A solid disk with a radius of #2 m# and mass of #8 kg# is rotating on a frictionless surface. If #480 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #18 Hz#?

1 Answer
Feb 28, 2018

Answer:

The torque is #=4.24Nm#

Explanation:

Apply the equation

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

The power is #P=480W#

The frequency is #f=18Hz#

The angular velocity is

#omega=2pif=2xxpixx18=(36pi)rads^-1#

Therefore,

The torque is

#tau=P/omega=480/(36pi)=4.24Nm#