A solid disk with a radius of $6 m$ and mass of $2 kg$ is rotating on a frictionless surface. If $3 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $4 Hz$ ?

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Answer 1 · 2017-11-16T06:13:16

The torque is $=0.12Nm$

Apply the equation

$P=tau omega$

The power is $P=3W$

The frequency is $f=4Hz$

The angular velocity is

$omega=2pif=2xxpixx4=(8pi)rads^-1$

Therefore,

The torque is

$tau=P/omega=3/(8pi)=0.12Nm$

A solid disk with a radius of 6 m6 m and mass of 2 kg2 kg is rotating on a frictionless surface. If 3 W3 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 4 Hz4 Hz?