# A solid disk with a radius of 5 m and mass of 2 kg is rotating on a frictionless surface. If 72 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 4 Hz?

$2.865 \setminus \setminus \textrm{N m}$

#### Explanation:

frequency of rotation of solid disk is $f = 4 \setminus H z$ hence its angular velocity $\setminus \omega$ is given as follows

$\setminus \omega = 2 \setminus \pi f = 2 \setminus \pi \setminus \cdot 4 = 8 \setminus \pi \setminus \textrm{r a \frac{d}{s}}$

Now, the power $P = 72 \setminus W$ imparted to the solid disk by applying torque $T$ to create final angular velocity $\setminus \omega = 8 \setminus \pi \setminus \textrm{r a \frac{d}{s}}$ is given as follows

$P = T \setminus \times \setminus \omega$

$T = \frac{P}{\setminus} \omega$

$= \frac{72}{8 \setminus \pi}$

$= \frac{9}{\setminus} \pi$

$= 2.865 \setminus \setminus \textrm{N m}$