# A solid disk with a radius of 8 m and mass of 9 kg is rotating on a frictionless surface. If 960 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 32 Hz?

Feb 19, 2017

The torque is $= 4.77 N m$

#### Explanation:

Power is related to torque, by the equation

$P = \tau \cdot \omega$

$P =$ power (W)

$\tau =$ torque (Nm)

$\omega =$ angular velocity (rads^-1)

Here,

$P = 960 W$

$\omega = 32 \cdot 2 \pi = 64 \pi r a {\mathrm{ds}}^{-} 1$

Therefore,

$\tau = \frac{960}{64 \pi} = 4.77 N m$