Posted by Danny on Friday, December 2, 2011 at 10:54am.
A merrygoround is a common piece of playground equipment. A 2.4 m diameter merrygoround with a mass of 300 kg is spinning at 22 rpm. John runs tangent to the merrygoround 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 merrygoround's angular velocity, in rpm, after John jumps on?

Physics  drwls, Friday, December 2, 2011 at 11:17am
The angular momentum of merrygoround PLUS John remains the same before and after he jumps on.
The moment of inertia of the merrygoround is
I = (1/2) M R^2 = 216 kg*m^2.
The initial angular velocity of the merrygoround 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 MerryGoRound speeds up.
Finally, convert the w2 value to rpm.
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