Question

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round, which can be modeled as a disk with a mass of 300 kg , is spinning at 24 rpm. John runs tangent to the merry-go-round at 4.6 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg.

Answer 1

By angular momentum conservation principle, the angular momentum of merry-go-round PLUS John remains the same before and after he jumps on.

The moment of inertia of the merry-go-round is
I = (1/2) M R² = (1/2)*300 * (1.5)² = 337.5 kgm²

The initial angular velocity of the merry-go-round is
1 = 24*2*/60 = 2.513 rad/s

The angular momentum conservation equation is:
I*1 + m*R*v = (I + mR^2)*2
where m is John's mass.
337.5* 2.513 + 30* (1.5) * 4.6 = ( 337.5 + 30*(1.5)² ) 2
848.23 + 207 = (405)*2
2 = 2.605 rad/s = 24.87 rpm

So the Merry-Go-Round speeds up.