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
P dilip_k
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#

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