Animated Solution for Physics - Rotational Motion: A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be (Take, $g=10{\text{ m/s}}^2$)

A Rollercoaster Ride

Imagine you are standing at the top of a smooth, frictionless rollercoaster track. A small particle of mass $m=20\text{ g}$ is released from point A with an initial push, giving it a velocity of $v_A=5\text{ m/s}$. It slides down the curve to the lowest point B. Our mission is to find the angular momentum of this particle about a specific point O, located high above the track, exactly when the particle reaches the bottom.

Conservation of Energy

Because the track is completely frictionless, there are no non-conservative forces doing work on our particle. This means we can confidently apply the Principle of Conservation of Mechanical Energy. The total energy (kinetic plus potential) at point A must perfectly equal the total energy at point B.

Let's set up the equation:

Notice how the mass $m$ appears in every term? We can elegantly cancel it out, showing that the final velocity is independent of the particle's mass. Rearranging for the final velocity squared, we get:

Now, let's substitute our known values. The initial velocity $v_A$ is $5\text{ m/s}$, the acceleration due to gravity $g$ is $10{\text{ m/s}}^2$, and the vertical drop $h$ from A to B is $10\text{ m}$.

Taking the square root of 225, we find that the particle is zooming past point B at a crisp $v_B=15\text{ m/s}$.

Angular Momentum

We aren't just looking for velocity; we need the angular momentum $\vec{L}$ about point O. The fundamental definition of angular momentum is the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p}$:

The magnitude of this cross product is given by $L=m{v}_B{r}_B\sin (\theta )$. At the lowest point B, the velocity vector is perfectly horizontal. The position vector from O to B is perfectly vertical. Therefore, the angle $\theta$ between them is exactly $90^{\circ }$, and $\sin (90^{\circ })=1$.

To finish the problem, we need the perpendicular distance $r_B$ from O to B. Looking at the diagram, point O is a distance $a=10\text{ m}$ above A's level, and A is $h=10\text{ m}$ above B. Thus, the total vertical distance is:

Now, we substitute everything into our angular momentum formula. Don't forget to convert the mass to standard SI units ($20\text{ g}=20\times 10^{-3}\text{ kg}$):

And there we have it! The angular momentum of the particle about point O is exactly $6{\text{ kg-m}}^2\text{/s}$.