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Applied torque/angular momemtum

  • #1
5
0
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

d)
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!
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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Looks good. As a check, what would you get for the torque using ##\tau = I\alpha##?
 
  • #3
5
0
Looks good. As a check, what would you get for the torque using ##\tau = I\alpha##?
Thanks, that's a great check!
 

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