A solid disk with a radius of $2 m$ and mass of $3 kg$ is rotating on a frictionless surface. If $36 W$ 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 · 2017-11-21T14:18:00

The torque is $=0.72Nm$

Apply the equation

$P=tau omega$

The power is $P=36W$

The frequency is $f=8Hz$

The angular velocity is

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

Therefore,

The torque is

$tau=P/omega=36/(16pi)=0.72Nm$

A solid disk with a radius of 2 m2 m and mass of 3 kg3 kg is rotating on a frictionless surface. If 36 W36 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?