# 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 7 s, what torque was applied to the disk?

Oct 11, 2017

The torque was $= 251.3 N m$

#### 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 disc is $m = 5 k g$

The radius is $r = 4 m$

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

So, $I = \frac{1}{2} \cdot 5 \cdot {\left(4\right)}^{2} = 40 k g {m}^{2}$

And the rate of change of angular velocity is

$\frac{d \omega}{\mathrm{dt}} = \frac{\Delta \omega}{t} = \frac{24 \pi - 10 \pi}{7} = 2 \pi r a {\mathrm{ds}}^{-} 2$

So,

The torque is

$\tau = 40 \cdot 2 \pi = 251.3 N m$