# 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 24 Hz?

Dec 30, 2015

$L = 81430$ $k g . {m}^{2} / s$

#### Explanation:

By definition, angular momentum is the cross product of the position vector from the axis of rotation to the rotating object and the linear momentum of the object.
That is, $\vec{L} = \vec{r} \times \vec{p}$ where $\vec{p} = m \vec{v}$.

So in this case, the magnitude of the angular momentum is
$L = r m v \sin {90}^{\circ}$
$= 6 \times 15 \times \left(\frac{2 \pi 6 \times 24}{1}\right)$, since $v = \frac{x}{t}$ and the object completes 24 circumferences or cycles per second).
$\therefore L = 81430 k g . {m}^{2} / s$.

The direction of $L$ will be obtained by the right hand rule as per normal vector cross-products - curl the fingers of your right hand from $\vec{r}$ to $\vec{p}$ and then your thumb will point in the direction of $\vec{L}$, which will be perpendicular to the plane formed by $\vec{r} \mathmr{and} \vec{p}$.
So for example if the circular rotation is in the xy-plane formed by the basis unit vectors $\hat{i} \mathmr{and} \hat{j}$, then the angular momentum will be along the basis unit vector $\hat{k}$.