Question

A particle of mass $5g$ is moving with a speed of $3\sqrt{2}cm{s}^{-1}$ in $xy$ plane along the line $y=x+4$. The magnitude of its angular momentum about the origin in $gc{m}^2{s}^{-1}$ is:

• A. $zero$
• B. $60$
• C. $30$
• D. $\frac{30}{\sqrt{2}}$

Answer 1

Answer: (B)

$60$

Angular momentum is defined as $L=\overrightarrow{r}$x $\overrightarrow{p}=m(\overrightarrow{r}$x $\overrightarrow{v})$ .....(1)
The particle moves along the line $y=x+4$
Slope of the line $y=x+4$ is $\tan \theta =1$.
Thus, the particle moves along a line which makes an angle of $\theta ={45}^0$ with $x-$axis.
Given,
$\left|\overrightarrow{v}\right|=3\sqrt{2}$
Hence,
$\overrightarrow{v}=3\sqrt{2}(\cos {45}^0\hat{i}+\sin {45}^0\hat{j})$
$\overrightarrow{r}=x\hat{i}+(x+4)\hat{j}$
Substituting $\overrightarrow{r}$ and $\overrightarrow{v}$ in eqn(1) we get:

$\overrightarrow{L}=m(x\hat{i}+(x+4\left)\hat{j}\right)$ x $\left(3\sqrt{2}\right(\cos {45}^0\hat{i}+\sin {45}^0\hat{j}\left)\right)$
$\therefore \overrightarrow{L}=5\times (x\hat{i}+(x+4\left)\hat{j}\right)$ x $\left(3\sqrt{2}\right(\cos {45}^0\hat{i}+\sin {45}^0\hat{j}\left)\right)$
$\therefore \overrightarrow{L}=5\times 3\sqrt{2}\times \frac{1}{\sqrt{2}}(x\hat{k}-(x+4\left)\hat{k}\right)=-60\hat{k}gmc{m}^{2}se{c}^{-1}$
$\therefore \left|\overrightarrow{L}\right|=60gmc{m}^{2}se{c}^{-1}$
We have used the following identities:
$\hat{i}$ x $\hat{j}$ $=\hat{k}$
$\hat{j}$ x $\hat{i}$ $=-\hat{k}$