Problem 43 - Physics

Question

Multiple-Concept Example 11 explores the concepts that are important in this problem. Pendulum A is a physical pendulum made from a thin, rigid, and uniform rod whose length is \(d\). One end of this rod is attached to the ceiling by a frictionless hinge, so the rod is free to swing back and forth. Pendulum B is a simple pendulum whose length is also \(d\). Obtain the ratio \(T_{\mathrm{A}} / T_{\mathrm{B}}\) of their periods for small-angle oscillations.

Step-by-Step Solution

Answer

The ratio \(\frac{T_A}{T_B} = \sqrt{\frac{2}{3}}\).

Step 1: Understanding the Pendulums

Pendulum A is a physical pendulum consisting of a uniform rod, while Pendulum B is a simple pendulum with a point mass. Both have the same length, denoted by \(d\), and execute small-angle oscillations.

Step 2: Period of a Physical Pendulum

For a physical pendulum, the period \(T_A\) is given by the formula \(T_A = 2\pi \sqrt{\frac{I}{mgd_c}}\), where \(I\) is the moment of inertia about the pivot point, \(m\) is the mass, \(g\) is the gravitational acceleration, and \(d_c\) is the distance from the pivot to the center of mass.

Step 3: Calculate the Moment of Inertia for Pendulum A

The moment of inertia \(I\) for a rod of length \(d\) pivoted at one end is \(\frac{1}{3}md^2\). The center of mass (COM) for this rod is located at \(\frac{d}{2}\) from the pivot.

Step 4: Substitute Values into Pendulum A's Period Formula

Substituting the moment of inertia we found into the formula, we have \(T_A = 2\pi \sqrt{\frac{\frac{1}{3}md^2}{mg\frac{d}{2}}} = 2\pi \sqrt{\frac{2d}{3g}}\).

Step 5: Period of a Simple Pendulum

For a simple pendulum, the period \(T_B\) is given by the formula \(T_B = 2\pi \sqrt{\frac{d}{g}}\), where \(d\) is the length of the pendulum and \(g\) is the gravitational acceleration.

Step 6: Calculate the Ratio of Periods

To find the ratio \(\frac{T_A}{T_B}\), divide the expression for \(T_A\) by \(T_B\): \(\frac{T_A}{T_B} = \frac{2\pi \sqrt{\frac{2d}{3g}}}{2\pi \sqrt{\frac{d}{g}}}\).

Step 7: Simplify the Ratio

Simplify the expression: \(\frac{T_A}{T_B} = \sqrt{\frac{2d}{3g}} \times \sqrt{\frac{g}{d}} = \sqrt{\frac{2}{3}}\).

Key Concepts

Physical Pendulum

A physical pendulum is a more complex system compared to a simple pendulum. It involves a rigid body that swings about a pivot point, rather than a simple mass on a string. In this case, Pendulum A is a physical pendulum made from a uniform rod. This means that every part of the rod contributes to its moment of inertia, making its behavior a bit different from a simple pendulum.

The period of a physical pendulum depends on its moment of inertia, the mass of the pendulum, the gravitational acceleration, and the distance from the pivot point to its center of mass. The formula used to determine the period of the physical pendulum is:

• \( T_A = 2\pi \sqrt{\frac{I}{mgd_c}} \)

where:

• \( I \) is the moment of inertia
• \( m \) is the mass
• \( g \) is the acceleration due to gravity
• \( d_c \) is the distance to the center of mass

This equation shows that the period is affected by the distribution of mass along the pendulum, in contrast to a simple pendulum.

Simple Pendulum

A simple pendulum is a fundamental concept in physics that consists of a small mass, called a bob, attached to the end of a lightweight string or rod. The length of this pendulum is the distance from the pivot point to the center of the mass. In the example, Pendulum B is a simple pendulum with the same length \(d\) as the physical pendulum.

The beauty of a simple pendulum is its straightforward equation for calculating the period of oscillation, especially under the assumption of small angle oscillations. The period \( T_B \) can be calculated using:

• \( T_B = 2\pi \sqrt{\frac{d}{g}} \)

Here, the period of the simple pendulum depends solely on the length of the pendulum and the gravitational acceleration \(g\). The mass of the pendulum and the amplitude (assuming small angles) do not affect the period, which simplifies analysis and predictions.

Moment of Inertia

The moment of inertia, often symbolized as \(I\), plays a crucial role in the dynamics of a physical pendulum. It is a measure of an object's resistance to changes in its rotation about a particular axis. For Pendulum A, which is a uniform rod, this concept is especially important.

The moment of inertia for a rod of length \(d\), pivoted at one end, is calculated using the formula:

• \( I = \frac{1}{3}md^2 \)

This value shows how the mass distribution along the length of the rod influences its motion. When the rod swings as a physical pendulum, every part of it contributes to the overall resistance to swinging, which is described by \(I\).

Understanding the moment of inertia is thus essential to calculating the period of a physical pendulum, revealing how each part of the pendulum's mass participates in the movement.

Small-Angle Oscillations

Small-angle oscillations are an approximation used in pendulum motion to simplify the mathematical treatment of the problem. This approximation is based on the assumption that the angle through which the pendulum swings is small (typically less than about 15 degrees).

This assumption allows the sine of the angle to be approximated as the angle itself (in radians), which reduces the complexity of the equations for the period of oscillation. For both the physical and simple pendulum, assuming small-angle oscillations means their periods can be calculated using the derived formulas without accounting for more complex trigonometric adjustments.

The small-angle approximation is crucial because it keeps the period independent of amplitude, meaning the period remains constant for small swings. This helps in applying uniform formulas across various scenarios, as shown in the problem where both pendulums use this approximation to find their periods efficiently.