A thin rod of length L and uniform cross-section is pivoted at its lowest point P inside a stationar

Q: Why do both the weight and the buoyant force act at the same point $G$?

Since the rod is uniform, its center of gravity is at its geometric center. Because the rod is completely submerged in a homogeneous liquid, the center of buoyancy (the centroid of the displaced liquid volume) coincides exactly with the center of gravity of the rod. Thus, both forces act at the same point $G$.

Q: Why is there a negative sign in the torque equation?

The negative sign indicates that the torque is a restoring torque. When the rod is displaced in the clockwise direction (positive $\theta$), the net upward force creates a counter-clockwise torque (negative direction) trying to bring the rod back to its vertical equilibrium position.

Q: What would happen if the density of the rod $d_1$ was greater than the density of the liquid $d_2$?

If $d_1 > d_2$, the weight would be greater than the buoyant force, resulting in a net downward force. In this case, the vertical downward position is already stable, and displacing it would lead to standard pendulum-like oscillations. However, if pivoted at the bottom, the rod would simply fall down to the bottom of the container and would not oscillate about the bottom pivot.

Q: How does the length of the rod affect the frequency of oscillation?

The angular frequency $\omega$ is inversely proportional to the square root of the length $L$ ($\omega \propto 1/\sqrt{L}$). This means a longer rod will oscillate more slowly (lower frequency, longer time period) due to its larger rotational inertia.

Q: Why did we use $I = \frac{1}{3}ML^2$ instead of $I = \frac{1}{12}ML^2$?

The moment of inertia of a uniform rod about an axis passing through its center of mass is $\frac{1}{12}ML^2$. However, our rod is pivoted at its lowest end $P$. By the parallel axis theorem, or standard derivation, the moment of inertia about one of its ends is $I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$.

Question

A thin rod of length $L$ and uniform cross-section is pivoted at its lowest point $P$ inside a stationary homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through $P$.

The density $d_1$ of the material of the rod is smaller than the density $d_2$ of the liquid. The rod is displaced by small angle $	heta$ from its equilibrium position and then released. Show that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters.

Sigma Percentile · Accepted Answer

### Introduction

Imagine a highly rigid, uniform rod submerged completely inside a stationary, non-viscous liquid.

Because the rod is less dense than the liquid ($d_1 < d_2$), it experiences a powerful upward buoyant force that exceeds its own weight.

Normally, this rod would shoot straight up to the surface and float.

However, because it is anchored at its lowest point $P$, it is forced to remain submerged.

This creates a fascinating mechanical system where any small angular displacement results in a beautiful, rhythmic oscillation.

Let's dive deep into the physics of this system and derive its angular frequency of oscillation.

### Analyzing the Equilibrium

In the vertical position, the rod is in a state of stable equilibrium.

Both the downward gravitational force (weight) and the upward buoyant force act along the exact same vertical line passing through the pivot $P$.

Because these forces are collinear and pass directly through the pivot, they produce zero net torque.

Thus, the rod remains perfectly upright.

### The Battle of Forces: Buoyancy vs. Gravity

When we displace the rod by a small angle $	heta$, the center of gravity $G$ (which is also the center of buoyancy since the rod is uniform and fully submerged) shifts horizontally.

Let's write down the expressions for the two competing forces acting at $G$:

1. **Weight of the rod ($W$):**

$$W = M g = (S L d_1) g$$

where $S$ is the cross-sectional area, $L$ is the length, and $d_1$ is the density of the rod.

2. **Buoyant Force ($F_B$):**

$$F_B = (S L d_2) g$$

where $d_2$ is the density of the liquid.

Since $d_2 > d_1$, the buoyant force is greater than the weight.

This results in a net upward force acting vertically at $G$:

$$F_{	ext{net}} = F_B - W = S L g (d_2 - d_1)$$

### Calculating the Restoring Torque

This net upward force $F_{	ext{net}}$ acts at the center of gravity $G$, which is at a distance of $\frac{L}{2}$ from the pivot $P$ along the rod.

