# Conservation of angular momentum when radius becomes very small?

This question here really interested me to think about what happens when radius decreases to a very small value, and how this contributes to an extreme increase in linear velocity. I'm not at all questioning the validity of angular momentum conservation, I'm just curious what I'm missing.

Let's say we have a point mass spinning on a string of radius $r_0$ away on a finger, like in many angular momentum demonstrations, with a velocity $v_0$. The initial angular momentum will be $mv_0r_0$. Let's say we bring the radius in really close, from, for example, 1 meter to .02 meters, by pulling down on the string to reduce the radius. This is a reduction of 50 times the original radius, which means that linear velocity must increase because moment of inertia decreases by a factor of 2500 ($M(R_0/50)^2)$ and therefore $\omega$ increases 2500 times its original value, and so $v = \omega / r$ increases to 50 times its original value. But if $v$ was originally, let's say $1m/s$, then going to $50m/s$ seems extremely unreasonable when the radius is only $2cm$. On top of that, the string should theoretically break due to the centripetal force required to keep it moving in the small circle, but this does not happen in a physics demonstration.

Also, the amount of work required to increase the velocity to 50 times is unreasonably high. There's something missing here. What is it exactly?

• What you are missing is the difference between 'ideal' and 'real'. – sammy gerbil Jan 23 '17 at 16:25

You're not missing much. First of all, if we do this experiment under ideal conditions (massless string, no aerodynamic drag, no torque), then the increase in $\omega$ will indeed follow angular momentum conservation, and, yes, the centrifugal force will become very large, so that work against it can indeed provide the needed kinetic energy. Back in the real world, meanwhile, that mass of yours will experience some very substantial aerodynamic drag at those kinds of velocities. In addition, you'll probably see other kinds of friction losses from the string (aerodynamic and mechanical), so in the end your velocities should be significantly lower.