Posted by **Danny** on Friday, December 2, 2011 at 10:54am.

A merry-go-round is a common piece of playground equipment. A 2.4 m diameter merry-go-round with a mass of 300 kg is spinning at 22 rpm. John runs tangent to 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 25 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

- Physics -
**drwls**, Friday, December 2, 2011 at 11:17am
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 R^2 = 216 kg*m^2.

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

w1 = 22*2*pi/60 = 2.30 rad/s

The angular momentum conservation equation is:

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

where m is John's mass.

498 + 150 = (216 + 36)*w2

w2 = 2.57 rad/s

So the Merry-Go-Round speeds up.

Finally, convert the w2 value to rpm.

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