A solid disk with a radius of #2 m# and mass of #2 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 #9 Hz#?

1 Answer
Aug 10, 2017

Answer:

The torque is #=16.98Nm#

Explanation:

The mass of the disc is #m=2kg#

The radius of the disc is #r=2m#

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=960W#

The frequency is #f=9Hz#

The angular velocity is #omega=2pif=2*pi*9=18pirads^-1#

Therefore,

the torque is #tau=P/omega=960/(18pi)=16.98Nm#