of contact. In particular, that would have a horizontal component in the direction of travel, but the only horizontal force component available is the normal force from the step, which would be acting the wrong way.
It might mean that rolling contact is maintained for an instant, after which it would follow the usual ballistic trajectory. However, it then gets quite hard to figure out whether the ball will end up on the step. A crude approximation could be had by considering only the vertical component of its KE after collision.
Or maybe we have to imagine a blob of superglue on the corner of the step?

6. Oct 27, 2015

### haruspex

Hang on while I get out my crystal ball so that I can see your working.

7. Oct 27, 2015

### AstroK

Never mind, I just have a talk with my professor about this, and it turned out that I had overlooked some obvious yet important details...

Last edited: Oct 27, 2015
8. Oct 28, 2015

### BvU

But we're still interested ! Can you enlighten us ?

9. Oct 28, 2015

### J Hann

I agree with this, that you need to take the initial angular momentum about the point of contact with the step.
As I described it, taking the initial angular momentum about the center of mass of the sphere (only) would omit the contribution
of angular momentum due to the linear motion of the sphere with regard to the point of contact with the step.
Thanks for the correction.