# What is the angular momentum of an object with a mass of 5 kg that moves along a circular path of radius 9 m at a frequency of  8 Hz ?

## I

Jan 13, 2016

$| \vec{L} | = 20357.325 k g . {m}^{2.} / s$

#### Explanation:

For a point object, magnitude of angular momentum may be given by $L = I \omega$, where I is the moment of inertia and $\omega$ is the angular velocity.

The moment of inertia for this point object may be given by
$I = M {R}^{2} = 5 \times {9}^{2} = 405 k g . {m}^{2}$.
(This is an approximation whose accuracy depends on how large the dimensions of the object is in comparison to the radius of its path, otherwise we must use the more precise definition of moment of inertia, ie $I = {\lim}_{n \to \infty} {\sum}_{j = 1}^{n} {m}_{j} {r}_{p e r p j}^{2} = {\int}_{M} {r}^{2} \mathrm{dm}$.)

Furthermore, angular frequency may be given by $\omega = 2 \pi f$
$= 2 \pi \times 8 = 50.265 r a d / s$.

Hence the magnitude of the angular momentum is $L = I \omega$
$= 405 \times 50.265 = 20357.325 k g . {m}^{2.} / s$.

Alternatively, we could also have used the definition of angular momentum : $\vec{L} = \vec{r} \times \vec{p}$, ie the cross product of the position vector from the axis of rotation to the object and its linear momentum vector $\vec{p} = m \vec{v}$. This would ultimately give the same final answer.

In either case, the direction of the angular momentum vector $\vec{L}$ may be obtained from the right hand rule by curling the fingers of the right hand either in the direction of $\omega$ then the thumb points to direction of $\vec{L}$ or curling right hand fingers from $\vec{r}$ to $\vec{p}$ then thumb still points to direction of $\vec{L}$.