Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `7 m`. If the object's angular velocity changes from `13Hz` to `17 Hz` in `5 s`, what torque was applied to the object?

Answer 1

The torque was `=738.9Nm`

Explanation:

The torque is the rate of change of angular momentum

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

For the object, the moment of inertia is

`I=mr^2`

The mass of the disc is `m=3 kg`

The radius of the disc is `r=7m`

The moment of inertia is

`I=3*7^2=147kg m^2`

The angular acceleration is

`alpha=2pi(f_2-f_1)/t=2pi(17-13)/5=1.6pirads^-2`

The torque is

`tau=I omega=147*1.6pi=738.9Nm`