Ball performing small oscillations within a hollow cylinder

[peripatein]

Homework Statement

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([itex]\dot{y}[/itex]² + [itex]\dot{x}[/itex]²) +(1/2)I[itex]\dot{ϕ}[/itex]² where I is the moment of inertia of the small ball. Hence, T is equal to (1/2)m[(R−r)²[itex]\dot{ϕ}[/itex]²] + (1/2)I[itex]\dot{ϕ}[/itex]². The potential energy, V, should be mgy. Now, the Lagrangian, L, should be T−V, hence (1/2)m[(R−r)²[itex]\dot{ϕ}[/itex]²] + (1/2)I[itex]\dot{ϕ}[/itex]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.

Homework Equations

The Attempt at a Solution

 

[voko]

That seems correct. What are you unsure about?

 

[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 [itex]\varphi[/itex]? Moreover, where exactly am I using the "rolling without slipping" fact, i.e. [itex]\dot{x}[/itex]=[itex]\varphi[/itex]r (or minus [itex]\varphi[/itex]r)? Is it by including the rotational kinetic energy? Where does that play a role in my solution, in other words?

 

[voko]

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 [itex]\varphi[/itex]?

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##?

Moreover, where exactly am I using the "rolling without slipping" fact, i.e. [itex]\dot{x}[/itex]=[itex]\varphi[/itex]r (or minus [itex]\varphi[/itex]r)? Is it by including the rotational kinetic energy? Where does that play a role in my solution, in other words?

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

 

[peripatein]

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

 

[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?

 

[peripatein]

You are absolutely correct, I meant to write θ

 

[voko]

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

 

[peripatein]

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

 

[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?

 

[peripatein]

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

 

[voko]

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

 

[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 :).

 

[voko]

The described behaviour would not involve slipping. For slipping to occur, static friction must play a role.

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.

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 :).

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

 

[peripatein]

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

 

[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?

 

[peripatein]

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

 

[voko]

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

 

[peripatein]

Are they equal? I am not sure.

 

[voko]

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

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 ##?

 

[peripatein]

Well, I have written down the coordinates of the vector connecting my origin to the CoM of the ball. Its differentiation wrt time should yield v, shouldn't it?

 

[voko]

It should, by definition of velocity.

 

[peripatein]

Alright, so |v|=(R-r)*(phi dot). And that should be equal to (theta dot)*(radius of ball), shouldn't it?

 

[voko]

Yes, that looks good.

 

[peripatein]

But the rotational kinetic energy should be (1/2)Iω². How is that equal to (1/2)I[itex]\dot{\varphi}[/itex] which I wrote in my attempt at solution? You did state my attempt was correct. Isn't there a discrepancy here?

 

[voko]

Yes indeed, I missed that mistake originally. So you were right doubting your result :)

Have you figured out how to deal with potential energy?

 

[peripatein]

Is the expression I obtained for potential energy erroneous too? :)

 

[voko]

No, but you had a question about it. Is your question resolved?

 

[peripatein]

Is it not obtained simply via the Taylor expansion of the potential and using the second order as the "effective" potential (under small oscillations approximation)?
By the way, what if I now wished to solve the same problem only with slipping without rolling? Do I simply nullify the rotational kinetic energy in my expressions?

 

[voko]

Is it not obtained simply via the Taylor expansion of the potential and using the second order as the "effective" potential (under small oscillations approximation)?

Do you understand why only the second order term is used?

By the way, what if I now wished to solve the same problem only with slipping without rolling? Do I simply nullify the rotational kinetic energy in my expressions?

I am not sure what you mean by that. Explain.

 

[peripatein]

I am not sure why only the second term counts/is used. I could make an educated guess, or try to at least. Is it because around the point of stable equilibrium the first derivative is zero as the potential obtains its minimum and the first term is subtracted from the potential to obtain the "effective" potential? Am I close?
The second part of that problem, with the hollow cylinder and the ball, asks for the angular frequency of the small oscillations supposing that this time the ball slips and doesn't roll.

 

[voko]

I am not sure why only the second term counts/is used. I could make an educated guess, or try to at least. Is it because around the point of stable equilibrium the first derivative is zero as the potential obtains its minimum and the first term is subtracted from the potential to obtain the "effective" potential? Am I close?

This is not just close, this is exactly it.

The second part of that problem, with the hollow cylinder and the ball, asks for the angular frequency of the small oscillations supposing that this time the ball slips and doesn't roll.

If the problem says "does not roll", then there no kinetic energy due to rolling. Even though I am not sure how one could achieve that in practice.

 

[peripatein]

It says "slips without rolling". Does that mean that I am solely left with the translational kinetic energy whereas the rotational equals zero, and all my expressions remain the same?

 

[voko]

I believe this is what is meant. But, as I said, I do not like this. Even if there is no friction, a freely moving ball will still rotate inside a cylinder. This is seen from considering the torques due to the weight of the cylinder and due to the normal force. I would bring this up to to your professor.