A thin circular ring of mass M and radius r is rotating about its axis with a constant angular veloc

Q: Why is the initial moment of inertia $M r^2$ and not $\frac{1}{2} M r^2$?

Because it's a **ring**, not a solid disc. All the mass of a ring is concentrated at its rim, at a distance $r$ from the central axis. The formula $\frac{1}{2} M r^2$ is for a uniform solid disc.

Q: Why do we add $2 m r^2$ to the final moment of inertia?

The two masses are placed at the opposite ends of a diameter, which means each mass is exactly at a distance $r$ from the axis of rotation. The moment of inertia of a point mass is $m r^2$, and since there are two of them, we add $2 m r^2$.

Q: What if the masses were dropped on the ring with some initial velocity?

If they had an initial velocity, they would bring their own initial angular momentum into the system. You would have to add their initial angular momentum to the ring's initial angular momentum before equating it to the final state.

Q: Does the kinetic energy of the system remain conserved in this process?

No! This is a perfectly inelastic collision (the masses stick to the ring). Rotational kinetic energy is lost in the form of heat or sound when the masses are "attached gently" and forced to accelerate to the ring's speed.

Q: Why doesn't gravity create a torque here?

Gravity acts downwards, parallel to the axis of rotation (assuming the ring is rotating in a horizontal plane). A force parallel to the axis of rotation cannot produce a torque about that axis.

Question

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects, each of mass $m$, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity

Options:

(A) $\frac{\omega M}{M + m}$

(B) $\frac{\omega (M - 2m)}{M + 2m}$

(C) $\frac{\omega M}{M + 2m}$

(D) $\frac{\omega (M + 2m)}{M}$

Sigma Percentile · Accepted Answer

### The Ballet of the Spinning Ring

Imagine you are watching a beautiful ballet. A dancer is spinning gracefully on the ice. Suddenly, she extends her arms, and her spin slows down. This elegant phenomenon is governed by one of the most profound laws of physics: the **Conservation of Angular Momentum**.

In our problem, we have a thin circular ring of mass $M$ and radius $r$, acting just like our ballet dancer. It is initially rotating with a constant angular velocity $\omega$.

### The Magic Word: "Gently"

The problem states that two objects, each of mass $m$, are attached **gently** to the opposite ends of a diameter. In the language of physics, "gently" is a massive clue. It means that the masses are placed without applying any external twisting force, or **torque**, to the ring.

When the net external torque on a system is zero, the total angular momentum of the system must remain perfectly conserved.

This gives us our master equation:
$L_i = L_f$

### Calculating the Moments of Inertia

To use our master equation, we first need to understand how the mass is distributed before and after the objects are added. This distribution is measured by the **moment of inertia** ($I$).

Initially, we only have the thin ring. All of its mass $M$ is concentrated at a distance $r$ from the central axis.
Therefore, the initial moment of inertia is:
$I_i = M r^2$

Now, what happens when we add the two masses? The ring is still there, but we have added two point masses, each of mass $m$, at the very edge of the ring (distance $r$ from the axis).

The moment of inertia of a single point mass is $m r^2$. Since we added two of them, the final moment of inertia becomes:
$I_f = M r^2 + m r^2 + m r^2 = (M + 2m) r^2$

### The Grand Finale

Now, we bring it all together. The angular momentum is the product of the moment of inertia and the angular velocity ($L = I \omega$).

Equating the initial and final angular momenta:
$I_i \omega_i = I_f \omega_f$

Substituting our expressions for the moments of inertia:
$(M r^2) \omega = (M + 2m) r^2 \omega_f$

Notice how the $r^2$ terms beautifully cancel out from both sides! This tells us that the final angular velocity doesn't even depend on the radius of the ring.

Solving for the final angular velocity $\omega_f$:
$\omega_f = \frac{M}{M + 2m} \omega$

Because we added mass to the outer edge, the moment of inertia increased. To keep the angular momentum conserved, the ring had to slow down, just like the ballet dancer extending her arms. The physics is not just mathematically elegant; it is physically intuitive!

Select Answer:

Visualized Solution

$Initial State$

$Initial Moment of Inertia$

$Adding Masses$

$Conservation of Angular Momentum$

$Final Moment of Inertia$

$Equating Angular Momenta$

$Final Calculation$

The Sigma Insight: Conservation of Angular Momentum

The Ballet of the Spinning Ring

The Magic Word: "Gently"

Calculating the Moments of Inertia

The Grand Finale

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