Question

What are the magnitude and direction of the angular momentum relative to the origin of the particle

Short answer

Hence the momentum of the particle is$-\left(1.273\mathrm{kg}\cdot {\mathrm{m}}^2/\mathrm{s}\right)\widehat{\mathbf{k}}$

Step-by-step solution

Step :1 Introduction

In the same manner that torque is the rotational equivalent of force, angular momentum is the rotational equivalent of linear momentum. In vector notation, angular momentum is given by

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

Here $\overrightarrow{r}$ is the position vector of the particle and$\overrightarrow{p}$ is the momentum vector of the given particle.

Step : 2 Explanation 

The travelling particle's direction is depicted in the diagram below.

Step :3 Moving in direction 

The particle is travelling with a velocity of $3.0\mathrm{m}/\mathrm{s}$in a direction ${45}^{\circ }$below the positive $x$-axis. The particle's horizontal and vertical components of velocity are then calculated.

$v_x=v\cos {45}^{\circ }$

$\begin{array}{r}=(3.0\mathrm{m}/\mathrm{s})\cos {45}^{\circ }\end{array}$

$\begin{array}{r}\\ =2.121\mathrm{m}/\mathrm{s}\end{array}$

And

$v_y=-v\sin {45}^{\circ }$

$=-(3.0\mathrm{m}/\mathrm{s})\sin {45}^{\circ }$

$=-2.121\mathrm{m}/\mathrm{s}$

Step : 4 Mass of the particle 

The mass $\left(m\right)$of the particle $200g$is $gtokg$

$m=200g$

$=\left(200\mathrm{g}\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{kg}}\right)$

$=0.200\mathrm{kg}$

The given particle lies in the $XY$plane. Then position vector of the given particle is

$\overrightarrow{r}=\left(1.0\mathrm{m}\right)\widehat{\mathbf{i}}+\left(2.0\mathrm{m}\right)\widehat{\mathbf{j}}$

And velocity vector of the given particle is

$\overrightarrow{v}=v_x\widehat{\mathbf{i}}+v_y\widehat{\mathbf{j}}$

$=(2.121\mathrm{m}/\mathrm{s})\widehat{\mathrm{i}}+(-2.121\mathrm{m}/\mathrm{s})\widehat{\mathrm{j}}$

$=(2.121\widehat{\mathrm{i}}-2.121\widehat{\mathrm{j}})\mathrm{m}/\mathrm{s}$

Step :5 Angular momentum

Since angular momentum is the rotational equivalent of linear momentum in much the same way that torque is the rotational equivalent of force. Angular momentum in vector notation is given by

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

$=\overrightarrow{r}\times m\overrightarrow{v}$

$=m(\overrightarrow{r}\times \overrightarrow{v})$

$=\left(0.20\mathrm{kg}\right)\left|\begin{array}{ccc}\widehat{\mathbf{i}}& \widehat{\mathbf{j}}& \widehat{\mathbf{k}}\\ 1.0\mathrm{m}& 2.0\mathrm{m}& 0\\ 2.121\mathrm{m}/\mathrm{s}& -2.121\mathrm{m}/\mathrm{s}& 0\end{array}\right|$

$=\left(0.20\mathrm{kg}\right)\left[\widehat{\mathbf{i}}\right(0-0)+\widehat{\mathbf{j}}(0-0)+\widehat{\mathbf{k}}(1.0\mathrm{m}\left)\right(-2.121\mathrm{m}/\mathrm{s})-(2.0\mathrm{m}\left)\right(2.121\mathrm{m}/\mathrm{s}\left)\right]$

$\overrightarrow{L}=-\left(1.273\mathrm{kg}\cdot {\mathrm{m}}^2/\mathrm{s}\right)\widehat{\mathbf{k}}$