A solid disk with a radius of #2 m# and mass of #3 kg# is rotating on a frictionless surface. If #36 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#?

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Answer 1 · 2017-06-12T15:15:37

The torque is #=0.95Nm#

Mass of disc #=3kg#

Radius of disc #=2m#

The power is related to the torque by the equation

#P=tau *omega#

#P=36W#

Frequency of rotation is #f=6Hz#

#omega=f*2pi#

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

torque, #tau=P/omega#

#=36/(12pi)=0.95Nm#