Animated Solution for Physics - Rotational Motion: A cubical block of side $a$ moving with velocity $v$ on a horizontal smooth plane as shown. It hits a ridge at point $O$. The angular speed of the block after it hits $O$ is

A Sliding Cube Meets a Ridge

Imagine a solid cubical block of mass $M$ and side length $a$ gliding effortlessly across a smooth, frictionless horizontal plane with a constant velocity $v$. Suddenly, its bottom-right corner strikes a small, immovable ridge at point $O$.

This isn't a gentle bump; it's a harsh, perfectly inelastic collision at that specific contact point. The cube's linear motion is abruptly halted at the corner, forcing the entire block to pivot and topple over the ridge. Our goal is to find the exact angular speed $\omega$ with which the cube begins to rotate immediately after the impact.

Conservation of Angular Momentum

When the cube hits the ridge, a massive impulsive force acts on it at point $O$. This force is what stops the corner from moving forward. If we were to analyze the linear momentum, we'd find it is definitely not conserved because of this external impulsive force.

However, physics offers us a beautiful loophole. If we calculate the torque of this impulsive force about point $O$ itself, what do we get? Zero! The force acts exactly at the pivot, meaning its lever arm is zero.

Since the net impulsive torque about point $O$ is zero, the angular momentum of the cube about point $O$ must remain strictly conserved just before and just after the collision.

Just before the impact, the cube is purely translating. It has no rotation. The angular momentum of a translating body about any point is simply its linear momentum multiplied by the perpendicular distance from the point to the line of action of the velocity.

The center of mass of the cube is located exactly at its geometric center, which is at a height of $\frac{a}{2}$ from the ground. The velocity vector $v$ passes horizontally through this center of mass.

Therefore, the perpendicular distance from the ridge $O$ to the velocity line is $\frac{a}{2}$.

The initial angular momentum $L_i$ about $O$ is:

Immediately after the impact, the cube is purely rotating about the ridge $O$ with an angular velocity $\omega$. The final angular momentum $L_f$ is given by the product of its moment of inertia about $O$ and its angular velocity:

To find $I_O$, we must first determine the moment of inertia of the cube about its center of mass, $I_{CM}$. For a solid cube of side $a$, the moment of inertia about an axis passing through its center and perpendicular to its face is:

Next, we need the distance $d$ from the center of mass to the pivot point $O$. Using the Pythagorean theorem on the half-sides of the cube:

Now, we deploy the Parallel Axis Theorem to shift our axis from the center of mass to the ridge $O$:

Substituting our values:

Equating and Solving

We have successfully gathered all the pieces of our puzzle. By the principle of conservation of angular momentum, we equate the initial and final states:

Now, it is just a matter of simple algebra. We can cancel the mass $M$ and one factor of the side length $a$ from both sides. Rearranging the equation to solve for the angular speed $\omega$:

And there we have it! The cube swings upward with this exact angular velocity. This problem beautifully demonstrates how choosing the right pivot point can make a seemingly complex collision problem elegantly simple.