A particle of mass m is projected at time t = 0 from a point P on the ground with a speed v0, at an

Q: Why does a projectile have angular momentum if it's not moving in a circle?

Angular momentum is defined relative to a specific origin as $\mathbf{r} \times \mathbf{p}$. As long as the particle's line of motion doesn't pass directly through the origin, it has a non-zero angular momentum. It doesn't need to be physically tied to a string or moving in a circle.

Q: Why did we use $t = v_0/g$? Is it a special point?

It's just a specific instant given in the problem. Interestingly, the total time of flight is $T = \frac{2v_0 \sin 45^\circ}{g} = \frac{\sqrt{2}v_0}{g} \approx 1.414 \frac{v_0}{g}$. So $t = v_0/g$ is a point past the maximum height, while the particle is on its way down.

Q: What does the negative sign in the final answer mean?

The negative sign on the $\hat{k}$ vector indicates that the angular momentum points into the page. By the right-hand rule, if you curl your fingers from the $\mathbf{r}$ vector towards the $\mathbf{v}$ vector, your thumb points inwards, corresponding to a clockwise rotation sense about the origin.

Q: Could we have used $\mathbf{L} = I\omega$?

No. The equation $\mathbf{L} = I\omega$ is generally reserved for rigid bodies rotating about a fixed axis or objects in pure circular motion. For a point particle in an arbitrary path like a parabola, always stick to the fundamental definition $\mathbf{L} = \mathbf{r} \times \mathbf{p}$.

Q: Is there a faster way to solve this without the messy cross product?

Yes! You can use the torque-angular momentum theorem: $\tau = \frac{d\mathbf{L}}{dt}$. The torque is $\mathbf{r} \times (-mg\hat{j}) = -mgx\hat{k}$. Since $x = (\frac{v_0}{\sqrt{2}})t$, you can simply integrate this torque from $t=0$ to $t=v_0/g$ to get the answer in three lines!

Question

A particle of mass $m$ is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_0$, at an angle of $45^\circ$ to the horizontal. Find the magnitude and direction of the angular momentum of the particle about $P$ at time $t = \frac{v_0}{g}$.

Sigma Percentile · Accepted Answer

### The Socratic Journey: Unraveling Angular Momentum in Projectile Motion

Imagine a particle launched from the ground, tracing a beautiful parabolic path through the air. When we hear the term "angular momentum," we usually picture a spinning top or a planet orbiting a star. But what does angular momentum mean for a projectile?

Angular momentum is simply the "moment of momentum," defined mathematically as $\mathbf{L} = \mathbf{r} 	imes \mathbf{p}$. As long as the particle's line of motion does not pass directly through our chosen origin, it possesses angular momentum relative to that origin. Let's explore two distinct ways to solve this classic JEE problem: the standard kinematic approach, and a brilliant "ninja technique" using torque.

### Phase 1: The Brute Force Kinematics (The Standard Path)

Our goal is to find the angular momentum about the launch point $P(0,0)$ at a very specific time, $t = \frac{v_0}{g}$.

First, we break the initial velocity into its horizontal and vertical components. Since the launch angle is $45^\circ$, both components are equal:

$u_x = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}$

$u_y = v_0 \sin 45^\circ = \frac{v_0}{\sqrt{2}}$

Next, we need the exact position vector $\mathbf{r}$ and velocity vector $\mathbf{v}$ at our target time. Using the equations of kinematics:

**Position Coordinates:**
$x = u_x t = \frac{v_0}{\sqrt{2}} \left(\frac{v_0}{g}ight) = \frac{v_0^2}{\sqrt{2}g}$

$y = u_y t - \frac{1}{2}gt^2 = \frac{v_0}{\sqrt{2}} \left(\frac{v_0}{g}ight) - \frac{1}{2}g \left(\frac{v_0}{g}ight)^2 = \frac{v_0^2}{\sqrt{2}g} - \frac{v_0^2}{2g}$

