A solid disk with a radius of $5 m$ and mass of $3 kg$ is rotating on a frictionless surface. If $72 W$ of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at $4 Hz$ ?

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Answer 1 · 2017-07-28T06:13:10

The torque is $=2.86Nm$

The mass of the disc is $m=3 kg$

The radius of the disc is $r=5m$

The power and the torque are related by the formula

$P=tau omega$

The angular velocity is $omega=2pif=2pi*4=8pirads^-1$

The power is $P=72W$

Therefore,

The torque is

$tau=P/omega=72/(8pi)=2.86Nm$

A solid disk with a radius of 5 m5 m and mass of 3 kg3 kg is rotating on a frictionless surface. If 72 W72 W of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at 4 Hz4 Hz?