Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `4 m`. If the object's angular velocity changes from `2 Hz` to `12 Hz` in `2 s`, what torque was applied to the object?

Answer 1

The torque is `=1005.3Nm`

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 object is `m=2 kg`

The radius of the circular path is `r=4m`

The moment of inertia is

`I=2*4^2=32kg m^2`

The angular acceleration is

`alpha=2pi(f_2-f_1)/t=2pi(12-2)/2=10pirads^-2`

The torque is

`tau=I omega=32*10pi=1005.3Nm`