A solid disk with a radius of #9 m# and mass of #8 kg# is rotating on a frictionless surface. If #360 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#?

1 Answer
Oct 15, 2017

Answer:

The torque is #=9.55Nm#

Explanation:

The mass of the disc is #m=8kg#

The radius of the disc is #r=9m#

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=360W#

The frequency is #f=6Hz#

The angular velocity is #omega=2pif=2*pi*6=(12pi)rads^-1#

Therefore,

the torque is #tau=P/omega=360/(12pi)=9.55Nm#