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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 -

    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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