Today in physics class we were talking about angular momentum and rotational kinetic energy. My teacher used the classic example of a figure skater spinning on ice - when she pulls her arms in, her angular momentum is conserved and her angular velocity increases, meaning that her rotational kinetic energy also increases. Of course, this increase in energy must come from somewhere - in this case, it comes from the figure skater doing work on her arms and pulling them in toward her body. Then I started wondering - if the figure skater slows her rotation by extending her arms, she decreases her rotational KE. Where is her energy going? Or to put it another way, what force is doing work on the figure skater in order to decrease her energy?

  • $\begingroup$ She will gain the energy while taking her arms away from her the same way she used it to bring her arms near her. $\endgroup$ Commented Jan 13, 2014 at 15:42
  • 4
    $\begingroup$ by pulling the arms towards the body, she does work against the centripetal force. you obviously understand that. why are you in doubt about the reverse situation? $\endgroup$
    – Nephente
    Commented Jan 13, 2014 at 15:48
  • $\begingroup$ @nephente no, centripetal force is the inward-acting force of tension on her arms. She's obviously not working against that when she pulls her arms in; you can't do work against a force when it's in same direction. $\endgroup$ Commented Jan 13, 2014 at 16:43
  • $\begingroup$ @Turtleweezard I beg to differ. Work is only done when the displacement is along the force $W=\int F\cdot ds$ No matter if in the same or opposite direction. If $W>0$ the system gains energy, in case $W<0$ it loses energy. A magnetic field for example does no work on a moving charge, because the Lorentzforce is always perpendicular to the direction of motion $F_L = q v\times B$ Since there is no torque, but work done, the only possible force is a radial one. The centripetal force is the only such in the problem. $\endgroup$
    – Nephente
    Commented Jan 13, 2014 at 17:22
  • $\begingroup$ @Turtleweezard I meant the centri_fugal_ force. Apologies. I cannot change this in my last comment anymore. But since centrifugal and centripetal force only differ by a sign, this is of minor concern (but still of importance) $\endgroup$
    – Nephente
    Commented Jan 13, 2014 at 17:35

2 Answers 2


I think there are 2 main sources of confusion:

First, because of gravity, extending your arms feels like work. We're only interested in the radial movement, though, and in this direction, the skater's arms are pulled by the centrifugal force (in the long tradition of spherical cows in vacuum, we could replace the figure skater with two beads on a spinning rod).

Second, the idea of rotational energy as kinetic energy. The relevant work variable is (as already mentioned) the radial extension of the skater's arms, and as far as that's concerned, rotational energy plays the part of potential energy.

Think of the skater pulling in her arms as compressing a spring, and extending the arms as its release.

Going by either the bead or spring model, the rotational energy gets converted into kinetic energy of the arms, accelerated by the centrifugal force in direction of the radial work variable and ultimately dissipating via vibrations when the arms abruptly reach maximal extension.

Of course, if the skater doesn't let her arms be accelerated and slowly extends them instead, the energy dissipates right away, which might be the more realistic approach.

  • $\begingroup$ The skater does some work by converting some of her energy into kinetic energy, that kinetic energy needs to be conserved. Now, if the skater stops converting her energy into kinetic, still the skater would had continued moving if the kinetic energy which was generated by the previous work was not dissipated. But the skater slowly decreases her kinetic energy, "where is her energy (kinetic) going?" Here it is to be noted that, due to gravity energy might dissipate or may not depending upon the relative value of force applied by the skater with respect to gravitational force .... $\endgroup$
    – Sensebe
    Commented Jan 13, 2014 at 23:00
  • $\begingroup$ ....if we assume that gravity is not opposing the skater's motion and thus not being the reason for kinetic energy dissipation, the kinetic energy would dissipate due to friction (major), air drag (minor), etc So, due to friction kinetic energy would convert into heat energy. Until kinetic energy is totally dissipated the skater moves for a while and stops. So, missing kinetic energy has been converted into heat energy or some other minor forms. $\endgroup$
    – Sensebe
    Commented Jan 13, 2014 at 23:06
  • $\begingroup$ @VINAY actually, I was referring to the apparent substantial and sudden energy loss that occurs when the skater's moment of inertia changes, not the trivial and gradual losses that occur due to friction with air and ice. $\endgroup$ Commented Jan 14, 2014 at 2:56
  • $\begingroup$ Thanks for the attempt, Christoph. I'll +1 it for spherical cows in a vacuum. $\endgroup$ Commented Jan 14, 2014 at 2:57
  • $\begingroup$ This question seems to lead to: What happens to the work done on a muscle, as it allows something to respond to an external force, but under control. Where does the energy go as I slowly walk down a flight of stairs? $\endgroup$
    – DJohnM
    Commented Jan 14, 2014 at 6:54

if the figure skater slows her rotation by extending her arms, she decreases her rotational KE. Where is her energy going? Or to put it another way, what force is doing work on the figure skater in order to decrease her energy?

In the rotating frame, as she extends arms, she gains work from the centrifugal force but does not use it in any useful way, so it transforms into internal energy of her body (her muscles will heat up). In the frame of spectators, rotational energy is transformed into internal energy as well.

  • $\begingroup$ So I suppose that we have to use the 'fictional force' of centrifugal force to make this situation fit the energy model. Is there any way to avoid using centrifugal force though? Can it be explained in terms of 'real' forces? $\endgroup$ Commented Jan 14, 2014 at 2:53
  • 1
    $\begingroup$ Centrifugal force is necessary if you want to explain it to the skater. If you want to explain it to the spectator, there is no centrifugal force; the kinetic energy is just transformed into internal energy of the muscles of the skater (dissipation is the natural end of most common manifestations of kinetic energy). It is quite common that mechanical energy degrades into internal form over time. $\endgroup$ Commented Jan 14, 2014 at 3:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.