# A solid disk with a radius of 2 m and mass of 6 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 9 Hz?

Feb 7, 2018

The torque is $= 8.49 N m$

#### Explanation:

Apply the equation

$\text{Power (W)" ="torque(Nm)"xx"angular velocity} \left(r a d {s}^{-} 1\right)$

The power is $P = 480 W$

The frequency is $f = 9 H z$

The angular velocity is

$\omega = 2 \pi f = 2 \times \pi \times 9 = \left(18 \pi\right) r a {\mathrm{ds}}^{-} 1$

Therefore,

The torque is

$\tau = \frac{P}{\omega} = \frac{480}{18 \pi} = 8.49 N m$