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A child exerts a tangential 43.5-N force on the rim of a disk-shaped merry-go-round with a radius of 2.41 m. If the merry-go-round starts at rest and acquires an angular speed of 0.086 rev/s in 3.54 s, what is its mass?

  • Physics -

    Applied torque = 43.5 * 2.41 = 104.84 Newton-meters

    Angular acceleration rate =
    0.086*2*pi/3.54 = 0.1526 rad/s^2

    The moment of inerti, I , is torque divided by angular acceleration.

    I = 687 kg*m^2

    For a uniform disc, The moment of inertia is

    I = (1/2) M R^2.

    Use that equation and the known value of I so solve for M.

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