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I am currently working on a practice problem for my upcoming exam and I have difficulties getting my head around moment of inertia.

If the ball has mass $m$ and is going around in a circle with velocity $v_0$ then I can determine it's angular momentum.

$\vec{L}=m(r_0\times v_0)$

Now suppose someone pulls on the string until the ball goes around the circle with radius $\frac{r_0}{2}$. What is it's new velocity?

I assumed angular momentum is conserved (I am not sure about this).

$\implies \vec{L_1}=\vec{L_2} \iff m(r_0\times v_0)=m(\frac{r_0}{2}\times v_1)$

$\iff r_0\times v_0=\frac{r_0}{2}\times v_1 \iff |r_0||v_0|\sin(\alpha)=|\frac{r_0}{2}||v_1|\sin(\alpha)$

I also assumed that the angle between any $r$ and $v$ would not change (also not sure about that one)

$\implies v_1=2v_0$

This would mean that half the radius implies double the velocity. I am somewhat puzzled by this because $v=\omega r$ tells me that the velocity should be half of the original velocity unless $\omega$ has changed. This brings me back to moment of inertia. Did moment of inertia change when the ball went from $r_0$ to $\frac{r_0}{2}$? This would explain the change in angular velocity since $\vec{L}=I\cdot \vec{\omega}$.

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The moment of inertia has indeed changed, since in general it is $mr^2$, so varying the distance $r$ from the axis varies the moment of inertia. This is what happens to ice skaters when they spin and contract their arms https://www.youtube.com/watch?v=AQLtcEAG9v0. They change their moment of inertia while conserving the acquired angular momentum, resulting in a different angular velocity.

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  • $\begingroup$ Thanks. I am not sure about this but don't I apply some sort of force(torque?) when pulling on the string $\implies$ angular momentum is not conserved? $\endgroup$ – qmd Feb 2 '15 at 12:24
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    $\begingroup$ not if the force is radial, for in that case $\mathbf F\times\mathbf r = 0$. $\endgroup$ – Phoenix87 Feb 2 '15 at 12:26
  • $\begingroup$ Ah I see. So angular momentum only changes if there is a torque? Also, don't I perform work if I pull on the string? If yes, wouldn't that also imply that there is an external force acting on the system? Sorry for asking so many questions but this topic confuses me so much. $\endgroup$ – qmd Feb 2 '15 at 12:33
  • $\begingroup$ yes $\mathbf M = \sum\mathbf F\times\mathbf r=\dot{\mathbf L}$. There is an external force pulling the string, but the system is constructed in such a way that this force will act radially, and this won't change the angular momentum. $\endgroup$ – Phoenix87 Feb 2 '15 at 12:39

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