Animated Solution for Physics - System of Particles and Rotational Motion: A small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an intial velocity $v_0$ in the $xy$ plane as shown in the figure. At a time $t<\frac{v_0\sin \theta }{g}$, the angular momentum of the particle is

The flight of a projectile is one of the most poetic phenomena in classical mechanics. When you throw a ball, it traces a perfect parabola, governed by the relentless, invisible hand of gravity. But while we often focus on its position or velocity, there's another hidden quantity silently evolving as the particle flies: its angular momentum.

In this problem, we are asked to find the angular momentum of a projectile about the origin at some time $t$. I know cross products and vectors can sometimes look intimidating, but let's take a breath. We are going to break this down step-by-step, and you'll see how beautifully the math unfolds to reveal the physics.

To find the angular momentum $\vec{L}$ about the origin, we need our master tool: the definition of angular momentum for a point particle.

This equation tells us a story. It says that to know the angular momentum, we need to know exactly where the particle is (its position vector, $\vec{r}$) and exactly how fast and in what direction it's moving (its velocity vector, $\vec{v}$).

Let's find these vectors. Imagine the particle at some time $t$. It has moved horizontally and vertically. The horizontal motion is uniform because there's no force acting in the $x$-direction. So, the $x$-coordinate is simply the horizontal velocity multiplied by time:

The vertical motion is influenced by gravity, pulling it down. Using our trusty kinematic equation $s=ut+\frac{1}{2}a{t}^2$, the $y$-coordinate is:

Combining these, our position vector $\vec{r}$ is:

Now, what about the velocity? The horizontal velocity remains constant, while the vertical velocity decreases due to gravity ($v=u+at$).

We have our $\vec{r}$ and our $\vec{v}$. Now, we bring them together in the cross product.

This looks like a massive wall of algebra, but don't make a silly mistake here. We just need to remember the rules of the cross product for unit vectors. Any vector crossed with itself is zero, so $\hat{i}\times \hat{i}=0$ and $\hat{j}\times \hat{j}=0$. We only care about the cross terms! We know that $\hat{i}\times \hat{j}=\hat{k}$ and $\hat{j}\times \hat{i}=-\hat{k}$.

Let's multiply the $x$-component of $\vec{r}$ with the $y$-component of $\vec{v}$, and subtract the product of the $y$-component of $\vec{r}$ and the $x$-component of $\vec{v}$.

Now, look closely at the expression inside the bracket. Both major terms share a common factor: $v_0\cos \theta t$. Let's factor it out to make our lives easier.

Notice how the $t$ from the second term was absorbed into the factored out $t$. Now, let's simplify the inside of the bracket.

Do you see it? The $v_0\sin \theta$ terms cancel each other out perfectly! This is the elegance of physics. We are left with:

Multiplying this back with our factored terms, we arrive at our final destination:

What does this mean physically? The negative $\hat{k}$ direction tells us that the angular momentum vector points into the page. If you use the right-hand rule from $\vec{r}$ to $\vec{v}$, your thumb points inwards. This makes perfect sense because gravity is constantly trying to rotate the particle clockwise about the origin. Furthermore, the magnitude of the angular momentum grows quadratically with time ($t^2$). It's not conserved because gravity is exerting a continuous torque on the system!