Frequency of Circular Motion in Ring Rolling Inside a Cone

[Saitama]

Homework Statement

(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.

Homework Equations

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!

 

[dauto]

The torque due to friction along the slant height would tend to rotate the ring and not allow the circular motion to take place

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

 

[haruspex]

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

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.

 

[dauto]

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.

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

 

[Saitama]

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.

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?

 

[dauto]

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?

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.

 

[Saitama]

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.

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.

 

[dauto]

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.

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.

 

[Saitama]

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.

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

 

[dauto]

Use τ = dL/dt

 

[Saitama]

Use τ = dL/dt

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.

 

[haruspex]

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.

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?

 

[Saitama]

In a small time dt, how far does the ring move around the cone?

$\omega r_e\,dt$?

What is therefore the change in direction of the spin?

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

 

[haruspex]

$\omega r_e\,dt$?

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

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

 

[Saitama]

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

What next?

 

[haruspex]

What next?

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

 

[Saitama]

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

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.

 

[haruspex]

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

The equation would be:
$$I\alpha=L\sin\theta \omega$$

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.

 

[Saitama]

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.

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!

 

[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.

 

[Saitama]

haruspex, can you please help me spot the error?

 

[haruspex]

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:
$$v_{cm}+\omega' h\tan\theta=\omega r$$
$$\Rightarrow \omega r_e+\omega h\tan\theta=\omega' r$$

I get $\omega r_e=\omega' r$
$v_{cm} = \omega r_e$, and the velocity of the point of contact is $v_{cm} - \omega' r = 0$.

 

[Saitama]

$v_{cm} - \omega' r = 0$.

Why?

The point of contact moves around the cone with velocity $\omega h\tan\theta$, why don't you take that into account?

 

[haruspex]

Why?

The point of contact moves around the cone with velocity $\omega h\tan\theta$, why don't you take that into account?

The point of contact moves, but I'm considering a point of the ring. Instantaneously that is at rest.

 

[Saitama]

The point of contact moves, but I'm considering a point of the ring.

But how does it explain that $\omega h\tan\theta$ doesn't come in play. I mean, if the point of contact is at rest, its net velocity must be zero i.e
$$v_{cm}+\omega h\tan\theta=\omega' r$$
I understand that the above is supposed to be incorrect but I just don't see why.

 

[haruspex]

But how does it explain that $\omega h\tan\theta$ doesn't come in play. I mean, if the point of contact is at rest, its net velocity must be zero i.e
$$v_{cm}+\omega h\tan\theta=\omega' r$$
I understand that the above is supposed to be incorrect but I just don't see why.

Need to distinguish between the point of contact and a point X on the ring that's currently in contact.
The velocity of X is the velocity of the centre of the ring plus the velocity of X relative to that. The velocity of the centre of the ring is ωr_e; the velocity of X relative to that is -ω'r. But X is instantaneously stationary, so the sum of these is zero.
The point of contact, on the other hand, travels faster than the centre of the ring, in the ratio h tan θ to r_e.

 

[Saitama]

Need to distinguish between the point of contact and a point X on the ring that's currently in contact.

Sorry if this sounds silly but aren't you talking about the same point? I don't see the difference between "point of contact" and "point X on the ring that's currently in contact".

 

[haruspex]

Sorry if this sounds silly but aren't you talking about the same point? I don't see the difference between "point of contact" and "point X on the ring that's currently in contact".

If a wheel rolls along level ground, the point of contact is always below the centre of the wheel, so moves at the same speed. The point on the wheel which is in contact with the ground at some instant is stationary at that instant; it is the instantaneous centre of rotation.

 

[Saitama]

If a wheel rolls along level ground, the point of contact is always below the centre of the wheel, so moves at the same speed. The point on the wheel which is in contact with the ground at some instant is stationary at that instant; it is the instantaneous centre of rotation.

I honestly don't see the difference yet. I have attached a sketch.

Is the blue part "the point of contact" or "the point X"?

 

[haruspex]

I honestly don't see the difference yet. I have attached a sketch.

Is the blue part "the point of contact" or "the point X"?

The "point of contact" of one object on another does not refer to a fixed piece of the first object; it refers dynamically to that part of the first object which is contact with the second at any given instant. When a wheel rolls along a road, the point of contact is always on the road directly under the centre of the wheel. Thus, it moves along at the same speed as the wheel.
If the blue part you have marked is intended as a mark on the wheel, that will descriibe a cycloid. When it makes contact with the road (becoming, transiently, the point of contact) it is instantaneaously at rest.
The equation you wrote taking the velocity of the centre of the ring, then adding to that the relative velocity of a point on the ring, gave you the velocity of that piece of the ring which was instantaneously in contact with the cone. That velocity was therefore zero.