I've learned in that in this experiment:

enter image description here

...the skater will start rotating faster when she brings her arms in and there is no net torque acting on her. But what would happen to her angular momentum and rotation if she only brings one of her arms inwards with the other arm sticking outwards?

edit: will a net torque act on the body? What will this cause?

  • $\begingroup$ Think about this for a moment: Why does $\omega$ increase in the original problem? $\endgroup$ – Kyle Kanos Oct 22 '13 at 16:08
  • $\begingroup$ I watched MIT's OCW video for this. Prof Lewin says that angular momentum, $L$ is conserved and $L = I\omega$. When she brings her arms in, her moment of inertia decreases, so her angular velocity goes up. The opposite happens when she takes her arms out so her angular velocity decreases. $\endgroup$ – shortstheory Oct 22 '13 at 16:14
  • $\begingroup$ Right. So would $I$ increase, decrease, or stay the same if she were to pull only one arm inwards? $\endgroup$ – Kyle Kanos Oct 22 '13 at 16:17
  • $\begingroup$ Um, it $I$ should decrease if she brings only one arm inwards, so I guess $\omega$ would still increase (though not as much)? But, my question wasn't pertinent only to the $\omega$. Shouldn't something else happen as well? $\endgroup$ – shortstheory Oct 22 '13 at 16:21
  • $\begingroup$ Okay, and what would be the consequence of this torque? Will angular momentum still be conserved? Will she move laterally wrt to the ground? I'm sorry for my ignorance, but I am new to rotational mechanics. $\endgroup$ – shortstheory Oct 22 '13 at 16:26

The asymmetric problem gets into more complicated aspects of the kinematics of rotating bodies than are usually the point when presenting this example.

In the initial, arms out symmetric configuration, the skater's center of mass is directly over the pivot point. If she brings one arm in, and still has the axis of her body strictly vertical, then her center of mass is no longer over the pivot point, and, were she not spinning, she would fall. Now, because she is spinning, you are dealing with a problem similar to that of a gyroscope in a gravitational field whose axis of rotation is non-vertical: the gyroscope precesses.

More step wise

1. (initial condition) the skater is spinning about a vertical axis, both arms outstretched.
2. skater starts pulling her left arm inward, this changes the location of her center of mass.
3. skater starts "falling" towards her outstretched right arm.
4. this is a torque (due to gravity and the friction on the ground that keeps her skate tip at a fixed point on the ice) on a spinning body.
5. this ends up causing her main axis of rotation to precess.

I believe that I can see these kinds of effects in some of the examples here. First thing to note is that the skate is always tracing a circle on the ice. The size of this circle is related to the degree of asymmetry in the skaters body position: more asymmetric - larger circle. This is consistent with the skater needing to manage the location of her center of mass by adjusting her body, in particular her legs, and/or needing to manange rotating about a non-vertical axis in such a way that her overall angular momentum is (very close to) exactly vertical.

This page has a nice summary of rigid body mechanics, which if worked through, could be applied to this situation.

| cite | improve this answer | |
  • $\begingroup$ Since she has not stopped spinning while watching from an external frame where would you say the center of her mass has shifted with respect to original centre of mass ? $\endgroup$ – Rijul Gupta Oct 22 '13 at 16:52
  • $\begingroup$ What if she tiptoes on one of her skate and then rotates about it, the centre of mass would be making a cone having its vertex at the tip of her skate but even for other cases like when she does not tip toes what cirle are you making ? $\endgroup$ – Rijul Gupta Oct 22 '13 at 17:13
  • $\begingroup$ Relative to the ground, her center of mass ends up doing a fairly complicated precssion motion; working through a gyroscope in a gravitational field is a typical problem for graduate level mechanics courses. $\endgroup$ – Dave Oct 22 '13 at 17:17
  • $\begingroup$ Wouldnt it simply be like a rotating top , if not then why ? $\endgroup$ – Rijul Gupta Oct 22 '13 at 17:23
  • $\begingroup$ Yes like a rigid top, which can execute complicated motion when its axis of symmetry is not vertical. $\endgroup$ – Dave Oct 22 '13 at 17:25

When she takes both her arms in, her moment of inertia decreases because the particles of her arms that were away from the axis of rotation have come closer to the axis. Now when she brings only one arm close to herself, obviously the same thing happens since again though half of the previous time but particles do come close to the axis of rotation.

Added after edit: why would internal forces cause torque, had there been a torque, net angular momentum would change but it does not, so no torque is applied.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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