JEE Main 2016 — Rotational Motion Question with Solution

The angular momentum is

$$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$$

For the particle moving from $$D\to A$$, we have

$$\overrightarrow{L}=m\left[\left(\frac{R}{\sqrt{2}}\hat{i}+\frac{R}{\sqrt{2}}\hat{j}\right)\times v(-\hat{j})\right]$$

$$=\frac{mvR}{\sqrt{2}}(-\hat{k})$$

For the particle moving from $$A\to B$$, we have

$$\overrightarrow{L}=m\left[\left(\frac{R}{\sqrt{2}}\hat{i}+\frac{R}{\sqrt{2}}\hat{j}\right)\times v\hat{i}\right]$$

$$=\frac{mvR}{\sqrt{2}}(-\hat{k})$$

For the particle moving from $$C\to D$$, we have

$$\overrightarrow{L}=mv\times$$ ($$\perp$$ distance) $$\times$$ $$(\hat{k})$$

$$=mv\left(\frac{R}{\sqrt{2}}+a\right)(\hat{k})$$

For the particle moving from $$B\to C$$, we get

$$\overrightarrow{L}=mv\times$$ ($$\perp$$ distance) $$\times$$ $$(\hat{k})$$

$$=mv\left(\frac{R}{\sqrt{2}}+a\right)(\hat{k})$$