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#?

Question

1198 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-25T06:54:21

The torque is #=9.55Nm#

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#