Animated Solution for Physics - Rotational Motion: Comprehension Passage

The beauty of rigid body dynamics lies in how elegantly it combines linear and rotational motion. In this problem, we are presented with a classic scenario: a fast-moving particle striking a pivoted disk. This isn't just a simple collision; it's a beautiful dance of angular momentum and energy conservation.

Imagine a heavy circular disk of mass $M=1\text{ kg}$ and radius $R=0.2\text{ m}$, hanging from a pivot at its top edge, point $C$. A tiny particle of mass $m=0.02\text{ kg}$ comes flying in horizontally from the right with a speed of $v_i=100\text{ m/s}$. It strikes the disk at point $P$, which is located at an angle of $45^{\circ }$ from the vertical.

Before we write any equations, we must decode the geometry. To calculate angular momentum about the pivot $C$, we need the perpendicular distances from $C$ to the lines of motion of the particle.

The vertical drop from $C$ to $P$ is the radius $R$ plus the vertical component of the radius $OP$, which is $R\cos 45^{\circ }$. So, the vertical distance is $y_P=R+\frac{R}{\sqrt{2}}$.

The horizontal shift from the center line to $P$ is simply $x_P=R\sin 45^{\circ }=\frac{R}{\sqrt{2}}$.

Angular Momentum

During the split-second of the collision, the pivot at $C$ exerts a massive impulsive force to keep the disk attached. Because this force acts exactly at point $C$, its lever arm is zero, meaning it creates zero torque about $C$. This is our golden ticket: the total angular momentum of the system about the pivot $C$ is perfectly conserved!

Let's set up the conservation equation: ${\vec{L}}_i={\vec{L}}_f$.

Initially, only the particle is moving horizontally. Its angular momentum about $C$ is its linear momentum multiplied by the vertical distance:

After the collision, the disk spins with an angular velocity $\omega$, and the particle deflects straight down with a speed $v_f=90\text{ m/s}$. The particle's new line of motion is vertical, so we use the horizontal distance from $C$:

Wait, what is $I_C$? The disk is rotating around its edge, not its center. We must use the Parallel Axis Theorem: $I_C=I_{\text{cm}}+M{d}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$.

Now, we carefully substitute our given values into the angular momentum equation:

Simplifying the terms:

Solving for $\omega$, we find the disk's angular velocity right after the hit is $\omega =7.138\text{ rad/s}$.

With the disk now spinning, it swings upward against gravity. Since the collision is over, the mechanical energy for the disk alone is conserved. The rotational kinetic energy it just gained will completely convert into gravitational potential energy at its highest point. We need to find how high the center $O$ rises, let's call it $h$.

Plugging in the numbers:

That's the answer to our first question! The center $O$ rises by $0.15\text{ m}$.

Energy Loss

Moving to the second part, we need to find the energy lost during the impact. This is an inelastic collision, so kinetic energy is not conserved. The loss is simply the initial kinetic energy of the incoming particle minus the total final kinetic energy of both the deflected particle and the spinning disk.

The initial energy is:

The final energy is the particle's new energy plus the disk's rotational energy:

Subtracting the two, we find the energy loss:

The problem is beautifully solved! By carefully tracking the geometry, applying the right conservation laws at the right times, and keeping our signs straight, a complex rotational collision becomes a straightforward sequence of logical steps.