The rate of rotation of a solid disk with a radius of 6 m and mass of 5 kg constantly changes from 32 Hz to 18 Hz. If the change in rotational frequency occurs over 8 s, what torque was applied to the disk?

Jun 29, 2017

The torque was $= 989.6 N m$

Explanation:

The torque is the rate of change of angular momentum

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

where $I$ is the moment of inertia

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

So, $I = 5 \cdot {\left(6\right)}^{2} / 2 = 90 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{32 - 18}{8} \cdot 2 \pi$

$= \left(\frac{7}{2} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 90 \cdot \left(\frac{7}{2} \pi\right) N m = 315 \pi N m = 989.6 N m$