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 `39 Hz`?

Answer 1

`70.573 kN\cdotm`

Explanation:

The angular momentum `L` may be equal to `mvr` or `mr^2\omega`

The given frequency `39Hz` can be expressed to `\omega` or the angular velocity where

`1Hz = (2\pirad)/(s)`

So, `39cancel(Hz)\cdot ((2\pirad)/s)/(1cancel(Hz))=78\pi (rad)/s`

`L = mr^2\omega=(8kg)(6m)^2(78\pi (rad)/s) = 70572.7373702 N\cdotm`