A solid disk with a radius of $9 m$ and mass of $2 kg$ 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 $15 Hz$ ?

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Answer 1 · 2017-02-25T09:23:44

The torque is $=5.1Nm$

The power is related to the torque, by the equation

$P=tau omega$

$P=$ power

$tau=$ torque

$omega=$ angular velocity

$P=480W$

$omega=15*2pi=(30pi)rads^-1$

$tau = 480/(30pi) =5.1Nm$

A solid disk with a radius of 9 m9 m and mass of 2 kg2 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 15 Hz15 Hz?