Question

A particle of mass $2kg$ is on a smooth horizontal table and moves in a circular path of radius $0.6m$. The height of the table from the ground is $0.8m$ If the angular speed of the particle is $12rad{s}^{-1}$. The magnitude of its angular momentum about a point on the ground right under the center of the circle is

Answer 1

Step 1. Given Data,

The mass of a particle $=2kg$
The radius of the circular path of motion of the body, $R$$=0.6m$
Height $=0.8m$
The angular speed of the particle, $\omega =12rad{s}^{-1}$

Step 2. Formula Used,

The angular momentum of an object having mass ($m$) and linear velocity ($v$) is:

$\left|\overrightarrow{L}\right|=m\left|\overrightarrow{v}\right|\left|\overrightarrow{r}\right|\sin \left(\theta \right)$

$\left|\overrightarrow{L}\right|$ is the magnitude of a vector quantity of the angular momentum of the body,

$\left|\overrightarrow{r}\right|$ is the magnitude of the radius of the circular path of motion of the body,

$\left|\overrightarrow{v}\right|$is the magnitude of another vector quantity, the velocity with which the body is travelling,

$\theta$ is the angle between the radius vector and the velocity vector.

$$\overrightarrow{L}$$= $\overrightarrow{r}\times \overrightarrow{p}$

$\overrightarrow{p}=mv$

$$\overrightarrow{L}$$= Angular Momentum

$m$ = mass of the particle

$$\overrightarrow{p}$$= linear momentum

$v$ is the velocity of the particle

From Pythagoras theorem,

$H=\sqrt{P^2+B^2}$

$H$ is the resultant radius vector, $B$ is the base, and $P$ is the height.

Step 3. Calculating the magnitude of its angular momentum,

We consider $H$as the resultant radius vector and we can denote it as $r_{perpendicular}$

The perpendicular distance between the point and mass $r_{perpendicular}$ is given by,

$r_{perpendicular}=\left|\overrightarrow{r}\right|=\sqrt{{\left(0.6\right)}^2+{\left(0.8\right)}^2}=1m$

The angle between the resultant radius vector and the linear velocity vector is $\theta ={90}^{\circ }$.
The angular momentum is,

$\left|\overrightarrow{L}\right|=m\left|\overrightarrow{v}\right|\left|\overrightarrow{r}\right|\sin \left(\theta \right)$

$\left|\overrightarrow{L}\right|=m\left|\omega R\right|\left|\overrightarrow{r}\right|\sin \left({90}^o\right)$
$\Rightarrow \left|\overrightarrow{L}\right|=2\times 12\times 0.6\times 1\times 1=14.4kg{m}^2{s}^{-1}$
Thus the magnitude of angular momentum, $L=14.4kg{m}^2{s}^{-1}$.

Therefore, the magnitude of its angular momentum about a point on the ground right under the center of the circle is$14.4kg{m}^2{s}^{-1}$.