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

Q: Why is angular momentum conserved here?

Because the masses are placed 'gently', meaning no external torque is applied to the system along the axis of rotation.

Q: What is the moment of inertia of a point mass?

For a point mass $M$ at a distance $R$ from the axis of rotation, its moment of inertia is $MR^2$.

Q: Why does the angular velocity decrease?

Since angular momentum ($L = I\omega$) is constant, an increase in the moment of inertia ($I$) must result in a proportional decrease in angular velocity ($\omega$).

Q: Does the radius of the ring matter in the final answer?

No, as you can see in the derivation, the $R^2$ term cancels out from both sides of the equation.

Q: What if the masses were not placed at the ends of a diameter?

As long as they are placed at a distance $R$ from the axis (on the ring), their moment of inertia would still be $MR^2$ each, so the result would be the same.

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 ring now rotates with an angular velocity $\omega'$ =

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a thin circular ring of mass $m$ and radius $R$ spinning smoothly about its central axis with an initial angular velocity $\omega$. This is our initial state. The ring possesses a certain amount of angular momentum due to its rotation.

Now, two objects, each of mass $M$, are gently placed at the opposite ends of its diameter. The word **gently** is crucial here. It implies that no external torque is applied to the system along the axis of rotation. Because of this addition of mass, the ring will now rotate with a new angular velocity, let's call it $\omega'$.

### The Master Equation
Since there is no external torque acting on the system ($	au_{ext} = 0$), the total angular momentum must remain conserved. This is a fundamental principle of rotational dynamics, often tested in JEE.

We can write the conservation of angular momentum as:
$$L_i = L_f$$
$$I_1 \omega = I_2 \omega'$$

### Calculating Moment of Inertia
Let's calculate the moment of inertia for both states. Initially, it's just the ring rotating about its central axis. The moment of inertia of a ring is given by:
$$I_1 = mR^2$$

Finally, we have the ring plus the two point masses. Each point mass $M$ is at a distance $R$ from the axis of rotation. The moment of inertia of a point mass is $MR^2$. Since there are two such masses, their combined moment of inertia is $2MR^2$. Therefore, the final moment of inertia of the system is:
$$I_2 = mR^2 + 2MR^2 = (m + 2M)R^2$$

### Final Calculation
Now, we substitute these values into our conservation equation:
$$mR^2 \omega = (m + 2M)R^2 \omega'$$

Notice how the $R^2$ terms cancel out beautifully from both sides. This tells us that the final angular velocity is independent of the radius of the ring! Rearranging for $\omega'$, we find:
$$\omega' = \frac{m}{m + 2M} \omega$$

**The angular velocity decreases** because the moment of inertia of the system increased while the angular momentum remained constant. This is the exact same principle that explains why an ice skater spins slower when they extend their arms!

Select Answer:

Visualized Solution

$L_{i} = I_{1} ω$

$L_{f} = I_{2} ω^{'}$

$τ_{e x t} = 0 \Rightarrow L_{i} = L_{f}$

$I_{1}$ and $I_{2}$

$m R^{2} ω = (m + 2 M) R^{2} ω^{'}$

$ω^{'} = \frac{m}{m + 2 M} ω$

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Setup Imagine a thin circular ring of mass $m$ and radius $R$ spinning smoothly about its central axis with an initial angular velocity $ω$

The Master Equation Since there is no external torque acting on the system ( $τ_{e x t} = 0$ ), the total angular momentum must remain conserved

Calculating Moment of Inertia Let's calculate the moment of inertia for both states

Final Calculation

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