A solid disk with a radius of #9 m# and mass of #4 kg# is rotating on a frictionless surface. If #360 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
Mar 25, 2017

Answer:

The torque is #=9.55Nm#

Explanation:

mass of disc #=4kg#

radius of disc #=9m#

The power is related to the torque by the equation

#P=tau *omega#

#P=360W#

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

torque, #tau=P/omega#

#=360/(12pi)=30/pi=9.55Nm#