A solid disk with a radius of $8 m$ and mass of $6 kg$ is rotating on a frictionless surface. If $60 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $12 Hz$ ?

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Answer 1 · 2017-03-28T08:20:36

The torque is $=0.8Nm$

mass of disc $=6kg$

radius of disc $=8m$

The power is related to the torque by the equation

$P=tau *omega$

$P=60W$

Frequency of rotation is $f=12Hz$

$omega=f*2pi$

angular velocity, $omega=12*2pi rads^-1$

torque, $tau=P/omega$

$=60/(24pi)=0.8Nm$

A solid disk with a radius of 8 m8 m and mass of 6 kg6 kg is rotating on a frictionless surface. If 60 W60 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 12 Hz12 Hz?