A solid disk with a radius of #6 m# and mass of #8 kg# is rotating on a frictionless surface. If #3 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #8 Hz#?

Question

1179 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T05:39:39

The torque is #=0.06Nm#

Mass of disc #=8kg#

Radius of disc #=6m#

The power is related to the torque by the equation

#P=tau *omega#

#P=3W#

Frequency of rotation is #f=8Hz#

#omega=f*2pi#

angular velocity, #omega=8*2pi rads^-1#

torque, #tau=P/omega#

#=3/(16pi)=0.06Nm#