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Angular Momentum of a sanding disk

  1. Nov 15, 2007 #1
    A sanding disk with rotational inertia 1.22x10^-3 kgm^2 is attached to an electric drill whose motor delivers a torque of 15.8 Nm
    a) find angular momentum
    b) find angular speed of the disk 33.0ms after the motor is turned on.

    L = Iω
    τ = Iα

    we can find angular acceleration with what we are given α = τ/I
    but I'm completely stumped on how to find L without having time.
    any hints?

    thanks
     
  2. jcsd
  3. Nov 15, 2007 #2

    Doc Al

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    Staff: Mentor

    Here's a hint: You are given the time! :smile:
     
  4. Nov 15, 2007 #3
    lol man I stared at that over and over again and thought, "I must be misreading this...wheres the trick?"

    But seriously, yes you are given the time. its "33.0ms" part
     
  5. Nov 15, 2007 #4
    Haha, thanks
    I realized from that that I can use the time to solve the first part of my problem.

    L = Iω

    L =τω/α
    ω = αt
    L =τt

    that works

    τt = Iω
    ω = τt/I
    which gives me 427m/s
    but i need the answer in rev/min
    im okay with seconds to minutes, but how do i convert meters to revolutions?
    2pi?
     
  6. Nov 15, 2007 #5
    If you can find its angular acceleration, then ( just like linear kinematics... ) its angular velocity is equal to its angular acceleration multiplied by the duration of the acceleration.
     
  7. Nov 15, 2007 #6

    Doc Al

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    Staff: Mentor

    The units for ω are radians/sec, not m/s. One revolution equals [itex]2 \pi[/itex] radians.
     
  8. Nov 15, 2007 #7
    woo i got it!

    thanks
     
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