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?

1 Answer

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