A solid disk with a radius of #8 m# and mass of #9 kg# is rotating on a frictionless surface. If #960 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #28 Hz#?

1 Answer
Jan 3, 2018

Answer:

The torque is #=5.46Nm#

Explanation:

Apply the equation

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

The power is #P=960W#

The frequency is #f=28Hz#

The angular velocity is

#omega=2pif=2xxpixx28=(56pi)rads^-1#

Therefore,

The torque is

#tau=P/omega=960/(56pi)=5.46Nm#