A solid disk with a radius of #8 m# and mass of #6 kg# is rotating on a frictionless surface. If #72 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #3 Hz#?

1 Answer
Sep 12, 2017

Answer:

The torque is #=3.82Nm#

Explanation:

The mass of the disc is #m=6kg#

The radius of the disc is #r=8m#

The power #=P# and the torque #=tau# are related by the following equation

#P=tau omega#

where #omega=# the angular velocity

Here,

The power is #P=72W#

The frequency is #f=3Hz#

The angular velocity is #omega=2pif=2*pi*3=6pirads^-1#

Therefore,

the torque is #tau=P/omega=72/(6pi)=3.82Nm#