A solid disk with a radius of #3 m# and mass of #4 kg# is rotating on a frictionless surface. If #60 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #12 Hz#?

1 Answer
May 15, 2017

Answer:

The torque is #=0.8Nm#

Explanation:

Mass of disc #=4kg#

Radius of disc #=3m#

The power is related to the torque by the equation

#P=tau *omega#

#P=60W#

Frequency of rotation is #f=12Hz#

#omega=f*2pi#

angular velocity, #omega=12*2pi rads^-1#

torque, #tau=P/omega#

#=60/(24pi)=0.8Nm#