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    ihatephysics111's Avatar
    ihatephysics111 Posts: 3, Reputation: 2
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    #1

    Dec 2, 2008, 04:41 PM
    Physics- Torque on a Merry-go-round
    A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 15 rpm in 9.6 s. Assume the merry-go-round is a uniform disk of radius 2.3 m and has a mass of 730 kg, and two children (each with a mass of 27 kg) sit opposite each other on the edge.Calculate the torque required to produce the acceleration, neglecting frictional torque. Then find the force required at the edge.

    What I have established / found (I think)

    Vi= 0
    Vf= .25 rps

    Moment of Inertia of a uniform disk is I=1/2 M R^2
    Moment of Inertia of the merry-go-round without children = 1930.85
    Moment of Inertia with children = 2073.68

    I know net torque is Tnet = I * alpha, but I don't know what alpha is...

    I also know that to find the force it is Fnet = (I * alpha)/r

    or so I think...

    so what is alpha and how do I find it? Help please! :confused:
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Dec 3, 2008, 02:30 PM

    Alpha is the rotational acceleration, typically in units of radians per second per second. You have found that the final rotational velocity is 0.25 rps, which is equivalent to pi/2 radians/sec. It took 9.6 s to accelerate up to this speed, so assuming constant acceleration:

    alpha = (pi/2 radians/sec) /9.6 sec

    Remember that for rotational veocity and acceleration:


    This is analogous to the linear acceleration, where for the case of constant acceleration velocity and acceleration are related by: v = at.

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