# Rotational Energy Question

1. Feb 15, 2017

### Arman777

1. The problem statement, all variables and given/known data
From an İnitial Lenght h,a solid ball rolls smoothly down one side of a U-shaped ramp and then moves up the other side,which is frictionless.What maximum height does the ball reach ?

2. Relevant equations
Energy conservation equations

3. The attempt at a solution
İnitial Kinetic energy is mgh and final energy wlll be mgh'...Is there some lost energy ?.When the object goes down it has also rotational energy,but when it climbs it cannot rotate so theres no rotational energy so
$mgh'=\frac 1 2mv^2$ and $mgh=\frac 1 2Iω^2+\frac 1 2mv^2$ (In bottom)

from that I gain $h'=h-\frac {v^2} {5g}$ but answer says $h'=\frac {5h} {7}$

2. Feb 15, 2017

### TSny

What would cause the ball to stop rotating when it starts up the other side?

3. Feb 15, 2017

### Arman777

no friction

4. Feb 15, 2017

### TSny

Why would the absence of friction stop the rotational motion of the ball?

5. Feb 15, 2017

### Arman777

there would be no torque to rotate the ball ?

6. Feb 15, 2017

### TSny

If something is already rotating, you don't need a torque to keep it rotating. Think of a spinning top. If no friction acts on the top, then it will keep spinning "forever". You would need a torque to stop it rotating.

7. Feb 15, 2017

### TomHart

I got 5h/7 as an answer. I couldn't follow what you were doing so I can't really help troubleshoot.

8. Feb 15, 2017

### Arman777

The main idea that I stucked is where is the lost energy ? initially it is energy is $mgh$ right ?

9. Feb 15, 2017

### TomHart

It looked like the two initial equations you wrote were correct. The initial potential energy is converted to linear and rotational kinetic energy. And then you showed the final potential (edit correction) energy equal to the linear kinetic energy. That is right. There is no longer any friction so the ball just keeps rotating with no friction to stop it. Therefore, that rotational kinetic energy is not lost (it just keeps going), but it does not get converted back to potential energy. So it looked to me like your thinking was correct.

10. Feb 15, 2017

### Arman777

then where am I going wrong ?

11. Feb 15, 2017

### haruspex

Is the ball rotating at the start? Is it rotating at the bottom of the U? Is it rotating when it reaches its highest point on the other side?

12. Feb 15, 2017

### TomHart

I don't know. I can't see any in between steps.

13. Feb 15, 2017

### haruspex

You have left v in your answer, which is not a given variable. You need to express v in terms of h.

14. Feb 15, 2017

### Arman777

I found it finally,Just one question,
I understand this part.
so the change in rotational energy is zero thats why it didnt converted to potantial energy ?

15. Feb 15, 2017

### Arman777

Even theres no friction ball with rotate but it will not converted to potantial energy ? .I thought it will not rotate and then derive those equations but I see that it will rotate...

16. Feb 15, 2017

### TomHart

Right. Once the ball hits the no-friction surface, it will just keep spinning forever - well, or until it slides back down and hits the surface with friction again.

17. Feb 15, 2017

### Arman777

I see now...withouth friction just gravaity will affect and it will came back with an acceleration (ıf its a ramp at angle $θ$ then $a=mgsinθ$)

18. Feb 15, 2017

### TomHart

Yes, if there was friction on both sides of the U, the rotational energy of the ball would help to drive it upward and it would reach the original height that it started from on the original side. Without friction on the upward climb, that rotational energy never gets engaged to help with the climb.