A person jumps on a merry go round, how much does the merry weigh?

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SUMMARY

The discussion focuses on calculating the mass of a merry-go-round after a person jumps onto it. The person has a mass of 52 kg and runs at a speed of 6.8 m/s before jumping onto the outer rim of the merry-go-round, which has a radius of 1.5 m. The final angular velocity of the merry-go-round is 1.3 rad/s. The initial momentum is calculated as Mp*Vp*R, while the final momentum is expressed as 1/2*Mm*R^2*w + Mp*R^2*w, confirming the setup for solving the mass of the merry-go-round (Mm).

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Homework Statement


A person with a mass of 52 kg runs with a speed of 6.8 m/s jumps onto the outer rim
of a merry-go-round. The merry-go-round has a radius of 1.5 m and can be modeled as a
large disk. The merry-go-round was initially at rest before the person jumped onto it and
rotates at 1.3 rad/s immediately after the person jumps on. What is the mass (in kg) of
the merry-go-round?


Homework Equations



Mm=Mass of merry go round
Mp=mass of person
R=radius of merry go round
Vp= velocity of person
w=angular velocity

The Attempt at a Solution


I just need someone to make sure I set this up right. I solved this using momentum. The initial momentum was Mp*Vp*r
The final momentum is 1/2*Mm*R^2*w+Mp*R^2*w
Solving I got 52*6.8*1.5=1/2*Mm*(1.5^2)*1.3+52*(1.5^2)*1.3

So does it look right to anyone?
 
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Looks right to me. (Assuming the person was moving tangent to edge of the merry go round when he jumped on.)
 

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