# Ball rolling and bouncing from a wall

1. Aug 16, 2015

### Karol

1. The problem statement, all variables and given/known data
A ball of mass m rotates without sliding at velocity v1 and hits a wall. it rotates backwards at velocity v2. what is the energy loss.

2. Relevant equations
Kinetic energy of a rigid body: $E=\frac{1}{2}I\omega^2$
Moment of inertia of a ball round it's center: $I=\frac{2}{5}mr^2$

3. The attempt at a solution

$$\Delta E=\frac{1}{2}m(v_2^2-v_1^2)+\frac{1}{2}I(\omega^2_2-\omega^2_1)$$
$$\omega r=v\rightarrow \omega^2=\frac{v^2}{r^2}$$
$$\Delta E=\frac{1}{2}m(v_2^2-v_1^2)+\frac{1}{2}\frac{2}{5}mr^2\cdot\left( \frac{v_2^2-v_1^2}{r^2} \right)$$
$$\Delta E=\frac{1}{2}m(v_2^2-v_1^2)\left( 1+\frac{2}{5} \right)$$

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Last edited: Aug 16, 2015
2. Aug 16, 2015

### Staff: Mentor

I'm not sure if you are supposed to assume $\omega r = v$. If there is no friction, why should $\omega$ change during the collision?
To make it worse, why should this relation hold before the collision?

3. Aug 16, 2015

### Karol

My mistake, i corrected in the OP, it rotates without sliding, sorry

4. Aug 16, 2015

### Staff: Mentor

Okay, then ignore the second comment, the first one is still valid.

5. Aug 16, 2015

### Hesch

I think it's more complicated to solve, because at higher speeds the ball will have so much rotational energy forward, that it will jump backwards, having hit the wall.
The rotational energy will be converted to "jumping energy".
Some moving pattern like this: ( moving forward / moving backward ):

|∩∩∩∩∩∩∩ >
|--------------------------- <

( wall to the left )

6. Aug 16, 2015

### Staff: Mentor

It won't jump without friction.

7. Aug 16, 2015

### Karol

But i still think, because of the wording of the question, that also in the first stage at v1 it rotates without sliding

8. Aug 16, 2015

### Staff: Mentor

You added that to the problem statement, so I guess it is true... yes.
It will slide after the collision, however.

9. Aug 16, 2015

### Karol

Why are you sure? if it will slide then it won't rotate fast enough for v2, but why do you say that?
Because of the collision i don't know if the angular momentum's magnitude is preserved and only the direction has changed, so i don't know nothing!

10. Aug 16, 2015

### Staff: Mentor

There is nothing that could change its rotation, as we don't have friction.
Edit: You removed the "without friction" from post 1.
Please post the full and exact problem statement.

11. Aug 16, 2015

### Karol

A ball of mass m rolls without sliding and hits a vertical wall with velocity v1. the ball's velocity after the collision is v2. what is the amount of energy that was emitted in the collision.

12. Aug 16, 2015

### haruspex

If there is unlimited friction with the ground but none with the wall, and we allow the rebound to take a little time instead of being instantaneous, that is all feasible and leads to your answer in the OP.

13. Aug 16, 2015

### Karol

And if there's friction with the wall? will the fact that the force from the wall can be in any angle make a difference? is that, the change in the direction of the reaction, the only effect of the friction in the wall?

14. Aug 16, 2015

### haruspex

If there is any friction with the wall then the ball will become airborne on the rebound.

15. Aug 16, 2015

### Karol

What is airborne? do you mean it will be a little time in the air like in the drawing? and why are sure that it will lift the ball off the ground, maybe the force won't be strong enough, or it doesn't matter if it will completely lift it or not? and what does it matter if the ball is airborne if we address to the later situation when it stabilizes and continues to roll without sliding?

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16. Aug 16, 2015

### haruspex

If the rebound takes a short time then the forces will be large, much larger than mg. If there is friction from the wall then its maximum value will be in proportion to the rebound force. Angular momentum about the point of impact on the wall will be preserved.
If the friction is so great that it does not slip against the wall during rebound then you can deduce the vertical velocity that results.
As long as v2 refers to the velocity when it has resumed rolling, it does not matter.

17. Aug 17, 2015

### Karol

Thanks but why, if the collision takes time, there isn't conservation of energy, am i right?
Does the stage of acceleration consume all the difference in energies?
So if the change in velocities is instantaneous there isn't acceleration and no losses?

Last edited: Aug 17, 2015