A particle of mass m is moving along the side of a square of side a, with a uniform speed v in the X

Q: Why is the position vector r not just the distance from the origin?

Angular momentum depends on the full vector $\mathbf{r}$, not just its magnitude. The cross product automatically extracts the perpendicular distance from the origin to the line of motion.

Q: How do we find the exact coordinates of point A?

The vector $\mathbf{R}$ has a length $R$ and an angle of $45^\circ$ with the X-axis. Using standard polar to Cartesian conversion, $x = R\cos 45^\circ$ and $y = R\sin 45^\circ$.

Q: Why does the x or y variable disappear when we take the cross product?

Because the cross product of parallel vectors is zero (e.g., $\hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0$). When velocity is purely in the X direction, only the Y component of the position vector contributes to the angular momentum.

Q: Is the angular momentum constant while the particle moves along a single side?

Yes! Because the perpendicular distance from the origin to the straight line of motion remains perfectly constant, the angular momentum $\mathbf{L}$ does not change until the particle turns the corner.

Q: Why are there two false statements? Can a question have multiple answers?

Yes, in JEE Advanced and certain sections of JEE Main, questions can be 'Multiple Correct' type. Here, both options (b) and (d) have incorrect mathematical signs, making them both false.

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 $\mathbf{L}$ about the origin?

Options:

(A) $\mathbf{L} = -\frac{mvR}{\sqrt{2}} \hat{\mathbf{k}}$, when the particle is moving from A to B.

(B) $\mathbf{L} = mv\left(\frac{R}{\sqrt{2}} - a\right) \hat{\mathbf{k}}$, when the particle is moving from B to C.

(C) $\mathbf{L} = mv\left(\frac{R}{\sqrt{2}} + a\right) \hat{\mathbf{k}}$, when the particle is moving from C to D.

(D) $\mathbf{L} = \frac{mvR}{\sqrt{2}} \hat{\mathbf{k}}$, when the particle is moving from D to A.

Sigma Percentile · Accepted Answer

### The Power of the Cross Product

Angular momentum is one of the most beautiful and conserved quantities in physics. When a particle moves in a straight line, it might seem counterintuitive that it possesses angular momentum. However, angular momentum is always defined relative to a specific origin.

The fundamental definition of angular momentum $\mathbf{L}$ for a point particle is the cross product of its position vector $\mathbf{r}$ and its linear momentum $\mathbf{p}$:

$$\mathbf{L} = \mathbf{r} 	imes \mathbf{p} = m(\mathbf{r} 	imes \mathbf{v})$$

In this problem, we have a particle tracing the perimeter of a square $ABCD$. To find the angular momentum along each side, we must carefully define the position vector $\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}$ and the velocity vector $\mathbf{v}$ at any instant.

### Analyzing the Setup

The problem gives us the position of point $A$ via a vector of length $R$ at an angle of $45^\circ$. Using basic trigonometry, the coordinates of $A$ are:

$$A = \left(R\cos 45^\circ, R\sin 45^\circight) = \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}ight)$$

Since the square has a side length of $a$, the other vertices are simply shifted by $a$ along the $X$ or $Y$ axes:
*   $B = \left(\frac{R}{\sqrt{2}} + a, \frac{R}{\sqrt{2}}ight)$
*   $C = \left(\frac{R}{\sqrt{2}} + a, \frac{R}{\sqrt{2}} + aight)$
*   $D = \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}} + aight)$

### The Master Equation in Action

Let's evaluate the cross product for each path. Remember the cyclic rules for unit vectors: $\hat{\mathbf{i}} 	imes \hat{\mathbf{j}} = \hat{\mathbf{k}}$ and $\hat{\mathbf{j}} 	imes \hat{\mathbf{i}} = -\hat{\mathbf{k}}$. Any cross product of a vector with itself is zero (e.g., $\hat{\mathbf{i}} 	imes \hat{\mathbf{i}} = 0$).

**Path A to B:**
The particle moves horizontally to the right, so $\mathbf{v} = v\hat{\mathbf{i}}$. The $y$-coordinate is constant at $\frac{R}{\sqrt{2}}$.
$$\mathbf{L}_{AB} = m\left[ \left(x\hat{\mathbf{i}} + \frac{R}{\sqrt{2}}\hat{\mathbf{j}}ight) 	imes v\hat{\mathbf{i}} ight] = m\left( \frac{R}{\sqrt{2}}v ight) (\hat{\mathbf{j}} 	imes \hat{\mathbf{i}}) = -\frac{mvR}{\sqrt{2}}\hat{\mathbf{k}}$$
This matches statement (a), so it is **True**.

**Path B to C:**
The particle moves vertically upwards, so $\mathbf{v} = v\hat{\mathbf{j}}$. The $x$-coordinate is constant at $\frac{R}{\sqrt{2}} + a$.
$$\mathbf{L}_{BC} = m\left[ \left(\left(\frac{R}{\sqrt{2}} + aight)\hat{\mathbf{i}} + y\hat{\mathbf{j}}ight) 	imes v\hat{\mathbf{j}} ight] = mv\left(\frac{R}{\sqrt{2}} + aight)(\hat{\mathbf{i}} 	imes \hat{\mathbf{j}}) = mv\left(\frac{R}{\sqrt{2}} + aight)\hat{\mathbf{k}}$$
Statement (b) claims the term is $\left(\frac{R}{\sqrt{2}} - aight)$, which makes it **False**.

**Path C to D:**
The particle moves horizontally to the left, so $\mathbf{v} = -v\hat{\mathbf{i}}$. The $y$-coordinate is constant at $\frac{R}{\sqrt{2}} + a$.
$$\mathbf{L}_{CD} = m\left[ \left(x\hat{\mathbf{i}} + \left(\frac{R}{\sqrt{2}} + aight)\hat{\mathbf{j}}ight) 	imes (-v\hat{\mathbf{i}}) ight] = m\left(\frac{R}{\sqrt{2}} + aight)(-v) (\hat{\mathbf{j}} 	imes \hat{\mathbf{i}}) = mv\left(\frac{R}{\sqrt{2}} + aight)\hat{\mathbf{k}}$$
This matches statement (c), so it is **True**.

**Path D to A:**
The particle moves vertically downwards, so $\mathbf{v} = -v\hat{\mathbf{j}}$. The $x$-coordinate is constant at $\frac{R}{\sqrt{2}}$.
$$\mathbf{L}_{DA} = m\left[ \left(\frac{R}{\sqrt{2}}\hat{\mathbf{i}} + y\hat{\mathbf{j}}ight) 	imes (-v\hat{\mathbf{j}}) ight] = m\left(\frac{R}{\sqrt{2}}ight)(-v) (\hat{\mathbf{i}} 	imes \hat{\mathbf{j}}) = -\frac{mvR}{\sqrt{2}}\hat{\mathbf{k}}$$
Statement (d) claims the value is positive, which makes it **False**.

### Final Conclusion

By systematically applying the cross product, we easily identified that statements (b) and (d) contain sign or algebraic errors. In multiple-correct questions, maintaining strict sign conventions is the key to securing full marks!

Select Answer:

Visualized Solution

Coordinates of the Square

Angular Momentum Formula

Path A \rightarrow B

Path B \rightarrow C

Path C \rightarrow D

Path D \rightarrow A

Conclusion

The Sigma Insight: Torque and Angular Momentum

The Power of the Cross Product

Analyzing the Setup

The Master Equation in Action

Final Conclusion

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