A solid disk with a radius of $18 m$ and mass of $6 kg$ is rotating on a frictionless surface. If $480 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $5 Hz$ ?

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Answer 1 · 2017-05-09T05:59:55

The torque is $=15.3N$

Mass of disc $=6kg$

Radius of disc $=18m$

The power is related to the torque by the equation

$P=tau *omega$

$P=480W$

Frequency of rotation is $f=5Hz$

$omega=f*2pi$

angular velocity, $omega=5*2pi rads^-1$

torque, $tau=P/omega$

$=480/(10pi)=15.3Nm$

A solid disk with a radius of 18 m18 m and mass of 6 kg6 kg is rotating on a frictionless surface. If 480 W480 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 5 Hz5 Hz?