Assistance with physics problem please!?

Question

A 3.0 m diameter merry-go-round with a mass of 250 kg is spinning at 20 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 30 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

Answer 1

This is conservation of angular momentum analogous to translational momentum, where there is an inelastic collision
I1*w1+I2*w2=(I1+I2)wf

The merry-go-round has I=.5*m*r^2
or .5*250*1.5^2
or 281.25
To convert 20 rpm to rad/sec multiply by 0.10472
20 rpm = 2.1 rad /sec

John can be considered as a point mass with I=m*r^2
at the edge his I is
30*1.5^2
67.5
His angular speed is w=v/r
w=5/1.5
or 10/3

Setting up the equations

281.25*2.1+67.5*10/3=(281.25+67.5)*wf

wf=2.339 rad/sec
or 22.33 rpm

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