# What is the angular momentum of a rod with a mass of  8 kg and length of 4 m that is spinning around its center at 12 Hz?

Dec 23, 2016

The answer is $256 \pi$ $k g {m}^{2} / s$ which to the nearest whole number is 804 $k g {m}^{/} s$

#### Explanation:

Angular momentum is similar in formula to linear momentum:

$L = I \times \omega$ as compared to $p = m \times v$

To find the moment of interia $I$, most students would check a list of known expressions, as every different geometry and even a different axis of rotation will play a part in determining the nature of $I$.

In the case of a rod being rotated about its centre, the moment of inertia is

$I = \frac{m {L}^{2}}{12}$

$L$ being the length of the rod, and $m$ its mass.

The angular velocity $\omega$ is equal to the number of radians swept out by the rod each second. In this case, as each revolution equals $2 \pi$ radians,

$\omega = 24 \pi$

Put it together:

$L = \left(\frac{m {L}^{2}}{12}\right) \times 24 \pi$ = $\frac{\left(8\right) \left({4}^{2}\right) \left(24 \pi\right)}{12}$ = $256 \pi \text{ } k g {m}^{2} / s$

which is approximately 804 $k g {m}^{2} / s$