A mass m is moving with a constant velocity along a line parallel to the X-axis, away from the origi

Q: Why doesn't the angular momentum increase as the particle gets further away?

While the distance $r$ increases, the angle $\theta$ decreases. The product $r \sin\theta$ represents the perpendicular distance to the line of motion, which remains perfectly constant.

Q: What if the particle was accelerating?

If the particle was accelerating, its velocity $v$ would change. Since $L = mvh$, a changing velocity would mean the angular momentum is no longer constant.

Q: Does the choice of origin matter?

The magnitude of angular momentum depends on the perpendicular distance $h$ from the chosen origin to the line of motion. While the value of $L$ might be different for a different origin, it will still remain constant as long as the velocity is constant.

Q: How do we determine the direction of the angular momentum?

Use the right-hand rule for the cross product $\vec{r} \times \vec{v}$. Point your fingers along $\vec{r}$ and curl them towards $\vec{v}$. Your thumb will point into the page, which is the negative Z-direction.

Q: Is angular momentum always conserved for a free particle?

Yes! A free particle experiences no net force, so there is no net torque about any fixed point. By Newton's second law for rotation ($\vec{\tau} = d\vec{L}/dt$), zero torque means angular momentum is conserved.

Question

A mass $m$ is moving with a constant velocity along a line parallel to the $X$-axis, away from the origin. Its angular momentum with respect to the origin

Options:

(A) is zero

(B) remains constant

(C) goes on increasing

(D) goes on decreasing

Sigma Percentile · Accepted Answer

Have you ever watched a train moving along a straight track in the distance and wondered about its angular momentum? It might seem counterintuitive to think about "angular" momentum for something moving in a straight line, but this is one of the most beautiful and frequently tested concepts in physics!

Let's embark on a journey to decode this elegant problem.

### Visualizing the Setup

Imagine you are standing at the origin of a coordinate system. A particle of mass $m$ is moving along a straight line parallel to the X-axis. It is moving away from you with a constant velocity $\vec{v}$.

Let's draw a perpendicular line from your position (the origin) to the path of the particle. We will call this perpendicular distance $h$. As the particle moves, its position vector $\vec{r}$ stretches and changes its angle $	heta$ with the X-axis.

At first glance, it feels like the angular momentum should change because the particle is getting further away. But physics has a beautiful surprise waiting for us!

### The Master Equation

To find the truth, we must return to the fundamental definition of angular momentum about a point. The angular momentum $\vec{L}$ is defined as the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p}$.

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

If we only look at the magnitude of this cross product, we get:

$$
|\vec{L}| = m r v \sin	heta
$$

Here, $	heta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$. Since $\vec{v}$ is parallel to the X-axis, $	heta$ is exactly the angle the position vector makes with the X-axis.

### The Geometric Trick

This is where the magic happens. As the particle moves forward, the distance $r$ increases, and the angle $	heta$ decreases. It seems like a complicated mess!

But look closely at the right-angled triangle formed by the position vector, the X-axis, and the perpendicular dropped from the particle. Using basic trigonometry, we can see that:

$$
\sin	heta = \frac{h}{r}
$$

Rearranging this gives us a profound geometric truth:

$$
r \sin	heta = h
$$

No matter where the particle is on that line, the product $r \sin	heta$ is always exactly equal to the constant perpendicular height $h$!

### The Final Revelation

Now, let's substitute this geometric truth back into our angular momentum equation:

$$
|\vec{L}| = m v (r \sin	heta) = m v h
$$

Look at that beautiful result! The mass $m$ is constant. The velocity $v$ is constant. And the perpendicular distance $h$ is constant. Therefore, the magnitude of the angular momentum is perfectly constant!

What about the direction? Using the right-hand rule, if you curl your fingers from $\vec{r}$ towards $\vec{v}$, your thumb points straight into the screen (the negative Z-direction). This direction never changes as the particle moves.

Since both the magnitude and the direction are constant, the angular momentum vector remains absolutely constant. This is a powerful universal principle: **a free particle moving with constant velocity has a constant angular momentum about any fixed point in space!**

Animated Solution for Physics - Rotational Motion: A mass $m$ is moving with a constant velocity along a line parallel to the $X$ -axis, away from the origin. Its angular momentum with respect to the origin

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Angular Momentum Formula$

$Magnitude of Angular Momentum$

$Geometric Substitution$

$Final Conclusion$

$The Way Forward$

The Sigma Insight: Torque and Angular Momentum

Visualizing the Setup

The Master Equation

The Geometric Trick

The Final Revelation

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