A point particle of mass M attached to one end of a massless rigid non-conducting rod of length L. A

Q: Why is the net force on the dipole zero, but the torque is non-zero?

In a uniform electric field, the forces on the positive and negative charges are equal in magnitude ($qE$) and opposite in direction. Therefore, they cancel each other out, resulting in zero net translational force. However, because these forces act at different points along the rod, they create a couple, which produces a non-zero rotational torque.

Q: Why did we calculate the moment of inertia about the midpoint of the rod?

Since the two masses at the ends of the rod are identical and the rod itself is massless, the center of mass of the system lies exactly at the midpoint of the rod. When the system is released in a uniform field, it rotates symmetrically about its center of mass, making the midpoint the natural axis of rotation.

Q: What is the physical meaning of the negative sign in the torque equation $\tau = -pE\sin\theta$?

The negative sign indicates that the torque is a restoring torque. It means that if you displace the rod in the clockwise direction, the torque will act in the counter-clockwise direction to pull it back to the equilibrium position, and vice versa.

Q: Why is the minimum time to become parallel equal to $T/4$ instead of some other fraction?

In simple harmonic motion, a particle starts from rest at maximum displacement (amplitude) and reaches the mean position (equilibrium) in exactly one-quarter of a full cycle. Since the parallel position ($\theta = 0$) is the equilibrium position, the time taken to reach it from the released position is $T/4$.

Q: What happens if the angle $\theta$ is large? Does the rod still execute simple harmonic motion?

No. If the angle $\theta$ is large, the approximation $\sin\theta \approx \theta$ is no longer valid. The restoring torque remains proportional to $\sin\theta$ rather than $\theta$, making the motion non-linear. While the motion is still periodic, it is no longer simple harmonic.

Question

A point particle of mass $M$ attached to one end of a massless rigid non-conducting rod of length $L$. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges $+q$ and $-q$ respectively. This arrangement is held in a region of a uniform electric field $E$ such that the rod makes a small angle $	heta$ (say of about 5 degrees) with the field direction. Find an expression for the minimum time needed for the rod to become parallel to the field after it is set free.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a delicate, massless, non-conducting rod of length $L$ suspended in a vast, uniform electric field $\vec{E}$. At the ends of this rod sit two identical point masses, each of mass $M$.

But there is a twist: one mass carries a positive charge $+q$, and the other carries a negative charge $-q$. Together, they form a classic **electric dipole** with a dipole moment vector $\vec{p}$ pointing from the negative charge to the positive charge. The magnitude of this dipole moment is:

$$p = qL$$

Now, let's tilt this rod slightly by a small angle $	heta$ (about $5^\circ$) relative to the direction of the electric field and release it from rest. What happens next?

### The Restoring Forces and Torque

The uniform electric field $\vec{E}$ immediately exerts electrostatic forces on both charges:

* A force $\vec{F}_+ = q\vec{E}$ pulls the positive charge in the direction of the field.
* An equal and opposite force $\vec{F}_- = -q\vec{E}$ pulls the negative charge opposite to the field.

Because these two forces are equal in magnitude and opposite in direction, the net translational force on the rod is zero. The rod will not fly away.

However, because these forces act at different points along the rod, they create a **restoring torque** that tries to rotate the rod back into alignment with the electric field. The magnitude of this restoring torque is given by:

$$	au = -p E \sin	heta$$

Substituting $p = qL$, we get:

$$	au = -q L E \sin	heta$$

The negative sign indicates that the torque acts in the direction opposite to the angular displacement $	heta$, pulling the system back toward equilibrium.

### Rotational Dynamics and Moment of Inertia

To find how the rod rotates, we must apply Newton's second law for rotation:

$$	au = I \alpha$$

where $I$ is the moment of inertia of the system and $\alpha$ is the angular acceleration. Since the rod itself is massless, the entire moment of inertia comes from the two point masses at the ends.

The system rotates about its center of mass, which lies exactly at the midpoint of the rod. The distance of each mass from this rotation axis is $L/2$. Therefore, the total moment of inertia is:

$$I = M\left(\frac{L}{2}ight)^2 + M\left(\frac{L}{2}ight)^2 = \frac{ML^2}{2}$$

Now, we substitute our expressions for torque and moment of inertia into the rotational equation of motion:

$$-q L E \sin	heta = \left(\frac{ML^2}{2}ight) \alpha$$

### The Small-Angle Approximation

Since the initial angle of displacement is very small ($	heta \approx 5^\circ$), we can safely apply the **small-angle approximation**:

$$\sin	heta \approx 	heta$$

This simplifies our equation of motion to:

$$-q L E 	heta = \frac{ML^2}{2} \alpha$$

Solving for the angular acceleration $\alpha$, we get:

$$\alpha = -\left(\frac{2qE}{ML}ight) 	heta$$

### Identifying Simple Harmonic Motion

Notice the form of this equation: the angular acceleration $\alpha$ is directly proportional to the angular displacement $	heta$ and is directed opposite to it. This is the hallmark of **angular Simple Harmonic Motion (SHM)**!

Comparing this with the standard angular SHM equation:

$$\alpha = -\omega^2 	heta$$

we can identify the angular frequency $\omega$ of the oscillation:

$$\omega^2 = \frac{2qE}{ML} \implies \omega = \sqrt{\frac{2qE}{ML}}$$

The total time period $T$ for one complete back-and-forth oscillation is:

$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{ML}{2qE}}$$

### Finding the Minimum Time to Become Parallel

When the rod is released from rest at its maximum displacement $	heta$, it begins to swing toward the parallel position ($	heta = 0$).

The minimum time $t$ needed for the rod to become parallel to the field for the first time is exactly **one-quarter of the total time period** of oscillation:

$$t = \frac{T}{4}$$

Substituting our expression for $T$:

$$t = \frac{2\pi}{4} \sqrt{\frac{ML}{2qE}} = \frac{\pi}{2} \sqrt{\frac{ML}{2qE}}$$

This is our final, elegant expression for the minimum time required for the rod to align with the electric field.

Visualized Solution

Visualizing the Dipole in a Uniform Field

Identifying the Restoring Forces

Calculating the Restoring Torque

Formulating the Restoring Torque Magnitude

Finding the Moment of Inertia

Setting Up the Equation of Motion

Applying the Small Angle Approximation

Solving for Angular Acceleration $α$

Comparing with Standard SHM

Finding the Time Period of Oscillation

Calculating the Minimum Time to Become Parallel

Exploring Further Scenarios

The Sigma Insight: Angular SHM

Analyzing the Setup

The Restoring Forces and Torque

Rotational Dynamics and Moment of Inertia

The Small-Angle Approximation

Identifying Simple Harmonic Motion

Finding the Minimum Time to Become Parallel

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