- #1

Salterium

- 3

- 0

**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

## 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([tex]\theta[/tex]) ([tex]\theta[/tex] the angle with the vertical) and in the x-direction, m d

^{2}x/dt

^{2}= N Sin([tex]\theta[/tex]).

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...