A solid disk with a radius of $6 m$ and mass of $3 kg$ is rotating on a frictionless surface. If $240 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $1 Hz$ ?

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Answer 1 · 2016-12-25T08:56:36

The torque is $=38.2Nm$

The Power is

$P=(dW)/dt=tau (d theta)/dt=tau omega$

Where $tau=$ torque

Here,

$P=240 W$

$omega=1*2pirads^(-1)$

So,

$tau=P/omega=240/(2pi)=120/pi=38.2Nm$

A solid disk with a radius of 6 m6 m and mass of 3 kg3 kg is rotating on a frictionless surface. If 240 W240 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 1 Hz1 Hz?