# A solid disk with a radius of 24 m and mass of 8 kg is rotating on a frictionless surface. If 480 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 18 Hz?

Mar 1, 2017

$\tau \approx 4.2$ $N$ $m$

#### Explanation:

One way to express torque is by the equation $\tau = \frac{P}{\omega}$. This is rearranged from $P = \tau \omega$.

Given that $P = 480 W$ and $f = 18 H z$, we can determine the angular velocity $\omega$ and use this to calculate the torque.

$\omega = 2 \pi f = 2 \pi \left(18 {s}^{-} 1\right) = 36 \pi \frac{r a d}{s}$

So, we have:

$\tau = \frac{480 W}{36 \pi \frac{r a d}{s}} \approx 4.2$ $N$ $m$