I'm working on a problem from YK Lim's "Problems and Solutions on Mechanics". It is Problem 1016. Here is the statement of the problem:
A mass $m$ moves in a circle on a smooth horizontal plane with velocity $v_0$ at a radius $R_0$. The mass is attached to a string which passes through a smooth hole in the plane.
a) What is the tension in the string? b) What is the angular momentum of $m$? c) What is the kinetic energy of $m$? d) The tension in the string is increased gradually and finally $m$ moves in a circle of radius $\frac{R_0}{2}$. What is the final value of the kinetic energy? e) Why is it important that the string be pulled gradually?
a) - c) are no problem. But d) befuddles me, because I see two plausible arguments that lead to contradictory results.
First, there is conservation of angular momentum, which tells me that as $m$'s 'orbital radius' gets smaller, its velocity has to get bigger, and therefore its kinetic energy gets bigger.
Second, there is the work-energy theorem, which tells me that for there to be a change in kinetic energy, there must be a force applied in the direction of displacement. But that is obviously not the case here. The only force is the tension in the string, which always acts orthogonally to the displacement. Therefore the velocity doesn't change, therefore the kinetic energy doesn't change.
My hunch here is that I'm misunderstanding the work-energy theorem, i.e. how or even whether it applies in this case. Perhaps it is that the work-energy theorem says that work done by a force leads to changes in kinetic energy, but it's not necessarily true that a change in kinetic energy was brought about by work done by a force?
EDIT: I foolishly left out part e) of the problem in my initial post, but have now edited to include it.
The author's solution to e) states:
The reason why the pulling of the string should be gradual is that the radial velocity of the mass can be kept small so that the velocity of the mass can be considered tangential. This tangential velocity as a function of $R$ can be calculated readily from the conservation of angular momentum.