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I came across a question regarding linear momentum $L$ and it's conservation, however I tried and got confused. It reads: A $40kg$ girl stands on the very edge of a rotating disc of mass $50kg$ and radius $2 meters$. She then throws a rock of mass $5kg$ at a speed of $10m/s$ which causes the rock to rotate in a clockwise direction. What is the angular velocity of the disc after the rock is thrown?

My attempt was to find the angular velocity caused by the rock using $v = wr$. I am unsure of how exactly to take into consideration the inertia of the girl disc system or if this is even the right approach. Also, since the disc would be rotating in a clockwise direction after the throw, would the angular velocity be negative? Please offer assistance.

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...causes the rock to rotate in a clockwise direction

I think that means the velocity vector of the rock is in a direction such as it were thrown out of a clockwise motion of the disc. In that case, the disc itself, along with the girl will rotate in an anti-clockwise direction.

The total angular momentum is $0$. The rocks angular momentum as it is thrown out of your hands is $mvr$, for mass and velocity of the stone and radius of the disc.

Divide the negative of this quantity with the moment of inertia of the girl + disc and you shall get your answer. The sign convention is pretty arbitrary. I would say you consider the anti-clockwise rotation as the one having positive angular velocity.

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I hope my interpretation is correct.

1) The girl/disc/rock system is initially at rest.

2) The girl throws the rock and it exits the girl/disc system, tangentially to the disc, at $10\frac{m}{s}$

3) The angular momentum of the rock equals $m*v*r=5*10*2$

4) The angular momentum of the whole system must be conserved at zero, so the girl/disc system must have an angular momentum that counters $5*10*2$.

The angular momentum of the girl/disc system will be:

$$H_{axis}=100=I_{disc}\omega_{disc} + I_{girl}\omega_{girl}$$

Are these enough hints?

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