# The rate of rotation of a solid disk with a radius of 9 m and mass of 5 kg constantly changes from 21 Hz to 12 Hz. If the change in rotational frequency occurs over 5 s, what torque was applied to the disk?

Nov 10, 2017

The torque was $= 2290 N m$

#### 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 $I$ is the moment of inertia

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

The radius is $r = 9 m$

For the solid disk, $I = \frac{1}{2} m {r}^{2}$

So, $I = \frac{1}{2} \cdot 5 \cdot {\left(9\right)}^{2} = 202.5 k g {m}^{2}$

The angular velocity is

$\omega = 2 \times \pi \times f$ where the frequency is $= f$

And the rate of change of angular velocity is

$\alpha = \frac{d \omega}{\mathrm{dt}} = \frac{\Delta \omega}{t} = \frac{42 \pi - 24 \pi}{5} = \left(\frac{18}{5} \pi\right) r a {\mathrm{ds}}^{-} 2$

The torque is

$\tau = 202.5 \cdot \frac{18}{5} \pi = 2290 N m$