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 $\vec{L}=I\cdot \vec{\omega}$.
Best Answer
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.