# The rate of rotation of a solid disk with a radius of 8 m and mass of 5 kg constantly changes from 16 Hz to 22 Hz. If the change in rotational frequency occurs over 7 s, what torque was applied to the disk?

Dec 15, 2016

The torque is $= 861.7 k g {m}^{2} {s}^{- 1}$

#### Explanation:

The torque is $\tau = I \alpha$

$\alpha = \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

$\alpha = \frac{6}{7} \cdot 2 \pi r a {\mathrm{ds}}^{- 1} = \frac{12}{7} \pi r a {\mathrm{ds}}^{- 1}$

The moment of inertia of a solid disc $I = \frac{1}{2} m {r}^{2}$

$= \frac{1}{2} \cdot 5 \cdot {8}^{2} = 160 k g {m}^{2}$

So,

$\tau = 160 \cdot \frac{12}{7} \pi k g {m}^{2} {s}^{- 1} = 861.7 k g {m}^{2} {s}^{- 1}$