Conservation of Angular Momentum?

[noagname]

Sorry about the crude drawing but here's the question. If you have a ball in a hollow pipe, which is being rotated around one of it's ends. Why does the ball leave the pipe?

I would think that the force the pipe applied to the point of contact would be in the tangent to the circle i.e. not in the direction of the length of pipe. The only other reason I can think of would be conservation of angular momentum, before the rotation the ball has an angular momentum of zero. So during the rotation the ball will increase it's radius to infinity in attempt to bring it's angular moment back to zero.

Am I anywhere close to the right answer?

image

 

[WannabeNewton]

To make the system simpler, let's say that the rod is rotating about one end (the fixed left end in your picture) with constant angular velocity $\omega$, that the unit mass particle fits exactly inside, and that it can slide freely without friction as in problem 2.33 from Kleppner's mechanics text. Fixing our origin to that end, the tangential part of Newton's 2nd law in polar coordinates then becomes $2\dot{r}\omega = N$ where $N$ is the normal (constraint) force on the particle from the rod. Now, at time $t_{0}$ we have $\dot{r}(t_{0}) = \frac{1}{2}\frac{N(t_{0})}{\omega}$ and at the very next instant $t_{0} + \Delta t$ we have $\dot{r}(t_{0} + \Delta t) = \frac{1}{2}\frac{N(t_{0} + \Delta t)}{\omega}$ so $\frac{\Delta\dot{r}}{\Delta t} = \frac{1}{2}\frac{\Delta N}{\omega \Delta t} = \frac{1}{2}\frac{\Delta N}{\Delta\theta}$ thus $\ddot{r} = \frac{1}{2}\frac{\mathrm{d} N}{\mathrm{d} \theta}$ (hopefully I didn't mess up any signs lol). So it would seem that for this simplified system, the rate of change of the normal force on the particle with respect to the polar angle (i.e. the changing direction of the normal force as the rod rotates about the fixed end) causes the particle to have radial motion.

EDIT: btw if you wanted to know the radial motion itself for the above simplified system, you could solve $a_{r} = \ddot{r} - r\omega^{2} = 0$ yielding the solution $r(t) = Ae^{\omega t} + Be^{-\omega t}$

 

[AlephZero]

I would think that the force the pipe applied to the point of contact would be in the tangent to the circle i.e. not in the direction of the length of pipe.

Correct. If the ball stayed inside the pipe, it would be accelerating towards the axis of rotation. But there is no way to apply a force to the ball to make that happen, so it doesn't happen.

The only other reason I can think of would be conservation of angular momentum, before the rotation the ball has an angular momentum of zero. So during the rotation the ball will increase it's radius to infinity in attempt to bring it's angular moment back to zero.

I don't like that argument for several reasons.
1. You need to say what point you are measuring angular momentum about. Let's assume you meant about the axis of rotation.
2. While the ball is in the pipe, there is a force acting on it (normal to the wall of the pipe) which has a moment about the axis. Therefore the angular momentum of the ball about the axis is increasing.
3. After the ball leave the pipe its angular momentum about the axis remains constant. It is moving at constant speed along a straight line that does not pass through the axis. It has constant linear momentum, and the moment of the linear momentum about the axis (= the angular momentum) is also constant.

 

[noagname]

Thanks guys! It makes sense now.