Angular momentum of merry go round

AI Thread Summary
The discussion revolves around calculating the final angular velocity of a merry-go-round after a runner jumps onto it. The runner has a mass of 36 kg and runs at 2.9 m/s, while the merry-go-round has a moment of inertia of 404 kgm^2 and is initially at rest. The key formula used is angular momentum conservation, expressed as mvr = (Imgr + Ir)w, where Imgr is the moment of inertia of the merry-go-round and Ir is the moment of inertia of the runner. The confusion arises regarding the correct calculation of the runner's moment of inertia, which should be determined based on his distance from the center of rotation. Understanding how to accurately compute the total moment of inertia for the system is essential to arrive at the final angular velocity of 0.381 rad/s.
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This question was asked a while ago by another user, but I do not understand how he got the answer:

"A runner of mass m=36 kg and running at 2.9 m/s runs and jumps on the rim of a playground merry-go-round which has a Moment Of Inertia of 404 kgm^2 and a radius of 2 m. Assuming the merry-go-round is initially at rest, what is its final angular velocity to three decimal places?

"According to the back of the book, the answer is 0.381 rad/s; however, I can never come up with that answer."

He says he used the formula mvr = (Imgr + Ir)w where Imgr is the inertia of the merry-go-round and Ir is the inertia of runner, then solved for w and was able to get the correct answer. When I plug in the numbers, I don't get the right answer at all. Is he using a different "I" value for the runner, if so how does he calculate it? Please help!
 
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angular momentum is given by L=I \omega = mvr

Angular momentum is always conserved. So that the angular momentum as the runner jumps on the merry go round should be the same as the angular momentum as merry go round and the runner.

So, the moment of inertia of the merry-go-round alone is 404 kgm^2, what is the moment of inertia of the runner at a distance of 2m from the centre of rotation of the merry-go-round? Then what would be the total moment of inertia of the entire system?
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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