A solid disk with a radius of #12 m# and mass of #5 kg# is rotating on a frictionless surface. If #450 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#?

1 Answer
Dec 3, 2017

Answer:

The torque is #=8.95Nm#

Explanation:

Apply the equation

#P=tau omega#

The power is #P=450W#

The frequency is #f=8Hz#

The angular velocity is

#omega=2pif=2xxpixx8=(16pi)rads^-1#

Therefore,

The torque is

#tau=P/omega=450/(16pi)=8.95Nm#