# Find the angular velocity

1. Apr 16, 2012

### november1992

1. The problem statement, all variables and given/known data

A ring (mass 2 M, radius 1 R) rotates in a CCW direction with an initial angular speed 1 ω. A disk (mass 2 M, radius 2 R) rotates in a CW direction with initial angular speed 4 ω. The ring and disk "collide" and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

What is the initial angular momentum Li of the ring+disk system? Write your answer in terms of MR2ω.

What is the final angular velocity ωf of the ring+disk system? Write your answer in terms of ω.

2. Relevant equations

I= βM$R^{2}$
L=I*ω
$L_{ring+disk}$=$I_{ring}$*$ω_{ring}$+$I_{disk}$*$ω_{disk}$
$L_{i}$=$L_{f}$=I*$ω_{f}$
β for disk = 1/2
β for ring = 1

3. The attempt at a solution

I got -14M$R^{2}$ for the initial angular momentum but I don't know how to find the final angular velocity

Last edited: Apr 16, 2012
2. Apr 17, 2012

### MrWarlock616

How did you get $-14M R^2$ ?
Here's what I did..I'm not sure it's right. Some expert will correct me..
$\omega_r = 1 \omega$

$\omega_d = - 4 \omega_r$ //opposite in direction to $\omega_r$

$I_r=M R^2$

$I_d=\frac{1}{2} M R^2$

$L_{ring+disc} = L_r + L_d = (M R ^2)(\omega_r) + (\frac{1}{2} M R^2)(-4 \omega_r)$

$\therefore L_{ring+disc} = M R^2 \omega_r - 2 M R^2 \omega_r = - M R^2 \omega_r$

$I_r + I_d = I_f=\frac{3}{2} M R^2$

$L_f=I_f \omega_f$

$\omega_f=\frac{L_f}{I_f} = \frac{- M R^2 \omega_r}{\frac{3}{2} M R^2} = - \frac{2}{3} \omega_r=- \frac{2}{3} \omega$

Someone correct me please if I went wrong..

Last edited: Apr 17, 2012
3. Apr 17, 2012

### Steely Dan

Now that you have the initial angular momentum, you've correctly realized that this equal to the product of the final angular velocity and the moment of inertia. Probably what's throwing you is how to calculate the moment of inertia. Well, if the two objects are sticking together, they will have the same angular velocity, so you can just add the two moments of inertia.

To MrWarlock: you've neglected in your moments of inertia that the masses and radii are not $M$ and $R$ for both objects.

4. Apr 17, 2012

### MrWarlock616

OHH sorry missed that ...ok i corrected it..

$\omega_r = 1 \omega$

$\omega_d = - 4 \omega_r$ //opposite in direction to $\omega_r$

$I_r = 2 M R^2$

$I_d = \frac{1}{2} 2 M (2 R)^2 = 4 M R^2$

$L_{ring+disc} = L_i = L_r + L_d = (M R ^2)(\omega_r) + (4 M R^2)(-4 \omega_r)$

$\therefore L_i = 2 M R^2 \omega_r - 16 M R^2 \omega_r = - 14 M R^2 \omega_r$

$I_r + I_d = I_f=\frac{3}{2} M R^2$

$L_f=I_f \omega_f = L_i$

$\omega_f=\frac{L_i}{I_f} = \frac{- 14 M R^2 \omega_r}{\frac{3}{2} M R^2} = - \frac{28}{3} \omega_r=- \frac{28}{3} \omega$

Is it right??? is it??

5. Apr 17, 2012

### Steely Dan

Not quite, you forgot to correct the total moment of the inertia in the final state with the updated information. But instead of posting the correct answer, let's leave something for the OP to do, eh?

6. Apr 17, 2012

### MrWarlock616

AARGGH $\frac{7}{3} \omega$, is it? xD