A solid disk with a radius of #6 m# and mass of #8 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
May 10, 2017

Answer:

The torque is #=0.48Nm#

Explanation:

Mass of disc #=8kg#

Radius of disc #=6m#

The power is related to the torque by the equation

#P=tau *omega#

#P=18W#

Frequency of rotation is #f=6Hz#

#omega=f*2pi#

angular velocity, #omega=6*2pi rads^-1#

torque, #tau=P/omega#

#=18/(12pi)=0.48Nm#