A solid disk with a radius of $2 m$ 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 · 2018-02-28T18:29:35

The torque is $=4.24Nm$

Apply the equation

$"Power (W)" ="torque(Nm)"xx"angular velocity" (rad s^-1)$

The power is $P=480W$

The frequency is $f=18Hz$

The angular velocity is

$omega=2pif=2xxpixx18=(36pi)rads^-1$

Therefore,

The torque is

$tau=P/omega=480/(36pi)=4.24Nm$

A solid disk with a radius of 2 m2 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?