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Problem: A particle of mass m is moving along the side of a square of side a, with a uniform speed v in the x-y plane as shown in the figure. Which of the following statements is false for the angular momentum $\vec{L}$ about the origin? (JEE Mains 2016)
Solution: The angular momentum of a particle of mass $m$ moving with a velocity $\vec{v}$ at the position $\vec{r}$ is given by \begin{align} \vec{L}=m\vec{r}\times\vec{v}=mvr_\perp \, \hat{n}, \nonumber \end{align} where $r_\perp$ is the component of $\vec{r}$ perpendicular to $\vec{v}$ (perpendicular distance or lever arm) and $\hat{n}$ is the unit vector in the direction of $\vec{r}\times\vec{v}$. Note that angular momentum is about the origin as $\vec{r}$ is a position vector from the origin to the current location of the particle. These quantities on given paths of the particle are \begin{align} &\mathrm{AB:} && r_\perp=\tfrac{R}{\sqrt{2}}, &&\vec{L}=mv\left(\tfrac{R}{\sqrt{2}}\right)(-\hat{k}),\nonumber\\ &\mathrm{BC:} && r_\perp=\tfrac{R}{\sqrt{2}}+a, &&\vec{L}=mv\left(\tfrac{R}{\sqrt{2}}+a\right)\hat{k},\nonumber \\ &\mathrm{CD:} && r_\perp=\tfrac{R}{\sqrt{2}}+a, &&\vec{L}=mv\left(\tfrac{R}{\sqrt{2}}+a\right)\hat{k},\nonumber\\ &\mathrm{DA:} && r_\perp=\tfrac{R}{\sqrt{2}}, &&\vec{L}=mv\left(\tfrac{R}{\sqrt{2}}\right)(-\hat{k}),\nonumber \end{align} Note that $\hat{n}$ is into the paper on the paths AB and DA.