Problem 22 - Classical Mechanics

Question

A uniform rod slides with its ends inside a smooth vertical circle. If the rod subtends an angle of \(120^{\circ}\) at the center of the circle, show that the equivalent simple pendulum has a length equal to the radius of the circle.

Step-by-Step Solution

Answer

The equivalent simple pendulum has a length equal to the circle's radius.

Step 1: Understanding the Problem

We have a rod which is subtending an angle of \(120^{\circ}\) at the center of a circle. This subtension creates an isosceles triangle with two equal sides being the radius of the circle. We need to prove that the equivalent length of a simple pendulum matching the dynamics of this rod is equal to the circle's radius.

Step 2: Geometry of the Rod and Circle

Since the rod subtends an angle of \(120^{\circ}\) at the circle's center, it divides the circle into two equal segments each subtending \(60^{\circ}\), forming an isosceles triangle with angle \(60^{\circ}\) at each base. The vertices of this triangle lie at the ends of the rod (on the circle) and the center.

Step 3: Calculate the Torque

For small oscillations, the torque due to gravity around the pivot point is proportional to the sin of the angle. For a pendulum, this is: \( \tau = -mg\sin(\theta) \). For the rod, we need to account for its center of mass.

Step 4: Calculate the Equivalent Length

The center of mass of a uniform rod lies at its midpoint. With the rod subtending \(120^{\circ}\) , its midpoint will subtend \(60^{\circ}\) at the center. Thus, it forms an equilateral triangle with each side being equal to the radius \(R\). The height of this triangle (or the equivalent length of a simple pendulum) lines up to be just the radius \(R\).

Step 5: Conclusion

Thus, since the center of mass falls directly below the pivot and along the radius, the equivalent simple pendulum for comparing oscillation periods and dynamics has the length equal to the circle radius.

Key Concepts

Simple Pendulum

The simple pendulum is a fundamental concept in mechanics. It comprises a mass, often called a "bob," attached to the end of a string or rod with negligible mass, that swings back and forth. The motion can be described by simple harmonic motion when the angle of displacement is small, meaning the motion repeats in a symmetric manner around its equilibrium position.

• The force causing the pendulum to swing is gravity, pulling on the mass.
• The restoring force can be expressed as: \(F = -mg \sin(\theta)\), where \( m \) is the mass and \( g \) is the acceleration due to gravity.
• For small angles, \( \sin(\theta) \approx \theta \), simplifying the analysis of pendulum motion.

The time period of a simple pendulum depends on its length and the acceleration due to gravity: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum.

Center of Mass

The center of mass is the average position of all the mass in an object. For a uniform rod, the center of mass is exactly at its midpoint. This concept is critical for analyzing systems like pendulums, where the center of mass influences the oscillation characteristics.

• In a uniform rod, the center of mass lies at its geometrical center.
• It determines how effectively a force, like gravity, can create torque and induce motion.
• Understanding the center of mass helps simplify complex systems into more basic models, like simple pendulums.

In this exercise, the rod subtends a total angle of \(120^{\circ}\). Its midpoint creates an equilateral triangle with sides equal to the circle's radius, lining the center of mass with the radius this way.

Torque

Torque is a measure of how much a force acting on an object causes it to rotate. In pendulum motion, torque is crucial as it determines how the system responds to gravitational forces.

• The formula for torque \( \tau \) is: \( \tau = r \times F \), or more specifically for pendulums: \( \tau = -mg \sin(\theta) \times L \) where \( L \) is the distance from the pivot to where the force is applied.
• Torque influences the angular acceleration \( \alpha \) of the system through Newton's second law for rotation: \( \tau = I \alpha \), with \( I \) being the moment of inertia.
• For small angles, \( \sin(\theta) \approx \theta \), which simplifies the analysis of motion.

In our example, the rod's midpoint aligns vertically with the center, simplifying the calculation of torque brought about by its center of mass.

Oscillation

Oscillation is a repetitive variation, usually in time, of some measure about a central value. In a mechanical system like a pendulum, oscillation refers to the regular, repeating motion initiated by displacement from equilibrium.

• When a pendulum swings, the potential energy at its highest point is converted to kinetic energy at the lowest.
• True oscillation approximates harmonic motion, repeating in cycles by swinging back and forth.
• The period of oscillation is the time taken for one complete cycle.

Simple pendulums exhibit harmonic oscillation when the angles are small, due to their restoring torque given by gravity. In this problem, the equivalent pendulum has a period of oscillation influenced only by the gravitational force and the pendulum length, which here is equal to the radius of the circle.