A solid disk with a radius of $12 m$ and mass of $5 kg$ is rotating on a frictionless surface. If $450 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $8 Hz$ ?

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Answer 1 · 2017-12-03T09:01:24

The torque is $=8.95Nm$

Apply the equation

$P=tau omega$

The power is $P=450W$

The frequency is $f=8Hz$

The angular velocity is

$omega=2pif=2xxpixx8=(16pi)rads^-1$

Therefore,

The torque is

$tau=P/omega=450/(16pi)=8.95Nm$

A solid disk with a radius of 12 m12 m and mass of 5 kg5 kg is rotating on a frictionless surface. If 450 W450 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 8 Hz8 Hz?