A solid disk with a radius of $24 m$ and mass of $8 kg$ 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 $18 Hz$ ?

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Answer 1 · 2017-03-01T21:09:53

$tau~~4.2$ $N$ $m$

One way to express torque is by the equation $tau=P/omega$ . This is rearranged from $P=tauomega$ .

Given that $P=480W$ and $f=18Hz$ , we can determine the angular velocity $omega$ and use this to calculate the torque.

$omega=2pif=2pi(18s^-1)=36pi(rad)/s$

So, we have:

$tau=(480W)/(36pi(rad)/s)~~4.2$ $N$ $m$

A solid disk with a radius of 24 m24 m and mass of 8 kg8 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 18 Hz18 Hz?