# 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 8 Hz?

Nov 19, 2016

The torque applied is 2.38 N-m.

#### Explanation:

This goes very simple, Power is the product of the torque and angular velocity, similar to $P = F v$.
$\implies P = \tau \omega$
So here angular acceleration comes with $\omega = 2 \pi \nu$.
Calculating the angular velocity with given frequency we get,
$\omega = 50.24 r a {\mathrm{ds}}^{-} 1$
Now the above power equation becomes
$\implies \tau = \frac{P}{\omega}$
Substituting the given values we get,
$\tau = \frac{120 W}{50.24 r a {\mathrm{ds}}^{-} 1}$.

$\therefore \tau = 2.38$Newtons-metre.