A solid disk with a radius of #12 m# and mass of #5 kg# is rotating on a frictionless surface. If #450 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #7 Hz#?

1 Answer
Oct 28, 2017

Answer:

The torque is #=10.23Nm#

Explanation:

The mass of the disc is #m=5kg#

The radius of the disc is #r=12m#

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=450W#

The frequency is #f=7Hz#

The angular velocity is #omega=2pif=2*pi*7=(14pi)rads^-1#

Therefore,

the torque is #tau=P/omega=450/(14pi)=10.23Nm#