# 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  3 6 Hz?

$1728 \setminus \pi = 5428.6$ $k g . {m}^{2} / s$
The moment-of-inertia of a rod of mass $M$ and length $L$ rotating about its centre is $I = \frac{1}{12} M {L}^{2} = \frac{1}{12} \left(8 k g\right) {\left(6 m\right)}^{2} = 24 k g . {m}^{2}$.
The angular frequency $\setminus \omega$ is related to the frequency $f$ as $\setminus \omega = 2 \setminus \pi f = 2 \setminus \pi \cdot \left(36 H z\right) = 72 \setminus \pi$ rad/s,
Angular Momentum : The magnitude of Angular Momentum vector is $L = I \setminus \omega = I .2 \setminus \pi f = \left(24 k g . {m}^{2}\right) \cdot \left(72 \setminus \pi\right) = 1728 \setminus \pi$ $k g . {m}^{2} / s$