# A solid disk with a radius of 12 m and mass of 5 kg is rotating on a frictionless surface. If 450 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 7 Hz?

Oct 28, 2017

The torque is $= 10.23 N m$

#### Explanation:

The mass of the disc is $m = 5 k g$

The radius of the disc is $r = 12 m$

The power $= P$ and the torque $= \tau$ are related by the following equation

$P = \tau \omega$

where $\omega =$ the angular velocity

Here,

The power is $P = 450 W$

The frequency is $f = 7 H z$

The angular velocity is $\omega = 2 \pi f = 2 \cdot \pi \cdot 7 = \left(14 \pi\right) r a {\mathrm{ds}}^{-} 1$

Therefore,

the torque is $\tau = \frac{P}{\omega} = \frac{450}{14 \pi} = 10.23 N m$