A solid disk with a radius of #2 m# and mass of #3 kg# is rotating on a frictionless surface. If #36 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #3 Hz#?

1 Answer
Jun 24, 2017

Answer:

The torque is #=1.91Nm#

Explanation:

The mass of the disc is #m=3 kg#

The radius of the disc is #r=2m#

The power and the torque are related by the formula

#P=tau omega#

The angular velocity is #omega=2pif=2pi*3=6pirads^-1#

The power is #P=36W#

Therefore,

The torque is

#tau=P/omega=36/(6pi)=1.91Nm#