A solid disk with a radius of 8 m and mass of 8 kg is rotating on a frictionless surface. If 480 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
Jul 2, 2016

=12.73Nm

Explanation:

Given

  • m-> "Mass of solid disk"=8kg

  • r-> "Radius of solid disk"=8m

  • P-> "Power used to solid disk"=480W

  • n-> "Frequency of rotaion of solid disk"=6Hz

We are to find out

  • tau->"Torque applied to solid disk"

Formula

"Power"(P)="Torque"(tau)xx"Angular velocity"(w)

=>"Power"(P)="Torque"(tau)xx2pixx"Frequency of rotation"(n)

=>480="Torque"(tau)xx2pixx6

"Torque"(tau)=480/(2pixx6)=12.73Nm