Ratio of masses of merry-go-round and the boy

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SUMMARY

The discussion focuses on calculating the ratio of the masses of a boy (mboy) and a merry-go-round (mmgr) based on their rotational dynamics. Given a radius of R = 3.4 m, the boy walks 25 m along the edge, causing the merry-go-round to rotate through an angle of 50°. Using the equations of rotational inertia and angular velocity, the final ratio of mboy to mmgr is determined to be -1, indicating that the masses are equal in magnitude but opposite in direction due to the nature of the system's rotation.

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Homework Statement


A boy of mass mboy stands at the edge of a merry-go-round of radius R = 3.4 m and mass mmgr, and both are initially at rest. The boy then walks along the edge of the merry-go-round. After walking a distance of 25 m relative to the merry-go-round, the boy finds that the merry-go-round has rotated through an angle of 50°. Find the ratio of mboy to mmgr.



Homework Equations


I_B*W_B=I_M*W_M
mr^2*delta theta/delta time=0.5*mr^2* delta theta/delta time


The Attempt at a Solution

 
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I_B=0.5mboy*(3.4m)^2I_M=0.5mmgr*(3.4m)^2W_B=2pi/25m *50 degreesW_M=-2pi/25m *50 degreesI_B*W_B=I_M*W_M(0.5mboy*(3.4m)^2)*(2pi/25m *50 degrees)=(0.5mmgr*(3.4m)^2)*(-2pi/25m*50 degrees)mboy/mmgr=(-3.4m)^2/(3.4m)^2mboy/mmgr=-1
 

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