Question

Two pendulums have the same dimensions (length $L$) and total mass ( $m$ ). Pendulum $A$ is a very small ball swinging at the end of a uniform massless bar. In pendulum $B$, half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

Short answer

Pendulum A takes longer for a swing.

Step-by-step solution

Define the Problem

We are tasked with finding the period of small oscillations for two pendulums with the same length $L$ and total mass $m$. Pendulum $A$ has all its mass concentrated at a point (a simple pendulum), and pendulum $B$ has its mass distributed equally between the ball and the bar (a physical pendulum).

Calculate the Period for Pendulum A

For pendulum $A$, which behaves like a simple pendulum, the formula for the period $T$ of small oscillations is given by:$$T_A=2\pi \sqrt{\frac{L}{g}}$$Where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Calculate the Moment of Inertia for Pendulum B

Pendulum $B$ is a physical pendulum with half the mass in the ball and the other half evenly distributed along the bar. First, calculate the moment of inertia $I$. The moment of inertia $I_p$ of the bar (length $L$, mass $m/2$, pivoted at one end) is given by:$$I_p=\frac{1}{3}\left(\frac{m}{2}\right)L^2=\frac{m{L}^2}{6}$$For the ball (mass $m/2$ at a vertex $L$ from the pivot):$$I_b=\left(\frac{m}{2}\right)L^2=\frac{m{L}^2}{2}$$Adding these: $$I=\frac{m{L}^2}{6}+\frac{m{L}^2}{2}=\frac{2m{L}^2}{3}$$

Calculate the Period for Pendulum B

The period $T$ of a physical pendulum is given by:$$T_B=2\pi \sqrt{\frac{I}{mgd}}$$where $d=L$ is the distance from the pivot to the center of mass of the ball. Substituting the calculated $I$ gives:$$T_B=2\pi \sqrt{\frac{\frac{2m{L}^2}{3}}{mgL}}=2\pi \sqrt{\frac{2L}{3g}}$$

Compare the Periods of Both Pendulums

We found that for pendulum $A$:$$T_A=2\pi \sqrt{\frac{L}{g}}$$And for pendulum $B$:$$T_B=2\pi \sqrt{\frac{2L}{3g}}$$Now, we compare:$$\sqrt{\frac{L}{g}}\text{and}\sqrt{\frac{2L}{3g}}$$Notice that $\sqrt{\frac{2}{3}}<1$, thus $T_B<T_A$.

Key concepts

Simple Pendulum

Let's delve into the concept of a simple pendulum, which is a classic example in physics for studying motion and oscillations. A simple pendulum consists of a small and heavy object, known as a "bob," attached to the end of a lightweight and rigid rod. The rod is assumed to be massless, so the bob alone contributes to the pendulum's inertia.

The motion of a simple pendulum can be described by its period, which is the time taken for one complete cycle of swing motion. For small angles, the period of a simple pendulum is approximately determined by the formula:

• $T=2\pi \sqrt{\frac{L}{g}}$

Here, $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity (approximately $9.81m{s}^{-2}$ on Earth's surface). This formula indicates that the period is independent of the mass of the bob and depends only on the length of the pendulum and the acceleration due to gravity.

In the problem example, pendulum A exemplifies a simple pendulum where the mass is concentrated entirely at the end. It shows typical behavior for periodic motion in physics.

Physical Pendulum

A physical pendulum is a generalized form of a pendulum where the mass is distributed along its length, rather than concentrated at a single point. This distribution of mass makes calculations slightly more complex than those for a simple pendulum.

For a physical pendulum, the period of oscillation depends on the moment of inertia and the distance between the pivot point and the center of mass.

In our specific example, pendulum B is a physical pendulum. Its mass is split evenly between a rod and the bob at the end. This results in different mechanics than a simple pendulum. To calculate its period, we use:

• $T=2\pi \sqrt{\frac{I}{mgd}}$

Where $I$ is the moment of inertia, $m$ is the total mass, $g$ is gravitational acceleration, and $d$ is the distance from the pivot to the center of mass.

This setup increases the complexity but gives insights into how mass distribution affects oscillation in physical objects.

Moment of Inertia

The moment of inertia is a fundamental concept that plays a crucial role when studying a physical pendulum. It is a measure of how much torque is needed for a desired angular acceleration around an axis. In essence, it quantifies how the mass is spread out around a pivot, affecting the pendulum's resistance to changes in motion.

Calculating the moment of inertia for different shapes and configurations is key to understanding and predicting the motion of physical pendulums. For pendulum B in our exercise:

• The rod's moment of inertia, when pivoted at one end, is $\frac{m{L}^2}{6}$ for its portion of the mass.
• The moment of inertia for the bob, concentrated at a distance $L$, is $\frac{m{L}^2}{2}$.

Adding these moments results in the total moment of inertia:

• $I=\frac{2m{L}^2}{3}$

Understanding and calculating the moment of inertia reveals how the same mass distributed differently in this scenario affects the pendulum's periodic motion. It explains why pendulum B swings faster compared to pendulum A, despite having the same total mass and length.