A solid disk with a radius of 8 m and mass of 5 kg is rotating on a frictionless surface. If 640 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 16 Hz?

Mar 1, 2017

The torque is $= 6.37 N m$

Explanation:

The power and the torque are related by the following equation

$P = \tau \omega$

Power $= 3 W$

The angular velocity $= 16 \cdot 2 \pi = 32 \pi r a {\mathrm{ds}}^{-} 1$

The torque is

$\tau = \frac{P}{\omega} = \frac{640}{32 \pi} = 6.37 N m$