A solid disk with a radius of #6 m# and mass of #2 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 #15 Hz#?

1 Answer
Jul 23, 2017

The torque is #=1.27Nm#

Explanation:

The mass of the disc is #m=2 kg#

The radius of the disc is #r=6m#

The power and the torque are related by the formula

#P=tau omega#

The angular velocity is #omega=2pif=2pi*15=30pirads^-1#

The power is #P=120W#

Therefore,

The torque is

#tau=P/omega=120/(30pi)=1.27Nm#