when an object, say, a remotely retractable rod is revolving about a fixed axis on a frictionless ice ( so Fn=mg are all the forces act on it )

if i press the button to remotely retract the rod from lenght l to lenght l1 (l1< l). Then momenta should be Iw = I1w1 where I1w.

the only force acting could only be the force to pull the rod inward which is directed towards the axis, hence, has zero torque is acting on the system

if there is zero torque, we got Σtorque = Iα = 0, hence, α=0

if angular acceleration = 0 then how could w change?

  • $\begingroup$ Conservation of angular momentum always holds, it might just be not that easy to see where it goes. In your case: Where does your 'retractable rod' go? Into the ice? Does it only get shorter? $\endgroup$ – Anedar Aug 17 '16 at 7:32
  • $\begingroup$ part of it goes inside the rest of the rod $\endgroup$ – emanon Aug 17 '16 at 7:47
  • $\begingroup$ The next question to ask is, of course, does the rotational energy rise, fall or remain the same and if it changes, where did it come-from/go-to? $\endgroup$ – dmckee Aug 17 '16 at 16:45

Angular momentum is conserved. Since the moment of inertia decreases, the angular velocity will increase. No external torque is necessary.

Internally there are several forces at work however. You already mentioned the force needed to pull the rod inwards. There is yet another force: Before the shortening, the outer pieces of the rod have a higher velocity (note: regular velocity, not angular velocity) than the the center piece. When contracting the rod, the outer pieces get slowed down by the center piece and the center parts get accelerated by the outer pieces. This is probably the force you were missing.

Note that figure skaters do this all the time to do a fast spinning pirouette.

  • $\begingroup$ seems to make sense. But if total torque is zero, shouldn't α be zero regardless of the moment of inertia : 0=Iα hence α=0 whatever I equals $\endgroup$ – emanon Aug 17 '16 at 8:39
  • 2
    $\begingroup$ No, because $$\frac{d L}{dt}=\frac{d (I \omega)}{dt}=I \frac{d\omega}{dt}+ \omega \frac{dI}{dt}$$ $\endgroup$ – Crimson Aug 17 '16 at 8:56

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