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Applied torque/angular momemtum

  1. Jun 9, 2016 #1
    This is rewording of a question on a test I've just done.

    A vinyl record on a turntable has radius R=0.15 m, mass M=1.5 kg. The angular speed is reduced from 33.3 rev per minute to zero as a result of an applied torque, in 7 seconds.

    Moment of inertia given as I=1/2 M R^2

    Calculate a) angular speed b) Magnitude of angular deceleration c)Moment of inertia d) Magnitude of applied torque.

    My attempt.

    a) angular speed, w = 33.3 x 2 pi / 60 s = 3.5 rad s^-1

    b) a= dw/dt = 3.5/7 = 0.50 rad s^-2

    c) I = 0.5 x 1.5 kg x 0.15^2 = 0.017 kg m^2

    angular momentum L = I w = 3.5 rad s^-1 x 0.017 kg m^2 = 0.059

    torque = dL/dt = 0.059/7 s = 0.0084 N m

    Does this look right (ignore any rounding errors). Seems quite a small magnitude for the torque.

    Anyway, thanks for any help!
  2. jcsd
  3. Jun 9, 2016 #2


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    Gold Member

    Looks good. As a check, what would you get for the torque using ##\tau = I\alpha##?
  4. Jun 9, 2016 #3
    Thanks, that's a great check!
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