A solid disk with a radius of $4 m$ and mass of $3 kg$ is rotating on a frictionless surface. If $72 W$ 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 · 2017-05-02T08:38:55

The torque is $=0.95Nm$

Mass of disc $=3kg$

Radius of disc $=4m$

The power is related to the torque by the equation

$P=tau *omega$

$P=72W$

Frequency of rotation is $f=12Hz$

$omega=f*2pi$

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

torque, $tau=P/omega$

$=72/(24pi)=0.95Nm$

A solid disk with a radius of 4 m4 m and mass of 3 kg3 kg is rotating on a frictionless surface. If 72 W72 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?