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Angular Acceleration of game show wheel

  1. Mar 15, 2005 #1
    A wheel on a game show is given an initial angular speed of 1.22 rad/s. It comes to rest after rotating through 3/4 of a turn. Find the average torque exerted on the wheel given that it is a disk of radius 0.71m and a mass 6.4kg.

    I'm lost. I don't really know where to begin. I know the answer is going to be 0.25Nm but I just need some help getting on the right track. Any suggestions?
     
    Last edited: Mar 15, 2005
  2. jcsd
  3. Mar 15, 2005 #2

    Andrew Mason

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    Use conservation of energy

    Work done = [itex]\tau \Delta \theta[/itex] = Energy available = [itex]\frac{1}{2}I\omega^2[/itex]

    AM
     
  4. Mar 15, 2005 #3

    xanthym

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    From the problem statement:
    {Disk Mass} = M = (6.4 kg)
    {Disk Radius} = R = (0.71 m)
    {Disk Moment of Inertia} = I = (1/2)*M*R2 = (1/2)*(6.4)*(0.71)2 = (1.613 kg*m2)
    {Initial Disk Angular Speed} = ω0 = (1.22 rad/s)
    {Final Disk Angular Speed} = ωf = (0.0 rad/s)
    {Disk Angular Rotation} = θ = (3/4 Turn) = (4.712 rad)
    {Torque Applied to Disk} = τ

    From Conservation of Rotational Kinetic Energy:
    {Rotational Work} = τ*θ = {Final Rotational Kinetic Energy} - {Initial Rotational Kinetic Energy}
    ::: ⇒ τ*θ = (1/2)*I*(ωf)2 - (1/2)*I*(ω0)2
    ::: ⇒ τ*θ = (1/2)*I*{(ωf)2 - (ω0)2}
    ::: ⇒ τ*(4.712 rad) = (1/2)*(1.613 kg*m2)*{(0.0 rad/s)2 - (1.22 rad/s)2}
    ::: ⇒ τ*(4.712) = (-1.2)
    ::: ⇒ |τ| = (0.255 N*m)


    ~~
     
  5. Mar 15, 2005 #4
    Thank you

    Thank you for the help. It's greatly appreciated :smile:
     
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