Question

What is the angular momentum of a rod with a mass of `8 kg` and length of `6 m` that is spinning around its center at `5 Hz`?

Answer 1

`vecL~~754(kgm^2)/s`

Explanation:

Angular momentum is given by `vecL=Iomega`, where `I` is the moment of inertia of the object, and `omega` is the angular velocity of the object.

The moment of inertia of a thin, rigid rod rotating about its center is given by `I=1/12ML^2` and angular velocity is given by `omega=2pif`, where `f` is the frequency.

Given that `f=5Hz`, we can calculate the angular velocity:

`omega=2pif=2pi(5s^-1)=10pi(rad)/s`

Given `M=8kg` and `L=6m`, we can calculate the moment of inertia:

`I=1/12ML^2=1/12(8kg)(6m)^2=24kgm^2`

Using our determined values for `I` and `omega`, the angular momentum is found by:

`vecL=Iomega=(10pi(rad)/s)(24kgm^2)=240pi(kgm^2)/s`

`=>vecL~~754(kgm^2)/s`