A ring of mass M and radius R is rotating with angular speed omega about a fixed vertical axis passi

Q: Why is the angular momentum conserved?

Angular momentum is conserved because there is no external torque acting on the system about the vertical axis of rotation. The forces moving the masses outwards are internal forces (or centrifugal pseudo forces if viewed from the rotating frame), which do not exert a net torque about the axis.

Q: Why is the initial moment of inertia only $MR^2$?

Initially, the two point masses are at the center of the ring ($O$). Their distance from the axis of rotation is zero ($r=0$). Since moment of inertia is $mr^2$, their contribution is zero. Thus, only the ring contributes to the initial moment of inertia.

Q: Does the radial velocity of the masses affect the angular momentum?

No, the radial velocity does not affect the angular momentum about the vertical axis. Angular momentum is $\vec{r} \times \vec{p}$. For radial motion, the velocity vector is parallel to the position vector, so their cross product is zero. Only the tangential velocity (due to rotation) contributes to the angular momentum.

Q: Is the kinetic energy of the system conserved?

No, the rotational kinetic energy is not conserved. As the masses move outwards, the moment of inertia increases, and the angular velocity decreases. The rotational kinetic energy $\frac{1}{2}I\omega^2 = \frac{L^2}{2I}$ decreases because $I$ increases while $L$ is constant. The 'lost' energy is converted into the radial kinetic energy of the masses or dissipated if there's friction along the rods.

Q: What if the masses were not equal?

The principle of conservation of angular momentum would still apply perfectly. You would just substitute their respective masses $m_1$ and $m_2$ into the final moment of inertia equation: $I_f = MR^2 + m_1 r_1^2 + m_2 r_2^2$.

Question

A ring of mass $M$ and radius $R$ is rotating with angular speed $\omega$ about a fixed vertical axis passing through its centre $O$ with two point masses each of mass $\frac{M}{8}$ at rest at $O$. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure.
At some instant, the angular speed of the system is $\frac{8}{9}\omega$ and one of the masses is at a distance of $\frac{3}{5}R$ from $O$. At this instant, the distance of the other mass from $O$ is

Options:

(A) $\frac{2}{3}R$

(B) $\frac{1}{3}R$

(C) $\frac{3}{5}R$

(D) $\frac{4}{5}R$

Sigma Percentile · Accepted Answer

The beauty of physics often lies in the invisible constraints that govern a system. When a spinning ice skater pulls their arms in, they spin faster. Why? Because of a profound law of nature: the conservation of angular momentum. In this JEE Advanced problem, we are presented with a mechanical analog of the ice skater, and our job is to decode the hidden geometry of its motion.

### Analyzing the Setup

Imagine a uniform ring of mass $M$ and radius $R$ rotating smoothly about its vertical central axis with an initial angular velocity $\omega$. Resting peacefully at the very center of this ring are two identical point masses, each with a mass of $\frac{M}{8}$.

These masses aren't glued down; they are free to slide radially outwards along two massless rods. As the system rotates, a tiny perturbation causes these masses to drift outwards. As they move away from the center, the mass distribution of the system changes.

### The Master Equation

The most crucial observation here is that **no external torque** is acting on the system about the vertical axis of rotation. The forces pushing the masses outwards are internal to the system.

According to Newton's laws applied to rotational motion, if the net external torque is zero, the total angular momentum ($L$) of the system must remain perfectly constant.

$$L_{initial} = L_{final}$$

This single equation is the master key to unlocking the entire problem.

### Setting up the States

Let's define our initial state. Initially, both point masses are sitting exactly at the center of rotation. Their distance from the axis is $r = 0$. Since the moment of inertia of a point mass is $mr^2$, their contribution to the initial moment of inertia is zero.

The initial moment of inertia is simply that of the ring:
$$I_{initial} = MR^2$$

Thus, the initial angular momentum is:
$$L_{initial} = I_{initial} \omega = (MR^2)\omega$$

Now, let's look at the final state. We are told that at a specific instant, the angular velocity has dropped to $\frac{8}{9}\omega$. One mass has slid out to a distance of $r_1 = \frac{3}{5}R$. Let's assume the second mass is at an unknown distance $x$.

The new moment of inertia of the entire system is the sum of the ring's moment of inertia and the moments of inertia of the two displaced masses:
$$I_{final} = MR^2 + \frac{M}{8}\left(\frac{3}{5}R\right)^2 + \frac{M}{8}x^2$$

The final angular momentum is:
$$L_{final} = I_{final} \left(\frac{8}{9}\omega\right)$$

### The Algebraic Battle

Equating the initial and final angular momenta, we get:
$$(MR^2)\omega = \left[ MR^2 + \frac{M}{8}\left(\frac{3}{5}R\right)^2 + \frac{M}{8}x^2 \right] \left(\frac{8}{9}\omega\right)$$

This might look intimidating, but let's take a breath and simplify. Notice that $M$ and $\omega$ appear on both sides. We can cleanly cancel them out!

$$R^2 = \left[ R^2 + \frac{1}{8}\left(\frac{9}{25}R^2\right) + \frac{x^2}{8} \right] \frac{8}{9}$$

Let's multiply the $\frac{1}{8}$ into the fraction:
$$R^2 = \left[ R^2 + \frac{9}{200}R^2 + \frac{x^2}{8} \right] \frac{8}{9}$$

To isolate the terms inside the bracket, let's multiply both sides by $\frac{9}{8}$:
$$\frac{9}{8}R^2 = R^2 + \frac{9}{200}R^2 + \frac{x^2}{8}$$

Now, let's group all the $R^2$ terms on the left side to solve for $x^2$:
$$\frac{x^2}{8} = \frac{9}{8}R^2 - R^2 - \frac{9}{200}R^2$$

Factoring out $R^2$, we have a simple fraction subtraction problem:
$$\frac{x^2}{8} = \left( \frac{9}{8} - 1 - \frac{9}{200} \right) R^2$$

To subtract these, we need a common denominator, which is 200:
$$\frac{x^2}{8} = \left( \frac{225}{200} - \frac{200}{200} - \frac{9}{200} \right) R^2$$

$$\frac{x^2}{8} = \left( \frac{225 - 200 - 9}{200} \right) R^2 = \frac{16}{200} R^2$$

Simplifying the fraction $\frac{16}{200}$ by dividing the numerator and denominator by 8 gives us $\frac{2}{25}$:
$$\frac{x^2}{8} = \frac{2}{25} R^2$$

Finally, multiply both sides by 8:
$$x^2 = \frac{16}{25} R^2$$

Taking the square root of both sides reveals the elegant final answer:
$$x = \frac{4}{5} R$$

### Conclusion

The math perfectly aligns with our physical intuition. As the masses moved outwards, the system's moment of inertia increased. To conserve angular momentum, the angular velocity had to decrease (from $\omega$ to $\frac{8}{9}\omega$). By tracking this exact exchange, we were able to pinpoint the exact location of the second mass.

This problem is a beautiful testament to how conservation laws act as an unbreakable accounting system for the universe!

Select Answer:

Visualized Solution

Visualizing the Setup

Conservation of Angular Momentum

Initial Angular Momentum

Final Angular Momentum Setup

Equating Initial and Final States

Simplifying the Equation

Rearranging Terms

Solving the Fractions

Final Calculation

Conclusion

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Setup

The Master Equation

Setting up the States

The Algebraic Battle

Conclusion

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