# The rate of rotation of a solid disk with a radius of 5 m and mass of 5 kg constantly changes from 12 Hz to 10 Hz. If the change in rotational frequency occurs over 9 s, what torque was applied to the disk?

Apr 25, 2018

The torque was $= 87.27 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

where $I$ is the moment of inertia

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

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

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

So, $I = 5 \cdot {\left(5\right)}^{2} / 2 = 62.5 k g {m}^{2}$

The initial angular velocity is

${\omega}_{0} = 2 \pi {f}_{0} = 2 \cdot 12 \cdot \pi = 24 \pi r a {\mathrm{ds}}^{-} 1$

The final angular velocity is

${\omega}_{1} = 2 \pi {f}_{1} = 2 \cdot 10 \cdot \pi = 20 \pi r a {\mathrm{ds}}^{-} 1$

The time is $t = 9 s$

The angular acceleration is

$\alpha = \frac{{\omega}_{1} - {\omega}_{0}}{t} = \frac{24 \pi - 20 \pi}{9} = \frac{4}{9} \pi r a {\mathrm{ds}}^{-} 2$

So,

The torque is $\tau = 62.5 \cdot \frac{4}{9} \pi = 87.27 N m$