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What is the required power?

  1. Mar 30, 2005 #1
    Hello, please help me with this problem. thanks :mad:

    A 22.0 kg wheel, essentially a thin hoop with radius 1.00 m, is rotating at 230 rpm. It must be brought to a stop in 10 s. How much work must be done to stop it? What is the required power?
     
  2. jcsd
  3. Mar 30, 2005 #2

    xanthym

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    Science Advisor

    From problem statement:
    {Wheel Mass} = M = (22.0 kg)
    {Wheel Radius} = R = (1.00 m)
    (Wheel Rotation Frequency} = f =(230 rpm) = (3.83333 rev/sec)
    {Wheel Angular Speed} = ω = 2*π*f = 2*π*(3.83333 rev/sec) = (24.0855 radians/sec)
    {Thin Hoop Moment of Inertia} = I = M*R2 = (22.0 kg)*(1.00 m)2 = (22 kg*m2)
    {Required Stopping Time} = ΔT = (10 sec)

    The wheel's Rotational Kinetic Energy is given by:
    {Wheel Rotational Kinetic Energy} = (1/2)*I*ω2 =
    = (1/2)*(22 kg*m2)*(24.0855 radians/sec)2 =
    = (6381.22 Joules)
    ::: ⇒ {Work Required To Stop} = (6381.22 Joules)

    {Ave Power Required To Stop} = {Work Required To Stop}/ΔT =
    = (6381.22 Joules)/(10 sec) =
    = (638.122 Watts)


    ~~
     
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