Question

A 2.8-m -diameter merry-go-round with a mass of 230 kg is spinning at 20 rpm . John runs around the merry-go-round at 5.0 m/s , in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 35 kg . What is the merry-go-round's angular speed, in rpm, after John jumps on?

Answer #1

intial angular speed of merry go round = 2pi*N/60

= 2pi*20 / 60 = 2.1 rad/s

initial moment of inertia of the merry go round = MR^2 / 2

= M*(D/2)^2 / 2

= 230*(2.8/2)^2 / 2

= 225.4 kg-m^2

Moment of inertia of John = mr^2

= 35*(2.8/2)^2

= 68.6 kg-m^2

angular speed of john = v/r = 5/(2.8/2) = 3.57 rad/s

By the conservation of angular momentum we have

initial angular momentum = final angular momentum

I1w1 + I2w2 = (I1 + I2)*w

225.4*2.1 + 68.6*3.57 = (225.4 + 68.6)*w

w = 2.443 rad/s

2pi*N/60 = 2.443

N = 23.3 rpm

so new angular speed is **23.3 rpm**

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