A solid disk with a radius of $8$ $m$ and mass of $2$ $kg$ is rotating on a frictionless surface. If $640$ $W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $16$ $Hz$ ?

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Answer 1 · 2017-03-28T12:23:01

The torque is $=6.4Nm$

mass of disc $=2kg$

radius of disc $=8m$

The power is related to the torque by the equation

$P=tau *omega$

$P=640W$

Frequency of rotation is $f=16Hz$

$omega=f*2pi$

angular velocity, $omega=16*2pi rads^-1$

torque, $tau=P/omega$

$=640/(32pi)=6.4Nm$

A solid disk with a radius of 88 mm and mass of 22 kgkg is rotating on a frictionless surface. If 640640 WW of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 1616 HzHz?