A solid disk with a radius of #8 m# and mass of #6 kg# is rotating on a frictionless surface. If #60 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #12 Hz#?

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Answer 1 · 2017-03-28T08:20:36

The torque is #=0.8Nm#

mass of disc #=6kg#

radius of disc #=8m#

The power is related to the torque by the equation

#P=tau *omega#

#P=60W#

Frequency of rotation is #f=12Hz#

#omega=f*2pi#

angular velocity, #omega=12*2pi rads^-1#

torque, #tau=P/omega#

#=60/(24pi)=0.8Nm#