# 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?

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 $\frac{1}{2} M {R}^{2} = 50 {\text{ kg m}}^{2}$. 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

$\Delta L = I {\omega}_{2} - I {\omega}_{1} = 2 \pi I \left({\nu}_{2} - {\nu}_{1}\right) = - 700 \pi {\text{ kg m"^2 "s}}^{- 1}$

Thus the torque is

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