Rotational Motion Question 6
Question 6 - 2024 (29 Jan Shift 2)
A body of mass $5 kg$ moving with a uniform speed $3 \sqrt{2} ms^{-1}$ in $X-Y$ plane along the line $y=x+4$. The angular momentum of the particle about the origin will be $\operatorname{kgm}^{2} s^{-1}$.
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Answer: (60)
Solution:
$y-x-4=0$
$d _1$ is perpendicular distance of given line from origin.
$d _1=\left|\frac{-4}{\sqrt{1^{2}+1^{2}}}\right| \Rightarrow 2 \sqrt{2} m$
So,
$$ \begin{aligned} |\overrightarrow{L}|=mvd _1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} kg m^{2} / s \\ & =60 kg m^{2} / s \end{aligned} $$
