# Ball performing small oscillations within a hollow cylinder

1. Jan 15, 2014

### peripatein

1. The problem statement, all variables and given/known data
A small ball of radius r performes small oscillations within a hollow cylinder of radius R. What would be the angular frequency of the oscillations given that the rolling is without slipping? The angle between the radius connecting the center of the hollow cylinder to the ground (my y axis) and the line connecting that center to the point of contact between the ball and the cylinder is ϕ. The positive x axis is to the right. The position of the ball may be written thus: y=R−(R−r)cosϕ; x=(R−r)sinϕ The ball has translational as well as rotational kinetic energy, hence, the total kinetic energy, T, should be: (1/2)m($\dot{y}$2 + $\dot{x}$2) +(1/2)I$\dot{ϕ}$2 where I is the moment of inertia of the small ball. Hence, T is equal to (1/2)m[(R−r)2$\dot{ϕ}$2] + (1/2)I$\dot{ϕ}$2. The potential energy, V, should be mgy. Now, the Lagrangian, L, should be T−V, hence (1/2)m[(R−r)2$\dot{ϕ}$2] + (1/2)I$\dot{ϕ}$2−(1/2)mg(R−r)ϕ2 (under small oscillations approximation). Now, from the Lagrangian, using the Euler-Lagrange formalism, the angular frequency could be easily determined. Is that correct? I believe it isn't yet am not really sure why. I'd be grateful for some comments on this solution.

2. Relevant equations

3. The attempt at a solution

Last edited: Jan 15, 2014
2. Jan 15, 2014

### voko

That seems correct. What are you unsure about?

3. Jan 15, 2014

### peripatein

Well, first of all, I am not quite sure why the potential energy under small oscillations would be as I wrote it. I know that under small oscillations it should be (1/2)kx2, where here k=mg(R-r), but why is x equal to $\varphi$? Moreover, where exactly am I using the "rolling without slipping" fact, i.e. $\dot{x}$=$\varphi$r (or minus $\varphi$r)? Is it by including the rotational kinetic energy? Where does that play a role in my solution, in other words?

4. Jan 15, 2014

### voko

Small oscillations happen about a stable equilibrium. If $U(q)$ is your potential energy, and $q_0$ is a stable equilibrium, what conditions does that impose on $U(q)$ at $q = q_0$?

How did you obtain the angular velocity for rotational kinetic energy?

5. Jan 15, 2014

### peripatein

I obtained the angular velocity via the relation $\varphi$=$\omega$t. Why does that necessarily stem from rolling without slipping?

6. Jan 15, 2014

### voko

You defined $\phi$ as the angle between the vertical and the radius from the cylinder's axis to the ball's point of contact with the cylinder. What does that have to do with the rotation of the ball around its center of mass?

7. Jan 15, 2014

### peripatein

You are absolutely correct, I meant to write θ

8. Jan 15, 2014

### voko

Do you mean you would have two independent angles in the Lagrangian?

9. Jan 15, 2014

### peripatein

Is the angle at which the ball rotates around its CM also phi?

10. Jan 15, 2014

### voko

Imagine that the contact between the ball and the cylinder is frictionless. Now detach the ball from the cylinder, make it rotate about its CM with any angular velocity, and place at the bottom of cylinder. Will the ball move anywhere?

11. Jan 15, 2014

### peripatein

Obviously it won't. Static friction must be present for the ball to move, or am I completely mistaken?

12. Jan 15, 2014

### voko

You are not mistaken, but the question is, does the described behavior involve "slipping"? What would happen if there were no slipping?

13. Jan 15, 2014

### peripatein

The described behaviour would not involve slipping. For slipping to occur, static friction must play a role. Were there no slipping, the CM of the ball would move at a velocity equal to the angular velocity of rotation around the ball's CM multiplied by the ball's radius. This is presumably not what you meant, but I am trying :).

14. Jan 15, 2014

### voko

You misunderstand what "slipping" means. It means merely that at the point of contact two bodies move with different velocities. Whether there is friction between them is not important per se. Of course, very large friction will make slipping almost impossible.

So when there is no slipping, can you relate the angular velocity of the ball with the velocity of its CoM?

15. Jan 15, 2014

### peripatein

Isn't (the size of) the angular velocity equal to the velocity of the center of mass divided by the ball's radius?

16. Jan 15, 2014

### voko

What will be the velocity of the ball's point of contact with the cylinder if its angular velocity and the velocity of its CoM are related as you indicated? Do you see what the condition of no slipping implies?

17. Jan 15, 2014

### peripatein

It implies the velocity of the ball's point of contact with the cylinder should be zero. Am I right?

18. Jan 15, 2014

### voko

It does. But it also means something about $\phi$ and $\theta$. How are they related because of this condition?

19. Jan 15, 2014

### peripatein

Are they equal? I am not sure.

20. Jan 15, 2014

### voko

Let the velocity of the ball's CoM be $v$. How is $v$ related to $\phi$?

Let the angular velocity of the ball about its CoM be $\omega$.

Let the velocity of the ball's point of contact be $u$. Express $u$ via $\phi$ and $\omega$.

Now you said $u = 0$. What condition does that give for $\phi$ and $\omega$?

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