A solid disk with a radius of $2 m$ and mass of $2 kg$ 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 $9 Hz$ ?

Question

1464 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-10T13:08:07

The torque is $=16.98Nm$

The mass of the disc is $m=2kg$

The radius of the disc is $r=2m$

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 $f=9Hz$

The angular velocity is $omega=2pif=2*pi*9=18pirads^-1$

Therefore,

the torque is $tau=P/omega=960/(18pi)=16.98Nm$

A solid disk with a radius of 2 m2 m and mass of 2 kg2 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 9 Hz9 Hz?