A solid disk with a radius of $8 m$ and mass of $9 kg$ is rotating on a frictionless surface. If $960 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $28 Hz$ ?

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Answer 1 · 2018-01-03T06:23:32

The torque is $=5.46Nm$

Apply the equation

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

The power is $P=960W$

The frequency is $f=28Hz$

The angular velocity is

$omega=2pif=2xxpixx28=(56pi)rads^-1$

Therefore,

The torque is

$tau=P/omega=960/(56pi)=5.46Nm$

A solid disk with a radius of 8 m8 m and mass of 9 kg9 kg is rotating on a frictionless surface. If 960 W960 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 28 Hz28 Hz?