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