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 25 rpm. John runs tangent to the merry-go-round at 5.8 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg.

Answer 1

w2 = 2.83 rad/s

Step-by-step explanation:

The moment of inertia of the merry-go-round is

I = (1/2) M R^2

I = 1/2 * 300 kg * 1.5 m^2

I = 337.5 kg*m^2

The initial angular velocity of the merry-go-round is

w1 = 25 rpm *2*pi /60

w1 = 2.6 rad/s

The angular momentum conservation equation is:

I*w1 + m*R*v = (I + mR^2)*w2

where m is John's mass.

337.5*2.6 + 30*1.5*2.6 = (337.5 + 30*(1.5)^2)

887.5 + 261 = (337.5 + 67.5)*w2

w2 = 2.83 rad/s