The rate of rotation of a solid disk with a radius of $4 m$ and mass of $5 kg$ constantly changes from $21 Hz$ to $12 Hz$ . If the change in rotational frequency occurs over $8 s$ , what torque was applied to the disk?

Question

1405 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-23T06:48:03

THe torque was $=282.7Nm$

The torque is the rate of change of angular momentum

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

The moment of inertia of a solid disc, is

$I=1/2*mr^2$

$=1/2*5*4^2= 40 kgm^2$

The rate of change of angular velocity is

$(domega)/dt=(21-12)/8*2pi$

$=(9/4pi) rads^(-2)$

So the torque is $tau=40*(9/4pi) Nm=(90pi)Nm=282.7Nm$

The rate of rotation of a solid disk with a radius of 4 m4 m and mass of 5 kg5 kg constantly changes from 21 Hz21 Hz to 12 Hz12 Hz. If the change in rotational frequency occurs over 8 s8 s, what torque was applied to the disk?