Animated Solution for Physics - Rotational Motion: An annular disk of mass M, inner radius a and outer radius b is placed on a horizontal surface with coefficient of friction $\mu$, as shown in the figure. At some time, an impulse $J_0\hat{x}$ is applied at a height h above the center of the disk. If $h=h_m$ then the disk rolls without slipping along the x-axis. Which of the following statement(s) is(are) correct ?

The physics of rigid body dynamics often presents us with beautiful symmetries between linear and rotational motion. This problem from JEE Advanced 2023 is a perfect example of how a single impulsive force can dictate the entire future state of a rolling object.

Imagine you are standing next to a massive, hollow wheel—an annular disk. You have a hammer, and you are going to strike it horizontally. Where you hit it determines whether it slides, spins in place, or perfectly rolls away. This "sweet spot" is what we call the center of percussion, and our goal is to find its exact height, $h_m$.

When the impulse $J_0$ strikes the disk, it does two things simultaneously. First, it delivers a massive, instantaneous push that changes the disk's linear momentum. According to the impulse-momentum theorem, the linear impulse equals the change in linear momentum. Since the disk starts from rest, we can write:

This equation tells us the velocity $v$ of the center of mass immediately after the strike. But the impulse isn't applied at the center; it's applied at a height $h$ above it. This offset creates a twisting effect—an angular impulse.

Just as linear impulse changes linear momentum, angular impulse changes angular momentum. The angular impulse is simply the linear impulse multiplied by its lever arm from the center of mass, which is $h_m$. This must equal the change in angular momentum, $I_c\omega$:

Now we have two equations governing the two types of motion. But they are independent until we link them with a physical constraint.

The problem states that if we hit the disk exactly at height $h_m$, it begins to roll without slipping immediately. Pure rolling means the point of the disk in contact with the ground has zero velocity relative to the ground. For this to happen, the forward velocity of the center $v$ must be perfectly canceled out by the backward velocity caused by rotation at the bottom edge. Since the outer radius is $b$, the condition is:

This is the bridge we needed! Let's substitute $v$ and $\omega$ into our angular impulse equation. We know $v=\frac{J_0}{M}$ and $\omega =\frac{v}{b}=\frac{J_0}{Mb}$. Plugging this into $J_0{h}_m=I_c\omega$, we get:

Notice how beautifully the impulse $J_0$ cancels out! The sweet spot $h_m$ depends only on the geometry and mass distribution of the disk, not on how hard you hit it:

To go further, we need the moment of inertia $I_c$ of our annular disk. Think of an annular disk as a large solid disk of radius $b$ with a smaller solid disk of radius $a$ removed from its center. Using standard derivations, the moment of inertia for an annular disk of mass $M$ is:

Substituting this back into our equation for $h_m$, the mass $M$ cancels out, leaving us with a purely geometric result:

Now, let's play with this result and test the given options.

Option A: What if the inner radius $a$ shrinks to zero? The hole vanishes, and we get a solid disk. Plugging $a=0$ into our formula gives $h_m=\frac{b^2}{2b}=\frac{b}{2}$. This matches Option A perfectly.

Option B: What if the inner radius $a$ expands until it's almost equal to the outer radius $b$? All the mass is pushed to the rim, creating a thin ring. Plugging $a=b$ gives $h_m=\frac{b^2+b^2}{2b}=b$. This matches Option B perfectly.

Option C: Let's look at the initial angular velocity $\omega$. We found earlier that $\omega =\frac{J_0}{Mb}$. Does this contain the inner radius $a$? No! The angular velocity depends only on the impulse, the mass, and the outer radius. It is completely independent of $a$. Option C is correct.

Option D: Finally, what if we hit the disk exactly at the center ($h=0$)? The lever arm is zero, so the angular impulse is zero. The disk gains linear velocity but no angular velocity ($\omega =0$). Without rotation, the bottom point is moving forward at velocity $v$, meaning it slides without rolling. Option D is correct.

In the end, all four statements reveal a deep, consistent truth about the mechanics of rolling bodies.