# 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?

Dec 28, 2016

$\vec{L} \approx 754 \frac{k g {m}^{2}}{s}$

#### Explanation:

Angular momentum is given by $\vec{L} = I \omega$, 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 $I = \frac{1}{12} M {L}^{2}$ and angular velocity is given by $\omega = 2 \pi f$, where $f$ is the frequency.

Given that $f = 5 H z$, we can calculate the angular velocity:

$\omega = 2 \pi f = 2 \pi \left(5 {s}^{-} 1\right) = 10 \pi \frac{r a d}{s}$

Given $M = 8 k g$ and $L = 6 m$, we can calculate the moment of inertia:

$I = \frac{1}{12} M {L}^{2} = \frac{1}{12} \left(8 k g\right) {\left(6 m\right)}^{2} = 24 k g {m}^{2}$

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

$\vec{L} = I \omega = \left(10 \pi \frac{r a d}{s}\right) \left(24 k g {m}^{2}\right) = 240 \pi \frac{k g {m}^{2}}{s}$

$\implies \vec{L} \approx 754 \frac{k g {m}^{2}}{s}$