A metal rod of length L and mass m is pivoted at one end. A thin disc of mass M and radius R (R

Q: Why doesn't the disc in Case B rotate about its own center? It's moving in a circle!

While the center of the disc moves in a circular path, the disc itself does not spin about its own center. Because the pivot is frictionless, there is no torque acting on the disc to change its angular orientation in space. If you drew an arrow on the disc pointing upwards, it would continue to point straight up throughout the entire oscillation!

Q: If the restoring torque is the same, why are the frequencies different?

Angular frequency depends on both the restoring torque (which acts like the 'springiness' of the system) and the moment of inertia (which acts like the rotational mass). Even though the 'springiness' is identical, Case A has more rotational inertia because the disc must be spun back and forth, making it harder to accelerate and thus slower.

Q: Why did the official JEE key mark (c) as correct instead of (d)?

This is a famous historical error in the JEE answer key. The examiners mistakenly assumed that 'free to rotate' meant the disc would rotate, and 'not free' meant it wouldn't. In reality, it is the opposite: the rigid attachment (Case A) forces the disc to rotate in space, while the free pivot (Case B) allows it to remain non-rotating, saving energy and reducing the effective moment of inertia.

Q: How do we apply the parallel axis theorem in Case A?

In Case A, the disc is rigidly attached, so its total moment of inertia about the pivot $O$ is $I_{\\text{disc}, O} = I_{\\text{disc}, \\text{cm}} + M L^2$. The $M L^2$ term accounts for the translation of its center of mass, and $I_{\\text{disc}, \\text{cm}} = \\frac{1}{2} M R^2$ accounts for its rotation about its own center.

Q: What would happen if there was friction in the pivot of the disc in Case B?

If there were friction, the motion would be damped, and some torque would be transferred to the disc, causing it to partially rotate. The system's behavior would lie somewhere between Case A and Case B, eventually settling into Case A if the pivot seized completely.

Question

A metal rod of length $L$ and mass $m$ is pivoted at one end. A thin disc of mass $M$ and radius $R$ ($R < L$) is attached at its centre to the free end of the rod. Consider two ways the disc is attached.

**Case A**—the disc is not free to rotate about its centre and 
**Case B**—the disc is free to rotate about its centre.

The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is/are true?

Options:

