# The rate of rotation of a solid disk with a radius of 9 m and mass of 5 kg constantly changes from 32 Hz to 18 Hz. If the change in rotational frequency occurs over 6 s, what torque was applied to the disk?

Feb 20, 2017

The torque was $= 2968.8 N m$

#### Explanation:

The torque is the rate of change of angular momentum

$\tau = \frac{\mathrm{dL}}{\mathrm{dt}} = \frac{d \left(I \omega\right)}{\mathrm{dt}} = I \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

The moment of inertia of a solid disc is $I = \frac{1}{2} \cdot m {r}^{2}$

$= \frac{1}{2} \cdot 5 \cdot {9}^{2} = \frac{405}{2} k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{32 - 18}{6} \cdot 2 \pi$

$= \left(\frac{14 \pi}{3}\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = \frac{405}{2} \cdot \frac{14 \pi}{3} N m = 2968.8 N m$