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Torque Req. for Fixed Rot. Acc.

  1. Jun 26, 2006 #1
    I'm having difficulty with this question:

    A day-care worker pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 10.1s. Assume the merry-go-round is a disk of radius 2.30m and has a mass of 830kg, and two children (each with a mass of 25.4kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

    Alright, this is my process; although, I know my end result is wrong:

    I used the angular kinematic equation to solve for the angular acceleration.

    f = ωi + αΔt)

    I got α = 0.186629 rad/s² (approx.) after converting from rpm's.

    The equation I have for torque is:

    τ = mr²α

    But, since we're working with a disk, I = ½MR².

    Therefore, I solved for torque using the equation:

    τ = ½MR²α

    I'm moderately confident with myself at this point, although I realize I can be completely off, but I think I'm screwing up what to use for mass.

    I plugged in the mass of the merry-go-round plus the mass of the two children.

    M = 880.8 kg

    Most likely, this is where my reasoning is flawed. I've just recently been introduced to torque, and it is honestly confusing me.

    Anyway. The answer I got:

    τ = ½MR²α = τ = ½(880.8 kg)(2.3)²(0.186629 rad/s²) =

    434.79 N*m

    This is the wrong answer, I know. But it is the best I could come up with based on the information my textbook is giving me. Any advice would be helpful.

     
  2. jcsd
  3. Jun 26, 2006 #2

    nrqed

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    For a point mass, I is MR^2 (where R is the distance from the point mass to the axis of rotation). For a disk, the moment of inertia is 1/2 MR^2. What you have to do here is to calculate the total moment of inertia, with is [itex] I_{total}=I_{child#1} + I_{child#2} + I_{disk} [/itex]

    Use this for the total moment of inertia. barring any algebra mistake, this should work.

    Patrick
     
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