Question 412: Centre of Mass and Rotational Motion

Question

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 _________ $$\mathrm{kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$$.

Accepted Answer

$$y-x-4=0$$
$$d_1$$ is perpendicular distance of given line from origin.
$$\mathrm{d}_1=\left|\frac{-4}{\sqrt{1^2+1^2}}ight| \Rightarrow 2 \sqrt{2} \mathrm{~m}$$
So
$$\begin{aligned}
|\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 	imes 3 \sqrt{2} 	imes 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \
& =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}
\end{aligned}$$

Centre of Mass and Rotational Motion

A body of mass $5 \mathrm{~kg} moving with a uniform speed 3 \sqrt{2} \mathrm{~ms}^{-1} in X-Y plane along the line y=x+4. The angular momentum of the particle about the origin will be _________ \mathrm{kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$.

Integer / Numeric Answer Type