How Does Pulling Arms In Affect Angular Velocity on a Rotating Stool?

[sauri]

A skinny student (40 kg) sits on a stool that can rotate, holding two 8 kg bowling balls with his arms out stretched. The stool weighs 20 kg. The student and stool are spun such that they have an angular velocity of 2 rad.s-1.
When the student pulls his arms in towards him, what happens to his angular velocity and what is his final angular velocity when they rest in his lap?
I know I have to find [tex]\omega[\tex] and I think using $$L=I\omega[\tex] should be the key. Only L and I are unknown..any ideas?$$[/tex]

 

[topsquark]

A skinny student (40 kg) sits on a stool that can rotate, holding two 8 kg bowling balls with his arms out stretched. The stool weighs 20 kg. The student and stool are spun such that they have an angular velocity of 2 rad.s-1.
When the student pulls his arms in towards him, what happens to his angular velocity and what is his final angular velocity when they rest in his lap?
I know I have to find [tex]\omega[\tex] and I think using $$L=I\omega[\tex] should be the key. Only L and I are unknown..any ideas?$$[/tex]

[tex]$$<br /> Well, there are no net external torques on the system so it sounds like angular momentum should be conserved.<br /> <br /> -Dan$$[/tex]

 

[sauri]

So L(initial)=L(final), so we find the kinertic energy (Iw^2)/2, find I and and substitute it to L=Iw and find w?

 

[Hootenanny]

That's about it. Just a quick note on latex, at the end you need to use [/tex] not [\tex], apart from that, your code is correct. If you want to put code on the same line use the [itex]tags instead of $$<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />$$[/itex]

 

[sauri]

When trying to find I from the k.e equation there was a prob that I ran into. The kinetic energy of the system is not given. K.E=(Iw^2). I don't see how (mv^2)/2 can be used as we do not know v, only w is known.

 

[BobG]

When trying to find I from the k.e equation there was a prob that I ran into. The kinetic energy of the system is not given. K.E=(Iw^2). I don't see how (mv^2)/2 can be used as we do not know v, only w is known.

It looks like you have to make some assumptions about the persons dimensions and the length of his arms.

Rotational kinetic energy is $$\frac{1}{2}I \omega^2$$

Moment of Inertia (I) is $$I = mr^2$$

Linear velocity (v) is $$v = r \omega$$

 

[sauri]

where r=2(pi) right?. Then we find v and sbstitute to (mv^2)/2=(Iw^2)/2 and find I which we sunstitute into L=Iw and find L right?. so what's $$I = mr^2$$ for?

 

[BobG]

where r=2(pi) right?. Then we find v and sbstitute to (mv^2)/2=(Iw^2)/2 and find I which we sunstitute into L=Iw and find L right?. so what's $$I = mr^2$$ for?

r is the radius.

$$KE = \frac{1}{2}m v^2$$
$$= \frac{1}{2}m (r \omega)^2$$
$$= \frac{1}{2}m r^2 \omega^2$$
$$= \frac{1}{2}I \omega^2$$

The $$I = m r^2$$ only applies for a point-mass or ring (since all the units of mass have the same I). The moment of inertia for anything else is a sum of the moments of inertia for all of the point-masses that make up the object.

For the original problem, you need to model the person-chair as a cylinder and find the moment of inertia for a cylinder and model the weights as point masses. You can derive the equation for the moment of inertia for a cylinder yourself ($$I = \Sigma m_i r_i^2$$) or look up the formula in a book or on-line.