# A solid disk with a radius of 3 m and mass of 4 kg is rotating on a frictionless surface. If 120 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 12 Hz?

##### 1 Answer
Jul 2, 2017

The torque is $= 1.59 N m$

#### Explanation:

The mass of the disc is $m = 4 k g$

The radius of the disc is $r = 3 m$

The power and the torque are related by the formula

$P = \tau \omega$

The angular velocity is $\omega = 2 \pi f = 2 \pi \cdot 12 = 24 \pi r a {\mathrm{ds}}^{-} 1$

The power is $P = 120 W$

Therefore,

The torque is

$\tau = \frac{P}{\omega} = \frac{120}{24 \pi} = 1.59 N m$