A solid disk with a radius of #2 m# and mass of #3 kg# is rotating on a frictionless surface. If #18 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #6 Hz#?

1 Answer
Jan 16, 2018

Answer:

The torque is #=0.48Nm#

Explanation:

Apply the equation

#"Power (W)" ="torque(Nm)"xx"angular velocity" (rad s^-1)#

The power is #P=18W#

The frequency is #f=6Hz#

The angular velocity is

#omega=2pif=2xxpixx6=(12pi)rads^-1#

Therefore,

The torque is

#tau=P/omega=18/(12pi)=0.48Nm#