What is the angular momentum of a rod with a mass of 15 kg and length of 6 m that is spinning around its center at 23 Hz?

Jan 19, 2016

$\vec{L} = 6503.085 k g . {m}^{2} / s$ in direction of $\vec{r} \times \vec{p}$.

Explanation:

The moment of inertia of a thin rod about an axis through its centre is given by (can be proven - ask if you want me to show you the derivation)

$I = \frac{1}{12} M {L}^{2}$

$= \frac{1}{12} \times 15 \times {6}^{2}$

$= 45 k g . {m}^{2}$.

The angular velocity of the rod is given by

$\omega = 2 \pi f$

$= 2 \pi \times 23 = 144.513 r a d / s$.

Hence the angular momentum is given by

$L = I \omega$

$= 45 \times 144.513$

$= 6503.085 k g . {m}^{2} / s$.

The direction hereof may be obtained by the right hand rule - curl the fingers of your right hand in the direction of the rotation and your thumb will point in the direction of the angular momentum $\vec{L}$.