# Proof involving a cylinder rolling off of a larger cylinder

1. Dec 9, 2009

### Salterium

1. The problem:
A Uniform right-cylinder of radius a is balanced on top of a perfectly rough (so that only pure rotation occurs) fixed cylinder of radius b (b>a), the axes of the two cylinders being parallel. If the balance is slightly disturbed, show that the rolling cylinder leaves the fixed one when the line which connects the centers of both cylinders makes an angle of Arccos(4/7) with the vertical.

Relevant equations
I=1/2 ma^2

3. The attempt at a solution

I really have no idea how to start. I understand that the cylinder will leave the circle when the curvature of the circle changes faster than the cylinder falls, so I used x^2 + y^2 = b^2 => y=(b^2-x^2)^(1/2) => y'=-x/(b^2-x^2)^(1/2). Then I set out to find dy/dt and dx/dt because I know that dy/dx= (dy/dt)/(dx/dt).

the force of gravity, mg is always downward, so all of the acceleration in the x direction comes from friction and the normal force, N. In the y direction, -ma= mg-N cos($$\theta$$) ($$\theta$$ the angle with the vertical) and in the x-direction, m d2x/dt2 = N Sin($$\theta$$).

This approach is not only wrong (notice the lack of consideration of a frictional force which is in both directions), but it is getting far too difficult in a hurry. I have absolutely no clue how else you could even begin to solve it. Mechanics is NOT my subject...

2. Dec 9, 2009

### AEM

Did you happen to solve the "When does a block sliding down a frictionless sphere leave the sphere" problem in Physics 101 (or whatever they called it in your school)? The block leaves the sphere when the normal force between the block and sphere goes to zero. That occurs when the 'centrifugal force'

$$F = m \frac{v^2}{r}$$

is equal and opposite to the component of the gravitational force normal to the surface of the sphere.

I think that the same criterion operates here and will allow you to solve the problem. The problem is phrased in such a way that you only need to consider a rolling cylinder and as a result you don't need to worry about the frictional force.

3. Dec 9, 2009

### Salterium

Thank you so much for that hint, it was exactly what I needed to get on track. In order to stay on the circular path, some force has to pull the cylinder towards the center of the big cylinder with force mv^2/r, so the only force that can do that here is gravity (so friction isn't really involved). So I set the force of gravity in the radial direction equal to the centripetal force, used conservation of energy to get rid of v, and solved for theta. Perfect. Thanks again, this one was giving me trouble.

4. Dec 9, 2009

### AEM

You're welcome. I'm happy to have helped.