Because the force is vertical and the rod is tilted, this force creates a torque that tries to pull the rod back to the vertical position.

The perpendicular distance from the pivot $P$ to the vertical line of action of $F_{	ext{net}}$ is:

$$r_{\perp} = \frac{L}{2} \sin	heta$$

Therefore, the restoring torque $	au$ about the pivot $P$ is:

$$	au = - F_{	ext{net}} \cdot r_{\perp} = - S L g (d_2 - d_1) \left(\frac{L}{2} \sin	hetaight)$$

The negative sign mathematically represents that the torque acts in the direction opposite to the angular displacement $	heta$, making it a **restoring torque**.

### The Small Angle Approximation

For small oscillations, we can use the Taylor series approximation for the sine function, where $\sin	heta \approx 	heta$ (in radians).

Substituting this into our torque equation yields:

$$	au \approx - \left[ \frac{S L^2 g (d_2 - d_1)}{2} ight] 	heta$$

Since the restoring torque is directly proportional to the negative of the angular displacement ($	au \propto -	heta$), the motion is confirmed to be **Simple Harmonic Motion (SHM)**!

### Rotational Inertia of the Rod

To find the angular acceleration, we must relate the torque to the rotational inertia of the system using Newton's second law for rotation:

$$	au = I \alpha = I \frac{d^2	heta}{dt^2}$$

For a uniform rod of mass $M$ pivoted at one of its ends, the moment of inertia $I$ is:

$$I = \frac{M L^2}{3}$$

Expressing the mass $M$ in terms of the rod's density $d_1$ and volume $S L$:

$$I = \frac{(S L d_1) L^2}{3} = \frac{S L^3 d_1}{3}$$

### The Master Equation of Motion

Now, let's equate our two expressions for the torque:

$$\left( \frac{S L^3 d_1}{3} ight) \frac{d^2	heta}{dt^2} = - \left[ \frac{S L^2 g (d_2 - d_1)}{2} ight] 	heta$$

We can beautifully simplify this equation by canceling the common terms $S$ and $L^2$ from both sides:

$$\left( \frac{L d_1}{3} ight) \frac{d^2	heta}{dt^2} = - \left[ \frac{g (d_2 - d_1)}{2} ight] 	heta$$

Solving for the angular acceleration $\alpha = \frac{d^2	heta}{dt^2}$:

$$\frac{d^2	heta}{dt^2} = - \left[ \frac{3g (d_2 - d_1)}{2 d_1 L} ight] 	heta$$

### Extracting the Angular Frequency

Comparing this with the standard differential equation for angular SHM:

$$\frac{d^2	heta}{dt^2} = - \omega^2 	heta$$

We can directly identify the square of the angular frequency:

$$\omega^2 = \frac{3g (d_2 - d_1)}{2 d_1 L}$$

Taking the square root of both sides gives us the final elegant expression for the angular frequency of oscillation:

$$\omega = \sqrt{\frac{3g (d_2 - d_1)}{2 d_1 L}}$$

This result shows that the frequency depends on the relative difference in densities, the acceleration due to gravity, and is inversely proportional to the length of the rod.

Visualized Solution

Understanding the Physical Setup

Displacing the Rod by a Small Angle $θ$

Forces Acting on the Rod

Net Upward Force at $G$

Restoring Torque about Pivot $P$

Applying Small Angle Approximation

Moment of Inertia of the Rod

Setting up the Differential Equation

Simplifying for Angular Acceleration $α$

Determining Angular Frequency $ω$

The Sigma Insight: Force and Energy Method in SHM

Introduction

Analyzing the Equilibrium

The Battle of Forces

Calculating the Restoring Torque

The Small Angle Approximation

Rotational Inertia of the Rod

The Master Equation of Motion

Extracting the Angular Frequency

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