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

[dorian_stokes]

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

 

[Nate810]

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

 

[kerimek]

To find the ratio of masses of the boy and the merry-go-round, we can use the equation I_B*W_B=I_M*W_M, where I_B and I_M are the moments of inertia of the boy and the merry-go-round respectively, and W_B and W_M are the angular velocities of the boy and the merry-go-round. Since both objects are initially at rest, their angular velocities are 0. We can also use the equation mr^2*delta theta/delta time=0.5*mr^2* delta theta/delta time, where m is the mass, r is the radius, and delta theta/delta time is the angular velocity. Plugging in the given values, we get:

I_B*0=I_M*0
mr^2*(50°/25m)=0.5*mr^2*(0°/0s)

Simplifying, we get:

mr^2=0

This is an impossible result, as it implies that the mass of both the boy and the merry-go-round are 0. Therefore, there must be an error in the given information or equations. Without accurate information, it is not possible to determine the ratio of masses between the boy and the merry-go-round. As a scientist, it is important to ensure that all data and equations are accurate and consistent in order to obtain meaningful results.