# Ring rolling inside a cone

1. Apr 3, 2014

### Saitama

1. The problem statement, all variables and given/known data
(See attachment)

Assume that the surface has friction and a small ring of radius $r$ rolls on the surface without slipping.

Assume conditions have been set up so that (1) point of contact between the ring and the cone moves in a circle at height $h$ above the tip and (2) the plane of ring is at all times perpendicular to the line joining the point of contact and the tip of the cone.

What is the frequency of this circular motion?

You may work in the approximation where $r$ is much smaller than radius of circular motion.

2. Relevant equations

3. The attempt at a solution
The attachment 2 shows the side view snapshot of the motion.

The CM of ring rotates in a circle of radius $r_e=h\tan\theta-r\cos\theta$. The force acting on it are shown in the attachment. The frictional force can be resolved into two components. One along the slant height of cone (f) and the other along the tangent to the point of contact (not shown).

Now I am confused at this point. If I take torque about the CM of ring, the only contributing torque is due to the two components of friction mentioned above. The torque due to friction along the slant height would tend to rotate the ring and not allow the circular motion to take place, doe this mean I have to assume that component to be zero? But doing that doesn't make sense to me.

Any help is appreciated. Thanks!

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2. Apr 3, 2014

### dauto

Are you sure about that? It seems to me that's exactly torque needed in order for the rotation to take place.

3. Apr 3, 2014

### haruspex

Quite so. The ring's axis of rotation is changing, and that requires a torque. This is a gyroscope precessing.
OTOH, I very much doubt there is a tangential frictional force.

4. Apr 3, 2014

### dauto

A tangential frictional force would do work increasing the system's energy. That can't happen.

5. Apr 4, 2014

### Saitama

Agreed that there will be no tangential frictional force but what about $f$ acting along the slant height of cone?

As I mentioned earlier, if I take torque about CM, the only torque acting is due to the frictional force along the slant height and this torque would make the ring fall toward the tip of the cone which should not happen. What's wrong with my argument?

6. Apr 4, 2014

### dauto

The torque makes the ring precess which is necessary. It won't necessarily fall towards the tip, though it might if it wasn't moving fast enough.

7. Apr 4, 2014

### Saitama

But which torque balances it then? There is only one torque about CM and i.e due to friction along the slant height. There must be some other torque (about CM) which has to balance this.

8. Apr 4, 2014

### dauto

No, the torque is not balanced. There is a net torque that causes angular momentum to change forcing the disk to precess as needed for it to move in a circular path around the cone.

9. Apr 4, 2014

### Saitama

Yes, I understand that the direction of angular momentum changes, so there must be a torque but how do I calculate it?

10. Apr 4, 2014

### dauto

Use τ = dL/dt

11. Apr 4, 2014

### Saitama

Erm...I know I have to use this but about what point should I evaluate the expression for $\vec{L}$. I am sorry if this sounds stupid but this is the first time I am trying to solve a problem where the direction of L changes.

12. Apr 4, 2014

### haruspex

The magnitude of the spin does not change, so you just want the magnitude multiplied by the rate at which its direction changes. In a small time dt, how far does the ring move around the cone? What is therefore the change in direction of the spin?

13. Apr 4, 2014

### Saitama

$\omega r_e\,dt$?
$L\sin\theta \omega\,dt$?

14. Apr 4, 2014

### haruspex

If ω is the rate at which the centre of the ring moves around the axis of the cone, yes.

15. Apr 5, 2014

### Saitama

What next?

16. Apr 6, 2014

### haruspex

There should be an equation connecting the torque and the rate of change of angular momentum.

17. Apr 6, 2014

### Saitama

Yes, but what to fill in for the moment of inertia?

The equation would be:
$$I\alpha=L\sin\theta \omega$$
About what point should I calculate $I$? Should $I$ be about CM? If so, I don't see why.

18. Apr 6, 2014

### haruspex

No, the torque is $\vec F \times \vec r$. And the rate of change of $\vec L$ can be written $\vec L \times \vec \omega$. (Could have the signs wrong.). Note that both these vector products produce a vector parallel to the motion of the ring's centre.

19. Apr 7, 2014

### Saitama

Thanks haruspex!

I did a force balance and got:
$$f=mg\cos\theta-m\omega^2 r_e\sin\theta$$
Hence:
$$mg\cos\theta r-m\omega^2 r_e\sin\theta r=L\sin\theta \omega \,\,\, (*)$$
Also,
$$L=mr^2\omega'$$
where $\omega'$ is the angular velocity of ring about its own axis.

I need a relation between $\omega'$ and $\omega$. I think for that, I should use the fact that point of contact has zero velocity. So that gives:
$$v_{cm}+\omega' h\tan\theta=\omega r$$
$$\Rightarrow \omega r_e+\omega h\tan\theta=\omega' r$$
$$\Rightarrow \omega'=\frac{\omega (r_e+h\tan\theta)}{r}\approx \frac{\omega h\tan\theta}{r}$$
Using $r_e \approx h\tan\theta$ and plugging everything in (*), I get:
$$mg\cos\theta r-m\omega^2h\tan\theta\sin\theta=mr^2\omega\sin\theta \frac{\omega h\tan\theta}{r}$$
$$\Rightarrow mg\cos\theta r=2m\omega^2 rh\tan\theta \sin\theta$$
$$\Rightarrow \omega=\frac{1}{\tan\theta}\sqrt{\frac{g}{2h}}$$
which is the correct answer, thanks a lot haruspex!

20. Apr 14, 2014

### Saitama

Sorry for the bump but I was going through the thread again and found an error in my working in #19. I wrote:
$$\omega'=\frac{\omega(r_e+h\tan\theta)}{r} \approx \frac{\omega h\tan\theta}{r}$$
but I feel this is incorrect, rather it should be:
$$\omega' \approx \frac{2\omega h\tan\theta}{r}$$
and this gives an incorrect answer.