A solid disk with a radius of $2 m$ and mass of $3 kg$ is rotating on a frictionless surface. If $18 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $6 Hz$ ?

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Answer 1 · 2018-01-16T06:22:50

The torque is $=0.48Nm$

Apply the equation

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

The power is $P=18W$

The frequency is $f=6Hz$

The angular velocity is

$omega=2pif=2xxpixx6=(12pi)rads^-1$

Therefore,

The torque is

$tau=P/omega=18/(12pi)=0.48Nm$

A solid disk with a radius of 2 m2 m and mass of 3 kg3 kg is rotating on a frictionless surface. If 18 W18 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 6 Hz6 Hz?