# Hoop rolling in a pipe

Gold Member

## Homework Statement

A hoop of mass ##m## and radius ##a## rolls without slipping inside a pipe of radius ##R## (the motion is 2D). Write down the kinetic and potential energy. Hence find the frequency of small oscillations about equilibrium.

## Homework Equations

Moment of inertia of a hoop: I=Mr2
Rotational kinetic energy

## The Attempt at a Solution

I envisaged the problem as a hoop rolling backwards and forwards in a pipe shaped like an arc of radius R.
Kinetic energy is sum of rotational and kinetic, so $$T = \frac{1}{2}mv^2 + \frac{1}{2} (ma^2) \left(\frac{v^2}{a^2}\right) = mv^2.$$

The path of the C.O.M of the hoop is arc shaped. Let ##\theta## be the angle between a vertical passing through the centre of the hoop and the C.O.M. Draw a horizontal at the base of the pipe. After a little time, the C.O.M is at angle ##\theta##. For small oscillations the increase in height of the C.O.M from its initial position (at a height a above the base of the pipe) is a + a(1-cosθ). So V = mg(a+a(1-cosθ)).

Is this correct? I am wondering if this can be solved more easily using the centre of momentum frame.

Saitama
V doesn't look correct. The CM of hoop is at a constant distance (R-a) from the centre of pipe. When you displace the hoop by angle ##\theta##, what is the distance of CM from the base?

Gold Member
Hi Pranav-Arora,
V doesn't look correct. The CM of hoop is at a constant distance (R-a) from the centre of pipe. When you displace the hoop by angle ##\theta##, what is the distance of CM from the base?

I believe that would be a + (R-a)(1-cosθ)

Saitama
I believe that would be a + (R-a)(1-cosθ)

Yes, looks right to me.

Gold Member
Yes, looks right to me.

Ok, so ##E = mv^2 + mg(a+(R-a)(1-\cos \theta))##. I think to find the freq of small oscillations, I should find a form ##\ddot{\theta}+ \omega^2 \theta = g##, where g is not a function of θ. Differentiating E would give me $$\ddot{\theta} + \left(\frac{g(R-a) - ga}{2a^2}\right)\theta = 0,$$from which I can extract ##\omega## and hence f.

Saitama
Ok, so ##E = mv^2 + mg(a+(R-a)(1-\cos \theta))##. I think to find the freq of small oscillations, I should find a form ##\ddot{\theta}+ \omega^2 \theta = g##, where g is not a function of θ. Differentiating E would give me $$\ddot{\theta} + \left(\frac{g(R-a) - ga}{2a^2}\right)\theta = 0,$$from which I can extract ##\omega## and hence f.

That doesn't look right.

Rewrite the potential energy as mg(R-(R-a)cosθ). Substitute in the energy equation and differentiate again.

What did you substitute for v?

Gold Member
That doesn't look right.
Could you please explain why? The equation is certainly dimensionally consistent.

What did you substitute for v?
Because of no slip, I said ##v_{COM} = a \dot{\theta}##.

Saitama
Because of no slip, I said ##v_{COM} = a \dot{\theta}##.

No, the hoop does not rotate about CM here, it rotates about the centre of pipe so ##v=(R-a)\dot{\theta}##.