A solid disk with a radius of #8 m# and mass of #6 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 #3 Hz#?

1 Answer
Apr 9, 2017

Answer:

The torque is #=3.18Nm#

Explanation:

mass of disc #=2kg#

radius of disc #=2m#

The power is related to the torque by the equation

#P=tau *omega#

#P=60W#

Frequency of rotation is #f=3Hz#

#omega=f*2pi#

angular velocity, #omega=3*2pi rads^-1#

torque, #tau=P/omega#

#=60/(6pi)=3.18Nm#