A body of mass $5 \mathrm{~kg}$ moving with a uniform speed $3 \sqrt{2} \mathrm{~ms}^{-1}$ in $X-Y$ plane along the line $y=x+4$. The angular momentum of the particle about the origin will be _________ $\mathrm{kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$.
Answer (integer)
60
Solution
<p>$y-x-4=0$</p>
<p>$d_1$ is perpendicular distance of given line from origin.</p>
<p>$$\mathrm{d}_1=\left|\frac{-4}{\sqrt{1^2+1^2}}\right| \Rightarrow 2 \sqrt{2} \mathrm{~m}$$</p>
<p>So</p>
<p>$$\begin{aligned}
|\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \\
& =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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