Question

The rate of rotation of a solid disk with a radius of `12 m` and mass of `5 kg` constantly changes from `4 Hz` to `6 Hz`. If the change in rotational frequency occurs over `4 s`, what torque was applied to the disk?

Answer 1

The torque is `=1131.0Nm`

Explanation:

The torque is the rate of change of angular momentum

`tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt=Ialpha`

For a solid disc, the moment of inertia is

`I=(mr^2)/2`

The mass of the disc is `m=5 kg`

The radius of the disc is `r=12m`

The moment of inertia is

`I=1/2*5*12^2=360kg m^2`

The angular acceleration is

`alpha=2pi(f_2-f_1)/t=2pi(6-4)/4=pirads^-2`

The torque is

`tau=I omega=360*pi=1131.0Nm`