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Angular momentum sanding disk

  1. Oct 24, 2006 #1
    A sanding disk with rotational inertia 1.2E-3 kg m^2 is attached to an electric drill whose motor delivers a torque of magnitude 16 N m about the central axis of the disk. About that axis, and with the torque applied for 33 ms,
    (a) What is the magnitude of the angular momentum of the disk?

    I have τ= 16 N m and t= 0.033 s
    I= 1.2e-3 kg m^2

    In my book first of all there are two equations given for angular momentum (L)
    Those are L= Iw and L=rp and I'm not sure which one I'm suppose to use. I have I but not w and I don't have r or p and p=mv correct? So I feel like this is actually a lot easier than I'm making it out to be but I can't get it. Any help will be greatly appreciated.
     
  2. jcsd
  3. Oct 24, 2006 #2

    radou

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    Homework Helper

    Use the fact that M*t = Iw2 - Iw1 = change in angular momentum, assuming that the disk started from rest, where M is the torque.
     
  4. Oct 24, 2006 #3

    Doc Al

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    Staff: Mentor

    One way is to use Newton's 2nd law for rotation to find the angular acceleration of the disk while the torque is applied:
    [tex]\tau = I \alpha[/tex]

    Once you have the angular acceleration, treat it as a kinematics problem.

    (Even easier is to use angular impulse, as radou suggests, if you've covered that.)
     
  5. Oct 25, 2006 #4
    Thanks for the help. There's a second part of the question that asks for the angular speed in rev/min. The angular speed I calculated was 438.9 rev/s and I just converted that. Just wanted to know if this is correct.
     
  6. Oct 26, 2006 #5

    Doc Al

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    Staff: Mentor

    Standard units for angular speed are radians/sec. How did you calculate your value of rev/s?
     
  7. Mar 31, 2007 #6
    if you treat it as a kinematics problem, how do you actually find the rotational momentum?
     
  8. Apr 1, 2007 #7

    Doc Al

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    Staff: Mentor

    Once you've found the angular acceleration, as I describe in post #3, use it to find the angular speed at the end of the given time interval:
    [tex]\omega = \alpha \Delta t[/tex]

    The angular momentum is just [itex]I \omega[/itex].
     
  9. Apr 1, 2007 #8
    thank you Doc Al
     
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