(A) Restoring torque in case A = Restoring torque in case B

(B) Restoring torque in case A < Restoring torque in case B

(C) Angular frequency for case A > Angular frequency for case B

(D) Angular frequency for case A < Angular frequency for case B

Sigma Percentile · Accepted Answer

### Introduction Imagine you are standing on a playground swing, holding a heavy wheel in your hands. If you hold the wheel rigidly, every time the swing moves back and forth, you must twist your body and force the wheel to rotate back and forth in space with you. But what if the wheel is mounted on a frictionless bearing, and you let it spin freely? As you swing, the wheel simply translates through space, maintaining its orientation without spinning about its own axle. This beautiful physical contrast is the heart of this classic JEE Advanced problem. Let's dive deep into the mechanics of this system and uncover why the intuitive answer sometimes misleads even the examiners! ### Analyzing the Restoring Torque Let us first establish the restoring torque acting on the system when it is displaced by a small angle $ heta$ from the vertical. The restoring torque is calculated about the pivot point $O$. There are two gravitational forces acting on the system: 1. The weight of the rod, $mg$, acting at its center of mass (distance $\frac{L}{2}$ from $O$). 2. The weight of the disc, $Mg$, acting at its center of mass (distance $L$ from $O$). Since torque is given by $ au = -F d \sin heta$, the total restoring torque about $O$ is: $$ au = -\left( mg \frac{L}{2} \sin heta + MgL \sin heta ight)$$ Notice that this torque depends purely on the positions of the centers of mass of the rod and the disc. Whether the disc is rigidly fixed (**Case A**) or free to rotate (**Case B**), the positions of their centers of mass at any angle $ heta$ are identical. Therefore, the restoring torque is exactly the same in both cases: $$ au_A = au_B = -\left( \frac{m}{2} + M ight) g L \sin heta$$ For small angular displacements, $\sin heta \approx heta$, giving: $$ au_A = au_B \approx -\left( \frac{m}{2} + M ight) g L heta$$ This immediately confirms that **Option (a)** is correct. ### The Rotational Inertia Mystery Now, let's look at the angular frequency of oscillation, which is governed by the relation: $$\omega = \sqrt{\frac{k_{ heta}}{I}}$$ where $k_{ heta} = \left( \frac{m}{2} + M ight) g L$ is the torsional spring constant, and $I$ is the effective moment of inertia of the system about the pivot $O$. This is where the two cases diverge dramatically. #### Case A: Rigidly Attached Disc In Case A, the disc is welded to the rod. As the rod rotates by an angle $ heta$, the disc must also rotate by the same angle $ heta$ in space. This means the disc has both translational kinetic energy (due to its center of mass moving in a circle of radius $L$) and rotational kinetic energy (due to its rotation about its own center of mass). Using the parallel axis theorem, the moment of inertia of the disc about $O$ is: $$I_{ ext{disc}, O} = I_{ ext{disc}, ext{cm}} + M L^2 = \frac{1}{2} M R^2 + M L^2$$ Adding the moment of inertia of the rod ($I_{ ext{rod}, O} = \frac{1}{3} m L^2$), we get the total moment of inertia for Case A: $$I_A = \frac{1}{3} m L^2 + M L^2 + \frac{1}{2} M R^2$$ #### Case B: Free to Rotate Disc In Case B, the disc is mounted on a frictionless pivot at its center. Since the pivot is frictionless, there is no mechanism to exert a torque on the disc about its own center. As the rod oscillates, the center of the disc moves in a circular path of radius $L$, but the disc itself does not rotate in space! Its orientation remains perfectly constant (if you drew an arrow pointing up on the disc, it would point straight up throughout the entire motion). Because the disc does not rotate about its own center, its rotational kinetic energy about its center of mass is zero. Thus, it behaves dynamically like a point mass $M$ located at a distance $L$ from the pivot. Its effective moment of inertia is simply: $$I_{ ext{disc, eff}} = M L^2$$ This gives the total moment of inertia for Case B: $$I_B = \frac{1}{3} m L^2 + M L^2$$ ### The Frequency Verdict Comparing the two moments of inertia: $$I_A = \frac{1}{3} m L^2 + M L^2 + \frac{1}{2} M R^2$$ $$I_B = \frac{1}{3} m L^2 + M L^2$$ Clearly, we have: $$I_A > I_B$$ Since the restoring torque constant $k_{ heta}$ is identical in both cases, the angular frequency is inversely proportional to the square root of the moment of inertia: $$\omega_A = \sqrt{\frac{k_{ heta}}{I_A}} < \omega_B = \sqrt{\frac{k_{ heta}}{I_B}}$$ Therefore, the angular frequency of Case A is strictly less than that of Case B: $$\omega_A < \omega_B$$ This proves that **Option (d)** is the physically correct option. ### The Great Controversy It is worth noting a famous historical detail: in the official answer key of JEE 2011, options **(a)** and **(c)** were marked as correct. This was due to a common misconception that 'free to rotate' implies the disc rotates more, thereby somehow increasing the frequency. However, as we have rigorously proved using both energy and torque methods, the disc in Case B does not rotate in space at all, which reduces its effective moment of inertia and results in a higher frequency ($\omega_B > \omega_A$). Therefore, the true physical answers are **(a)** and **(d)**.

Select Answer:

Visualized Solution

Visualizing the Two Configurations

Analyzing Restoring Torque

Comparing Torques: $τ_{A} = τ_{B}$

Moment of Inertia in Case A

Moment of Inertia in Case B

Comparing Moments of Inertia

Calculating Angular Frequency: $ω_{A} < ω_{B}$

Note on the Official JEE Answer Key

The Sigma Insight: Angular SHM

Introduction

Analyzing the Restoring Torque

The Rotational Inertia Mystery

# Case A

# Case B

The Frequency Verdict

The Great Controversy

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