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A particle of mass ' $\mathrm{m}$ ' is moving along a line $\mathrm{y}=\mathrm{x}+\mathrm{a}$ with a constant velocity ' $v$ '. The angular momentum of the particle about the origin is
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Verified Answer
The correct answer is:
$\frac{\text { mva }}{\sqrt{2}}$
Given, $\mathrm{y}=\mathrm{x}+\mathrm{a}$
compairing with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
$$
\begin{aligned}
& \mathrm{m}=1, \mathrm{c}=\mathrm{a} \\
& \tan \theta=1=\tan 45^{\circ} \\
& \theta=45^{\circ}
\end{aligned}
$$
The velocity in vector form
$$
\begin{aligned}
\vec{v} & =v \cos \theta \hat{\mathrm{i}}+v \sin \theta \hat{\mathrm{j}}=v \cos 45 \hat{\mathrm{i}}+v \sin 45 \hat{\mathrm{j}} \\
& =\frac{v}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})
\end{aligned}
$$
Let us assume particle is at point $(0, a)$ $\overrightarrow{\mathrm{r}}=(0-0) \hat{\mathrm{i}}+(\mathrm{a}-0) \hat{\mathrm{j}}=\mathrm{a} \hat{\mathrm{j}}$
Angular momentum,
$$
\begin{aligned}
& \overline{\mathrm{L}}=m(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{v}}) \\
& =m\left(a \hat{\mathrm{j}} \times \frac{v}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\right)=\frac{m v a}{\sqrt{2}}(-\hat{\mathrm{k}})
\end{aligned}
$$
compairing with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
$$
\begin{aligned}
& \mathrm{m}=1, \mathrm{c}=\mathrm{a} \\
& \tan \theta=1=\tan 45^{\circ} \\
& \theta=45^{\circ}
\end{aligned}
$$
The velocity in vector form
$$
\begin{aligned}
\vec{v} & =v \cos \theta \hat{\mathrm{i}}+v \sin \theta \hat{\mathrm{j}}=v \cos 45 \hat{\mathrm{i}}+v \sin 45 \hat{\mathrm{j}} \\
& =\frac{v}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})
\end{aligned}
$$
Let us assume particle is at point $(0, a)$ $\overrightarrow{\mathrm{r}}=(0-0) \hat{\mathrm{i}}+(\mathrm{a}-0) \hat{\mathrm{j}}=\mathrm{a} \hat{\mathrm{j}}$
Angular momentum,
$$
\begin{aligned}
& \overline{\mathrm{L}}=m(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{v}}) \\
& =m\left(a \hat{\mathrm{j}} \times \frac{v}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})\right)=\frac{m v a}{\sqrt{2}}(-\hat{\mathrm{k}})
\end{aligned}
$$
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