# At wits ends please someone help

1. Nov 22, 2004

### americanangel

At wits ends please someone help!!!

Okay I've been going at this problem for the last three days...I went to the physics tutors at school and they were no help!!! Please someone help me with this one!!!!
Here's the problem: A woman of mass m stands at the edge of a solid cylinderical platform of mass M and radius R. At t=0, the platform is rotating with negligible friction at an angular velocity w about a fixed vertical axis through its center and the person begins walking with speed v toward the center of the platform. Determine the angular velocity of the system as a function of time and what will be the angular velocity when the women reaches the center?

Now i know this has to deal with momentum conservation and the change in the moment of inertia for each situation. But I can't get that down on paper!!!

2. Nov 22, 2004

### AKG

Let L represent the angular momentum for the system
Let I represent the moment of inertia for the system
Let w represent the angular momentum of the sytem

L = Iw
w = L/I
By conservation of angular momentum, L = L1, the initial angular momentum (which you should be able to easily compute)
w = L1/I
I is a simple function of time. It has two terms that are added together: one that remains constant corresponding to the platforms moment of inertia, and a second corresponding the the woman's moment of inertia, which varies with time. Basically, the woman's moment of inertia (treat her as a point mass) is a function of the distance from the center, and since you know her speed, you can express her distance from the center as a function of time.

3. Nov 22, 2004

### americanangel

What I'm not understanding is the idea of the angular velocity of the system? Let me show you what I have so far and maybe it can help!

I said that the L(person)=(mR^2)(v/R) and L(merry go round)=(1/2MR^2)w. Now from momentum conservation, since there is no outside torque and it started from rest (correct?) then this should be equal to zero. But there is no time involved? and how do you get an additive angular velocity for the whole system when I have it in two different pieces?

4. Nov 22, 2004

### AKG

Variables

$L(t)$ represents angular momentum of the system
$I_w(t)$ represents moment of inertia for the woman
$I_p(t)$ represents moment of inertia for the platform
$I(t)$ represents moment of inertia for the system
$\omega (t)$ represents initial angular velocity of the system
$d(t)$ represents the woman's distance from the center

The above are all functions of time

$t$ is the time elapsed from the start

Now, note that $L(t) = L(0)$ and $I_p(t) = I_p(0)$, i.e. they are constant.

So:

$$L(t) = I(t)\omega (t)$$

$$L(0) = (I_p(0) + I_w(t))\omega (t)$$

$$\omega (t) = \frac{L(0)}{I_p(0) + I_w(t)}$$

$$\omega (t) = \frac{I(0)\omega (0)}{I_p(0) + I_w(t)}$$

$$\omega (t) = \frac{(I_p (0) + I_w (0))\omega }{I_p(0) + md^2}$$

The above should really be more than enough. You should be able to compute $I_p (0)$ and $I_w (0)$, and should be able to easily express $d$ in terms of $t$.

5. Nov 22, 2004

### americanangel

okay i see what you did now! I think i was missing the connection...i knew what you were talking about but putting it onto paper was the difficult part...I figured out Iw(o) and Ip(O) and just plugged everything back in and I actually got what i had before (from the second time three days ago) which is odd because I did it a lot different!
Thanks for the help though...i really appreciated it