# The rate of rotation of a solid disk with a radius of 1 m and mass of 5 kg constantly changes from 16 Hz to 27 Hz. If the change in rotational frequency occurs over 12 s, what torque was applied to the disk?

Mar 27, 2017

The torque was $= 14.4 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

mass, $m = 5 k g$

radius, $r = 1 m$

The moment of inertia of a solid disc is

$I = \frac{1}{2} \cdot m {r}^{2}$

$= \frac{1}{2} \cdot 5 \cdot {1}^{2} = 2.5 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{\left(27 - 16\right)}{12} \cdot 2 \pi$

$= \left(\frac{11}{6} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 2.5 \cdot \left(\frac{11}{6} \pi\right) N m = 14.4 N m$