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.


2 Answers 2


The only force is the tension in the string, which always acts orthogonally to the displacement.

This is the key. The tension in the string is always orthogonal to the displacement only if the radius is constant. As the radius changes the path of the mass is no longer tangent to the circle and therefore no longer orthogonal to the string.

Angular momentum is conserved, work is done, and KE increases.

  • $\begingroup$ I foolishly left out part e) of the question, which I think is relevant to your explanation here. The authors state that the radius is stipulated to be decreased gradually specifically to keep the velocity (and hence the displacement) tangential. I see what you are saying though: there is just no way the mass can get from the larger radius to the smaller without radial displacement. However, wouldn't that mean the final kinetic energy would have a radial component of velocity in it? $\endgroup$ Commented Sep 12, 2020 at 5:16
  • $\begingroup$ @GhostRepeater This is why it is important that the string is pulled gradually. if the string is pulled gradually then we are meant to assume that the radial velocity at any time is negligible compared to the tangential velocity. $\endgroup$
    – gandalf61
    Commented Sep 12, 2020 at 7:21
  • $\begingroup$ @Ghost Repeater the slow decrease makes the KE from the radial component of velocity arbitrarily low, but it cannot reduce the work done decreasing the radius. $\endgroup$
    – Dale
    Commented Sep 13, 2020 at 0:37

I'm adding this answer in support of the answer by Dale.

Work done by a centripetal force when an object moves along an inward spiral is effectively the same case as work done when an object is moving down a ramp.

Work done when moving along an inward spiral

In the idealized case of a frictionless ramp the change of kinetic energy as an object slides down a ramp is independent of the inclination of the ramp. When you decrease the inclination of the ramp the ramp becomes longer, but the same heigth difference is still there. The height difference alone determines the amount of change of energy.

The same logic applies in the case of a centripetal force doing work. If you make the inward spiral twice as gradual then the overall process takes twice as long to complete; in the end the centripetal force has done the same amount of work.

Incidentally, for the force profile you can use an easing function.

Implementation of easing function in the case of this particular image: at the start the centripetal force is the required centripetal force for circular motion. Then the centripetal force is gradually increased. When the object is close to the end radius the centripetal force is gradually adjusted towards the required centripetal force for the final radial distance.

More general

A ramp doesn't need to have a constant inclination, it can be any profile. Only the height difference counts.

An inward spiral can have any profile, the work done by the centripetal force is independent of the shape of the spiral. You can make it a wild ride; it makes no difference. Of course, as you mention: you do need to avoid that there is a residual radial velocity. As you approach the desired final radial distance you need to ease into circular motion.


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