The rate of rotation of a solid disk with a radius of $8 m$ and mass of $5 kg$ constantly changes from $2 Hz$ to $17 Hz$ . If the change in rotational frequency occurs over $3 s$ , what torque was applied to the disk?

Question

1413 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-13T16:56:26

The torque was $=5026.55Nm$

The torque is the rate of change of angular momentum

$tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

For a solid disc, $I=(mr^2)/2$

So, $I=5*(8)^2/2=160kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(17-2)/3*2pi$

$=(10pi) rads^(-2)$

So the torque is $tau=160*(10pi) Nm=(1600pi)Nm=5026.55Nm$

The rate of rotation of a solid disk with a radius of 8 m8 m and mass of 5 kg5 kg constantly changes from 2 Hz2 Hz to 17 Hz17 Hz. If the change in rotational frequency occurs over 3 s3 s, what torque was applied to the disk?