# The rate of rotation of a solid disk with a radius of 1 m and mass of 5 kg constantly changes from 6 Hz to 24 Hz. If the change in rotational frequency occurs over 6 s, what torque was applied to the disk?

Oct 4, 2017

The torque was $= 47.1 N m$

#### 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

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

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

The radius of the disc is $r = 1 m$

So, the moment of inertia is $I = 5 \cdot {\left(1\right)}^{2} / 2 = 2.5 k g {m}^{2}$

The rate of change of the angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{24 - 6}{6} \cdot 2 \pi = 6 \pi r a {\mathrm{ds}}^{-} 2$

So,

The torque is

$\tau = 2.5 \cdot 6 \pi = 47.1 N m$