The rate of rotation of a solid disk with a radius of 4 m4m and mass of 5 kg5kg constantly changes from 12 Hz12Hz to 5 Hz5Hz. If the change in rotational frequency occurs over 12 s12s, what torque was applied to the disk?

1 Answer
Ananda Dasgupta
Jul 18, 2016

#183.3 "N m" clockwise.

Explanation:

Although it doesn't say so in the problem, it seems safe to assume that the disk is rotating about an axis passing through its center and normal to it, In this case the moment of inertia about this axis is given by 1/2 M R^2 = 50 " kg m"^212MR2=50 kg m2. Also we are going to use the standard convention that counter-clockwise rotation is positive.

To calculate the angular momentum we use L=I omegaL=Iω where omega=2 pi nuω=2πν is the angular frequency (nuν being the frequency). Thus, change in the angular momentum of the disk is

Delta L = I omega_2 - I omega_1 = 2 pi I (nu_2-nu_1) = -700 pi " kg m"^2 "s"^{-1}

Thus the torque is

tau = {Delta L}/{Delta t} = - -700/12 pi " kg m"^2 "s"^{-2} ~~ -183.3 "N m"

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