Debate in simple mechanics

1. Nov 13, 2009

marcusmath

Me and my mathematics teacher are in a (trivial) debate as follows;

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.
$$f(x) = -ax^{2}+r$$

Also define a semicircle with radius r.
$$g(x) = \sqrt{r^{2}-x^{2}}$$

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 :)

2. Nov 13, 2009

tiny-tim

Hi marcusmath!

If the cylinder came to an end at the top, then the object would continue in a parabola.

The only question is whether that parabola starts outside the circle or inside it … in other words, which has the greater radius of curvature.

Can you work that out?

3. Nov 13, 2009

marcusmath

If you're thinking of the object moving along horizontally inside the cylinder and coming out the end, I think I may have explained it badly. The object is moving in a perfect circle vertically, it never leaves the cylinder. I called it a cylinder rather than just a circle to show that the object is moving inside the circle rather than outside, so the reaction force is acting inward.

Yes, I think this could be a better way of explaining my argument taking into account the radius of curvature.
If the parabola is completely outside the circle, the object will move in circular motion as it will be kept in by the 'cylinder'. However if the circle and the parabola meet at a single point at the top of the circle but the parabola is wider than the circle (which it must be for the object to reach the top in circular motion), I say the object will instantaneously leave and rejoin the surface of the top of the cylinder and at that point R must equal 0 but the object will continue in circular motion back down the circle. (R being normal reaction)
If this is correct, for an object to move in circular motion R must be greater than zero, but R can equal zero for an instant in time at the top of the circle.

Thanks tiny-tim :)

4. Nov 13, 2009

uart

Hi marcus, I know what you mean.

If at the top of the cylinder you have $g = v^2/r$ then that is the lowest velocity possible to maintain circular motion. At this velocity the reaction force is zero, but only at the one single point - precisely at the top of the cycle.

Interestingly at this point (x=0) the equation of the circle, $y=\sqrt{(r^2 - x^2)}$, and the equation of the parabola, $y = r - x^2/(2r)$, that the particle would otherwise follow (if not constrained by the cylinder) each have identical values of y, y', y'' and y''' (position, first, second and third derivatives). They differ at fourth and higher derivatives btw.

Last edited: Nov 13, 2009
5. Nov 14, 2009

tiny-tim

Hi uart!

Of course, you could also expand r(1 - x2/r2)1/2, or differentiate wrt x2 instead of x.

6. Nov 14, 2009

marcusmath

Any chance you could prove this ($g = v^2/r$)? I tried proving it myself but kept getting slightly differing results. My teacher won't admit he's wrong on the basis of a statement I can't prove. :)

And that is quite interesting. Is it not simply because the parabola and the circle share a common tangent at x=0? For instance, could I not pick any two curves with a common tangent and find the derivatives at that point to see the same result?

7. Nov 14, 2009

tiny-tim

You should get g = v2/r from good ol' Newton's second law …

Fnet = ma, and if there's no reaction force, then Fnet is simply gravity.

8. Nov 14, 2009

vanesch

Staff Emeritus
You are entirely correct, actually, although I've never seen the argument in this form

There's another argument. I take it that your teacher accepts that the ball will remain in contact with the cylinder as long as the reaction force R > 0. Now, for a given velocity on top, which we can take as the "initial velocity" v0, the reaction force will be a function of this velocity v0, of the radius rho of the cylinder, and of the earth's gravity g:

R(v0, rho, g).

It is a continuous function of v0, rho and g as long as R > 0 (it will be an analytic expression in them, a formula, right ?).

Now, consider what you are talking about: the "free flight time" over which the object will NOT be in contact with the cylinder when starting out. Clearly, this is a number between 0 and a finite time (time to fall down I guess). This will also be a continuous function of v0, rho and g: T(v0, rho, g).

Let us consider that we change, say, g while keeping v0 and rho constant, in the neighbourhood of R >~0. By the inverse function theorem, we can consider that g(R) is a continuous function of R in that neighbourhood.

Plug this into T, and keeping v0 and rho constant, we don't think of them. We now have:

T(g(R)) or rather directly T(R). So the "free flight time" is a continuous function of R for R >~0.

We know that T = 0 as long as R > 0 (as long as R > 0, the object remains in contact with the cylinder, and the free flying time is hence 0).

So we have a neighbourhood R >~0 where T(R) equals 0. The limit R - > 0 of this function T, which is continuous, must hence also be 0.

QED.

(to piss off a math teacher )

(of course, it might be a pain to demonstrate continuity, but the point is in general important: if you have a mechanical yes/no constant condition for an open interval of a parameter, then it will also be the case on the border of that interval, unless strange, discontinuous dynamics)

9. Nov 14, 2009

marcusmath

-.- faceplants. I overcomplicated that way too much.

vanesch, your proof makes it seem certain now, whereas when I first posted the thread, I wasn't entirely sure it was true myself. Thanks :)