**Velocity Components:**
$v_x = u_x = \frac{v_0}{\sqrt{2}}$

$v_y = u_y - gt = \frac{v_0}{\sqrt{2}} - g\left(\frac{v_0}{g}ight) = \frac{v_0}{\sqrt{2}} - v_0$

Now, we deploy the cross product formula for angular momentum in two dimensions: $\mathbf{L} = m(x v_y - y v_x)\hat{k}$. Let's carefully substitute our expressions. It looks intimidating, but watch the magic of algebra unfold:

$x v_y = \frac{v_0^2}{\sqrt{2}g} \left(\frac{v_0}{\sqrt{2}} - v_0ight) = \frac{v_0^3}{2g} - \frac{v_0^3}{\sqrt{2}g}$

$y v_x = \left(\frac{v_0^2}{\sqrt{2}g} - \frac{v_0^2}{2g}ight) \frac{v_0}{\sqrt{2}} = \frac{v_0^3}{2g} - \frac{v_0^3}{2\sqrt{2}g}$

Subtracting the two yields our final determinant:

$x v_y - y v_x = \left(\frac{v_0^3}{2g} - \frac{v_0^3}{\sqrt{2}g}ight) - \left(\frac{v_0^3}{2g} - \frac{v_0^3}{2\sqrt{2}g}ight) = -\frac{v_0^3}{2\sqrt{2}g}$

Multiplying by mass $m$, we get $\mathbf{L} = -\frac{mv_0^3}{2\sqrt{2}g} \hat{k}$. The magnitude is $\frac{mv_0^3}{2\sqrt{2}g}$, and the negative $\hat{k}$ direction means it points perpendicularly into the page.

### Phase 2: The Ninja Technique (Integration of Torque)

Is there a faster way? Absolutely. Let's invoke Newton's Second Law for rotation: $	au = \frac{d\mathbf{L}}{dt}$.

If we can find the torque acting on the particle, we can simply integrate it over time to find the change in angular momentum. What forces are acting on our projectile? Only gravity, acting straight down! So, $\mathbf{F} = -mg\hat{j}$.

The torque about the origin is:

$	au = \mathbf{r} 	imes \mathbf{F} = (x\hat{i} + y\hat{j}) 	imes (-mg\hat{j}) = -mgx (\hat{i} 	imes \hat{j}) = -mgx\hat{k}$

Notice something beautiful? The torque depends *only* on the $x$-coordinate! We already know that $x = u_x t = \frac{v_0}{\sqrt{2}} t$. Substituting this into our torque equation gives:

$	au = -mg \left(\frac{v_0}{\sqrt{2}} tight) \hat{k}$

Now, we just integrate this torque from $t = 0$ to $t = \frac{v_0}{g}$:

$\mathbf{L} = \int_0^{\frac{v_0}{g}} 	au dt = \int_0^{\frac{v_0}{g}} -\frac{mgv_0}{\sqrt{2}} t \, dt \, \hat{k}$

$\mathbf{L} = -\frac{mgv_0}{\sqrt{2}} \left[ \frac{t^2}{2} ight]_0^{\frac{v_0}{g}} \hat{k} = -\frac{mgv_0}{\sqrt{2}} \left( \frac{v_0^2}{2g^2} ight) \hat{k} = -\frac{mv_0^3}{2\sqrt{2}g} \hat{k}$

In just three lines of calculus, we arrived at the exact same answer! This is the elegance of physics—different mathematical paths converging on the same universal truth.

Visualized Solution

Initial Setup

Initial Velocity Components

Position Vector at $t = \frac{v _{0}}{g}$

Velocity Vector at $t = \frac{v _{0}}{g}$

Angular Momentum Formula

Computing the Cross Product

Final Calculation

Conclusion

The Sigma Insight: Torque and Angular Momentum

The Socratic Journey

Phase 1

Phase 2

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