The rate of rotation of a solid disk with a radius of #4 m# and mass of #5 kg# constantly changes from #12 Hz# to #5 Hz#. If the change in rotational frequency occurs over #12 s#, what torque was applied to the disk?

1 Answer
Jul 18, 2016

Answer:

#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"^2#. Also we are going to use the standard convention that counter-clockwise rotation is positive.

To calculate the angular momentum we use #L=I omega# where #omega=2 pi nu# is the angular frequency (#nu# 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" #