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Angular work problem

  1. Mar 27, 2005 #1
    A 42.0 kg wheel, essentially a thin hoop with radius 0.90 m, is rotating at 250 rpm. It must be brought to a stop in 10 s. How much work must be done to stop it.

    I used the equation delta K = 1/2 I w^2_f - 1/2 I w^2_i = W and got - 11676 J, but this is not right. Help?
     
  2. jcsd
  3. Mar 27, 2005 #2

    xanthym

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    Science Advisor

    From problem statement:
    {Hoop Mass} = (42.0 kg)
    {Hoop Radius} = (0.9 m)
    {Moment of Inertia for Thin Hoop} = M*R^2
    {Frequency of Rotation} = f = (250 RPM) = (4.1667 rev/sec)
    {Angular Velocity} = ω = 2*π*f = 2*π*(4.1667) = (26.18 radians/sec)

    {Rotational Kinetic Energy} = (1/2)*I*ω^2 =
    = (1/2)*(M*R^2)*ω^2 =
    = (1/2)*(42.0)*{(0.9)^2}*{(26.18)^2} =
    = (11659 J)


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