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A 36.0 kg child runs with a speed of 2.80 m/s tangential to the rim of...

A 36.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round(Figure 1) . The merry-go-round has a moment of inertia of 518 kg⋅m2 and a radius of 2.31 m . When the child jumps onto the merry-go-round, the entire system begins to rotate. Calculate the final kinetic energy of the system.

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Answer #1

angular momentum conservation

Li = Lf

mvR = IW ( I = 518+ mR^2)

36*2.80*2.31 = ( 518 + 36*(2.31)^2)*W

=> W = 0.327 rad/s

KE = 1/2*IW^2 = 1/2*( 518 + 36*(2.31)^2)*(0.327)^2 = 37.96 J

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