A solid disk with a radius of $12 m$ and mass of $8 kg$ 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 $12 Hz$ ?

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Answer 1 · 2017-07-18T11:41:51

The torque is $=8.49Nm$

The mass of the disc is $m=8 kg$

The radius of the disc is $r=12m$

The power and the torque are related by the formula

$P=tau omega$

The angular velocity is $omega=2pif=2pi*12=24pirads^-1$

The power is $P=640W$

Therefore,

The torque is

$tau=P/omega=640/(24pi)=8.49Nm$

A solid disk with a radius of 12 m12 m and mass of 8 kg8 kg is rotating on a frictionless surface. If 640 W640 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 12 Hz12 Hz?