Spinning Ice Skater Question

[xXhumans0monstersXx]

So basically, I was doing my AP Physics 1 homework and came across the spinning ice skater question yet again.

The question states, "An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?"

I already know that the kinetic energy increases, but can someone please explain to me why the angular momentum stays constant? I've done my research and I can't seem to find an explanation that I understand. I'm not the brightest when it comes to physics so if anyone could help that would be great! I have my first AP exam next week so a reply ASAP would be most appreciated. :)

 

[SteamKing]

So basically, I was doing my AP Physics 1 homework and came across the spinning ice skater question yet again.

The question states, "An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?"

I already know that the kinetic energy increases, but can someone please explain to me why the angular momentum stays constant? I've done my research and I can't seem to find an explanation that I understand. I'm not the brightest when it comes to physics so if anyone could help that would be great! I have my first AP exam next week so a reply ASAP would be most appreciated. :)

You should study more about angular momentum, specifically, conservation of the same:

https://en.wikipedia.org/wiki/Angular_momentum

 

[haruspex]

why the angular momentum stays constant? I've done my research and I can't seem to find an explanation

Let's start with linear momentum. When a force $\vec F(t)$ acts from body A on body B for a time the contribution to B's momentum is $\int \vec F.dt$. By the law of action and reaction, B exerts force $-\vec F(t)$ on A, so alters its momentum by $-\int \vec F.dt$ Thus the combined momentum is constant.
We can use the same for angular momentum. If the point of contact is at displacement $\vec r$ from the reference axis then taking the cross product of that with F(t) yields the angular moment and, on integrating, the change in angular momentum.