Question

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 $\overrightarrow{L}$ about the origin?

• A. $L=-\frac{mv}{\sqrt{2}}R$ $\hat{k}$ when the particle is moving from A to B.
• B. $L=mv\left[\frac{R}{\sqrt{2}}-a\right]\hat{k}$ when the particle is moving from C to D.
• C. $L=mv\left[\frac{R}{\sqrt{2}}+a\right]\hat{k}$ when the particle is moving from B to C.
• D. $L=\frac{mv}{\sqrt{2}}R$ $\hat{k}$ when the particle is moving from D to A.

Answer 1

The correct option is B $L=mv\left[\frac{R}{\sqrt{2}}-a\right]\hat{k}$ when the particle is moving from C to D.

The angular momentum of a particle of mass $m$ with a velocity $\overrightarrow{v}$ and position vector $\overrightarrow{r}$.
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
The magnitude can be given as, $L=mvr$
From C to D, $\overrightarrow{r}$ is $(\frac{R}{\sqrt{2}}+a)$
So, $\overrightarrow{L}$ from C to D is $=mv(\frac{R}{\sqrt{2}}+a)\hat{k}$
Hence option B is incorrect.