A solid disk with a radius of $6 m$ and mass of $8 kg$ is rotating on a frictionless surface. If $18 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $6 Hz$ ?

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Answer 1 · 2017-05-10T06:34:41

The torque is $=0.48Nm$

Mass of disc $=8kg$

Radius of disc $=6m$

The power is related to the torque by the equation

$P=tau *omega$

$P=18W$

Frequency of rotation is $f=6Hz$

$omega=f*2pi$

angular velocity, $omega=6*2pi rads^-1$

torque, $tau=P/omega$

$=18/(12pi)=0.48Nm$

A solid disk with a radius of 6 m6 m and mass of 8 kg8 kg is rotating on a frictionless surface. If 18 W18 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 6 Hz6 Hz?