A solid disk with a radius of #8 m# and mass of #9 kg# is rotating on a frictionless surface. If #960 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #32 Hz#?

1 Answer
Feb 19, 2017

Answer:

The torque is #=4.77Nm#

Explanation:

Power is related to torque, by the equation

#P=tau* omega#

#P=# power (W)

#tau=# torque (Nm)

#omega=# angular velocity (rads^-1)

Here,

#P=960W#

#omega=32*2pi=64pi rads^-1#

Therefore,

#tau=960/(64pi)=4.77Nm#