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Determining torque required

  1. Nov 12, 2006 #1
    Problem 2) What is the angular momentum of a 2.8 kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1500 rpm?
    b) How much torque is required to stop it in 7.0 s?

    1500 rpm * 2pi rad = 9424.78 rad/m
    (9424.78 rad/m) /60 s = 157.08 rad/s

    I = 1/2mr2
    I = 1/2(2.8 kg)(.18 m)2 = 0.04536 kgm2

    L = Iw = (0.04536 kgm2)(157.08 rad/s) = 7.13 kgm2/s

    I have determined the angular momentum, but I am unsure of how to determine the torque required to stop it in 7 s

    How do you determine this with the torque formula and time involved? Any help?
  2. jcsd
  3. Nov 12, 2006 #2
    The definition of torque is

    [tex] \mathbf\tau = \frac{d\mathbf{\mathrm L}}{dt} [/tex]

    If you want to change the angular momentum from your value to 0, then the average torque over 7 s most then be?
    Last edited: Nov 12, 2006
  4. Nov 12, 2006 #3
    t=(7.13 kgm^2/s)/7.0 s=1.02 kgm^2

  5. Nov 12, 2006 #4
    hmm....well you know the initial angular velocity ''157.08 rad/s''. I would say you need to calculate the angular deceleration of the spinning wheel from 157.08 r/s to 0 r/s in 7 secs using W(final)=W(initial)+a^t. I make the angular deceleration 22.44 rad s-2.

    Then using the formula T=I x a......cha ching !
  6. Nov 12, 2006 #5
    a=-22.44 rad/s^2 since stopping

    T=0.04536 kgm^2*-22.44
    T=-1.02 m*N
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