Conservation of Angular Momentum with Changing mass

[cozzbp]

I am completely stuck on this problem, and was wondering if anyone could point me in the right direction.

Homework Statement

In the spinning fireworks display shown in the figure the ignited gun powder is ejected from the jets at a speed of 24.3 m/s.
http://volta.byu.edu/ph121/homework/hw18f2.png

The jets are a distance of 0.462 m from the center of the fixture. If the total mass of gunpowder from all four jets is 250 g and the total mass of the jets themselves is 472 g, what will the angular velocity in radians per second of the fixture be when the gunpowder is exhausted? Neglect air resistance and assume that the mass of the arms of the fixture is inconsequential. Take the counter-clockwise direction (out of the page) as positive.

Homework Equations

L = mvr
L = I$$\omega$$

The Attempt at a Solution

Here is what I have tried thus far.
I thought that I could simply calculate the Angular Momentum (L) by doing the following:
L = 4*((m_fuel/4 + m_rocket/4) * v * r)
4*(.1805 * 24.3 * .462) = 8.105
then I simply subtracted the mass of the fuel for one rocket, and solved for v
8.105=4*(.118*v*.462)
This apparently does not yield the correct answer.

Any help would be greatly appreciated!

 

[disillusion]

first line L=mvr. v is the velocity of the gas, not the jet.

Also you need to consider the total angular momentum before and after.

 

[LowlyPion]

Welcome to PF.

Consider that there is a conservation of angular momentum. Initially, it is 0, so that suggests that after burn it will be the same won't it?

So won't that mean that

m_p*v*r = I * ω

where I = m_r*r²

m_p would be the mass of the gunpowder ejected
m_r would be the mass of the rocket pods.

Simplifying then doesn't that give

ω = (m_p * v)/(m_r * r)

 

[cozzbp]

Thank you both very much. That is exactly what I needed! My only question would be how did you know what the moment of inertia would be for that system?

 

[disillusion]

I=mr² for every point mass on the system. r is the distance between the point mass and the axis of rotation.

In here you can just assume the outlets are point masses since the arms are light.
But in general you need to sum up all the point masses so
$$I=\sum m_{i}r_i^{2}$$

 

[cozzbp]

Oh ok, that makes sense. Thanks again!