A solid disk with a radius of $24 m$ 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 $3 Hz$ ?

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Answer 1 · 2017-03-03T09:33:55

The torque is $=25.5Nm$

mass of disc $=6kg$

radius of disc $=24m$

The power is related to the torque by the equation

$P=tau *omega$

$P=480W$

angular velocity, $omega=3*2pi rads^-1$

torque, $tau=P/omega$

$=480/(6pi)=25.5Nm$

A solid disk with a radius of 24 m24 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 3 Hz3 Hz?