You do not say what the moment of inertia of the Merry Go Round is.
I will call it I
39.4 revs/min (2 pi rad/rev)(1 min/60s) = 4.13 radians /s = W1
omega of John = v/r = 6.36/3.21
= 1.98 rad/s
Initial angular momentum
= I (4.13) + 48.4(3.21)^2 (1.98)
Final I = I + 48.4(3.21)^2
final omega = Initial angular momentum / Final I
A merry-go-round is a common piece of playground equipment. A 6.42m diameter merry-go-round with a mass of 648.0kg is spinning at 39.4rpm. John runs tangent to the merry-go-round at 6.36m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 48.4kg. What is the merry-go-round's angular velocity, in rad/s, after John jumps on?
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