# 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?

$70.573 k N \setminus \cdot m$

#### Explanation:

The angular momentum $L$ may be equal to $m v r$ or $m {r}^{2} \setminus \omega$

The given frequency $39 H z$ can be expressed to $\setminus \omega$ or the angular velocity where

$1 H z = \frac{2 \setminus \pi r a d}{s}$

So, $39 \cancel{H z} \setminus \cdot \frac{\frac{2 \setminus \pi r a d}{s}}{1 \cancel{H z}} = 78 \setminus \pi \frac{r a d}{s}$

$L = m {r}^{2} \setminus \omega = \left(8 k g\right) {\left(6 m\right)}^{2} \left(78 \setminus \pi \frac{r a d}{s}\right) = 70572.7373702 N \setminus \cdot m$