I've read all the questions with similar titles but I couldn't find an answer.

Suppose I'm rotating with my arms extended on a frictionless surface. I have angular momentum and energy:

\begin{equation} L_0=I_0\ \omega_0 \end{equation} \begin{equation} E_0=\frac{1}{2}I_0\ \omega_0^{2} \end{equation}

Where $I_0$ is my moment of Inertia with my arms extended and $\omega_0$ is my initial angular velocity.

Suddenly, I decide to flex my arms to decrease my moment of Intertia. Then my angular momentum and energy are:

\begin{equation} L_f=I_f\ \omega_f \end{equation}

\begin{equation} E_f=\frac{1}{2}I_f\ \omega_F^{2} \end{equation}

If I use conservation of energy to calculate final angular velocity I get:

\begin{equation} \omega_f = \sqrt{\frac{I_0}{I_f}}\omega_0 \end{equation}

But if I use conservation of angular momentum:

\begin{equation} \omega_f = \frac{I_0}{I_f}\omega_0 \end{equation}

Both can't be right... Is energy not conserved in this problem? why?

Edit: Many anwers have pointed out that I'm actually doing work when I pull my arms back. Thanks for that clarification! What would happen if the system is a disk rotating with two persons in each side and they start walking towards the center? They walk using static frictional force which does not do work. Would energy be conserved then?

  • $\begingroup$ Try including the work you did pulling in your arms... $\endgroup$ – DJohnM Sep 10 '17 at 21:36
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    $\begingroup$ The key is in your question: the word flex. $\endgroup$ – garyp Sep 10 '17 at 21:41
  • $\begingroup$ Why do you think that friction does not do work? Its a force like any other, and the objects in contact are moving... $\endgroup$ – Eddy Sep 10 '17 at 21:50
  • $\begingroup$ Because static friction is applied on a point that doesn't move so there isn't any displacement along its direction and hence no work. $\endgroup$ – P. C. Spaniel Sep 10 '17 at 21:53
  • $\begingroup$ Ah, so your confusion is that you think someone can stand on a spinning disk but there foot not move. How exactly? $\endgroup$ – Eddy Sep 10 '17 at 22:09

You do work on your arms as you pull them in, thus your energy has increased. The correct conserved quantity is angular momentum, as you deduce. The amount of work done on your arms can either be computed directly (force times distance type approach) or by using the solution from angular momentum (final energy minus initial energy type approach).

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  • $\begingroup$ Thanks for the answer! I edited the question in view of your answer. $\endgroup$ – P. C. Spaniel Sep 10 '17 at 21:46

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