Question

A particle of mass $m$ is moving along the side of a square of side ${}^{\prime }{a}^{\prime }$. with a uniform speed $\nu$ in the $x-y$ plane as shown in the figure:
Which of the following statement is false for the angular momentum $L$ about the origin?

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

Answer 1

The correct option is A $L=\frac{mv}{\sqrt{2}}R\hat{k}$ when the particle is moving from $A$ to $B$

Angular momentum,

$\overrightarrow{L}=m\left(R\sin {45}^o\hat{i}\right)\times v(-\hat{j})=\frac{mv\overrightarrow{R}}{\sqrt{2}}(-\hat{k})$ When particle move from D to A.

Hence, $\overrightarrow{L}=\frac{mv\overrightarrow{R}}{\sqrt{2}}\hat{k}$ is wrong.