A solid disk with a radius of $6 m$ 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 $32 Hz$ ?

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Answer 1 · 2017-10-13T08:59:10

The torque is $=4.78Nm$

The mass of the disc is $m=9kg$

The radius of the disc is $r=6m$

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 $P=960W$

The frequency is $f=32Hz$

The angular velocity is $omega=2pif=2*pi*32=(64pi)rads^-1$

Therefore,

the torque is $tau=P/omega=960/(64pi)=4.78Nm$

A solid disk with a radius of 6 m6 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 32 Hz32 Hz?