# Rotational Dynamics on a stool

1. Mar 13, 2006

### 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 $$\omega[\tex] and I think using [tex]L=I\omega[\tex] should be the key. Only L and I are unknown..any ideas? 2. Mar 13, 2006 ### topsquark Well, there are no net external torques on the system so it sounds like angular momentum should be conserved. -Dan 3. Mar 13, 2006 ### sauri So L(initial)=L(final), so we find the kinertic energy (Iw^2)/2, find I and and substitute it to L=Iw and find w? 4. Mar 13, 2006 ### Hootenanny Staff Emeritus That's about it. Just a quick note on latex, at the end you need to use$$ 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 $$5. Mar 13, 2006 ### sauri When trying to find I from the k.e equation there was a prob that I ran into. The kinetic energy of the system is not given. K.E=(Iw^2). I donot see how (mv^2)/2 can be used as we do not know v, only w is known. 6. Mar 13, 2006 ### BobG It looks like you have to make some assumptions about the persons dimensions and the length of his arms. Rotational kinetic energy is [tex]\frac{1}{2}I \omega^2$$

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

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

7. Mar 13, 2006

### 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 whats $$I = mr^2$$ for?

8. Mar 13, 2006

### BobG

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