A solid disk with a radius of $8 m$ and mass of $6 kg$ 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 $3 Hz$ ?

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Answer 1 · 2017-09-12T18:19:21

The torque is $=3.82Nm$

The mass of the disc is $m=6kg$

The radius of the disc is $r=8m$

The power $=P$ and the torque $=tau$ are related by the following equation

$P=tau omega$

where $omega=$ the angular velocity

Here,

The power is $P=72W$

The frequency is $f=3Hz$

The angular velocity is $omega=2pif=2*pi*3=6pirads^-1$

Therefore,

the torque is $tau=P/omega=72/(6pi)=3.82Nm$

A solid disk with a radius of 8 m8 m and mass of 6 kg6 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 3 Hz3 Hz?