A solid disk with a radius of 6 m and mass of 3 kg is rotating on a frictionless surface. If 240 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 1 Hz?

Dec 25, 2016

The torque is $= 38.2 N m$

Explanation:

The Power is

$P = \frac{\mathrm{dW}}{\mathrm{dt}} = \tau \frac{d \theta}{\mathrm{dt}} = \tau \omega$

Where $\tau =$ torque

Here,

$P = 240 W$

$\omega = 1 \cdot 2 \pi r a {\mathrm{ds}}^{- 1}$

So,

$\tau = \frac{P}{\omega} = \frac{240}{2 \pi} = \frac{120}{\pi} = 38.2 N m$