Frequencies of Small Oscillations for a Hoop Rolling in a Pipe

[CAF123]

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?

 

[CAF123]

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.

 

[CAF123]

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?

 

[CAF123]

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 $v=(R-a)\dot{\theta}$.