# 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  2/3 Hz ?

## I

Dec 17, 2015

We will first convert $H z$ into $\frac{r a d}{\sec}$ using the formula below:
$\setminus \omega = 2 \setminus \pi f$
$\setminus \omega = 2 \setminus \times \setminus \pi \setminus \times \frac{2}{3}$
$\setminus \omega = 4.188 \frac{r a d}{\sec}$

We then use the formula for Angular momentum
$L = \setminus \vec{r} \setminus \times \setminus \vec{p}$= $L = \setminus \vec{r} \setminus \times m \setminus \vec{v}$
Since we are calculating Angular momentum wrt to the center of the circle, and during circular motion $\setminus \vec{v}$ is perpendicular to $\setminus \vec{r}$ hence the cross product simply becomes a simple multiplication, with the angular momentum pointing along Z-axis( assuming motion is happening in the X-Y plane)

Velocity in a circular motion, given its angular velocity is given by:
$v = \setminus \omega r$, where $r$ is the radius of circle

Therefore,
$L = m \setminus \omega {r}^{2} \setminus \hat{z}$
$L = 5 \setminus \times 4.188 \setminus \times {9}^{2}$= $1696.46 K g$ ${m}^{2} / \sec$ $\setminus \hat{z}$

Dec 17, 2015

$L = 1696.7 {\text{ ""kg.m"^2"s}}^{- 1}$

#### Explanation:

The expression for angular momentum is:

$L = m \times v \times r$

$m$ = mass

$v$ = velocity

$r$ = radius

We can use the information given to find the velocity $v$:

The frequency of rotation $f$ is $\frac{2}{3} \text{Hz}$

This means that the time period $T$ is $\frac{1}{f}$ which equals $\frac{3}{2} = 1.5 \text{s}$

This means that it takes $1.5 \text{s}$ to make one complete revolution.

We know the circumference of the circle is $2 \pi r$.

This is the distance travelled in $1.5 \text{s}$

So the velocity is given by:

$v = \frac{2 \pi 9}{1.5} = 12 \pi \text{m/s}$

So we can now get $L$:

$L = 5 \times 12 \pi \times 9$

$\therefore L = 1696.7 {\text{ ""kg.m"^2"s}}^{- 1}$