- #1

- 16

- 0

My argument is that an object moving in vertical circular motion on the inside of a cylinder can instantaneously experience a reaction force of zero at the top of the circle.

He argues that as soon as reaction force from the cylinder is gone, even for an instant, the object will begin to fall in a parabolic path away from the surface, claiming that for an object to move in vertical circular motion R>0.

I believe that R>=0 is sufficient for circular motion given R tends to zero on the upward climb and is only zero for an infinitesimal period of time.

The reason for this is,

the object in motion will be attempting to form a parabolic path but the cylinder it is in is providing a normal reaction force to make the path circular.

Now define a parabolic curve that crosses the y-axis at it's maximum point r.

[tex]f(x) = -ax^{2}+r[/tex]

Also define a semicircle with radius r.

[tex]g(x) = \sqrt{r^{2}-x^{2}}[/tex]

Now if the curves have two common tangents, the object will leave at the first (when R=0) and rejoin the circle at the second to continue in circular motion.

Of course it is impossible for these two curves to have two common tangents (which can be proved by differentiating each and equating).

However it is possible for them to have a single tangent, which they do given any value of a, at the top of the circle.

Now, any quadratic must have 2 solutions and in this case the solutions are equal. This means there are two tangents to the circle at a single point (I know this sounds silly but I have not better way of saying it)

So when R=0, the object will leave the circle, following the parabolic path at the tangent and rejoins the circular path at the next tangent, which is at the same point.

Therefore the object will continue to move in vertical circular motion after R hits 0.

I know this is hardly a rigorous proof but I'm sure one can be produced.

This probably makes no sense, but if anyone understands and can tell me if this is correct, I'd be grateful :)