# The rate of rotation of a solid disk with a radius of 5 m and mass of 5 kg constantly changes from 18 Hz to 24 Hz. If the change in rotational frequency occurs over 1 s, what torque was applied to the disk?

Jun 8, 2017

The torque was $= 2356.2 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

Where the moment of inertiais $= I$

and $\omega$ is the angular velocity

The mass of the disc is $m = 5 k g$

The radius of the disc is $r = 5 m$

For the solid disc, $I = \left(\frac{1}{2} m {r}^{2}\right)$

So, $I = \frac{1}{2} \cdot 5 \cdot {\left(5\right)}^{2} = \frac{125}{2} k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{24 - 18}{1} \cdot 2 \pi$

$= \left(12 \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = \frac{125}{2} \cdot \left(12 \pi\right) N m = 2356.2 N m$