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Chapter 13: Problem 53

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

Expert verified
Pendulum B takes longer to swing than pendulum A.

Step by step solution

01

Define the Pendulum Period

For a simple pendulum with a point mass at the end, the period is given by the formula: \( T = 2\pi \sqrt{\frac{L}{g}} \) where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. For pendulum A, which is a simple pendulum, we will use this formula.
02

Calculate the Period of Pendulum A

Since pendulum A is a simple pendulum, we apply the formula for the period: \( T_A = 2\pi \sqrt{\frac{L}{g}} \). Since this involves only the length and gravity, we have calculated the period of pendulum A as \( T_A = 2\pi \sqrt{\frac{L}{g}} \).
03

Consider Pendulum B as a Physical Pendulum

Pendulum B is not a simple pendulum because the mass is distributed along its length. We must treat it as a physical pendulum. For a physical pendulum, the period is given by: \( T = 2\pi \sqrt{\frac{I}{mgh}} \) where \( I \) is the moment of inertia about the pivot and \( h \) is the distance from the pivot to the center of mass.
04

Determine the Moment of Inertia for Pendulum B

For pendulum B, half the mass \( \left(\frac{m}{2}\right) \) is in the ball, and half is in the bar. The moment of inertia is the sum of inertia of the ball and bar: \( I = \frac{m}{2}L^2 + \frac{1}{3}\left(\frac{m}{2}\right)L^2 = \frac{mL^2}{3} + \frac{mL^2}{6} = \frac{mL^2}{2} \).
05

Calculate the Distance to the Center of Mass for Pendulum B

The center of mass is located at \( \frac{2}{3}L \) from the pivot, as \( h = \frac{\frac{m}{2} \times L^0 + \frac{1}{2} m \times \frac{L}{2}}{m} = \frac{2}{3}L \).
06

Calculate the Period of Pendulum B

Using the formula for a physical pendulum, and the values derived, \( T_B = 2\pi \sqrt{\frac{\frac{mL^2}{2}}{mg\frac{2L}{3}}} = 2\pi \sqrt{\frac{L}{g}\left(\frac{3}{4}\right)} = \pi \sqrt{\frac{3L}{2g}} \).
07

Comparison of Periods

Finally, compare the periods: \( T_A = 2\pi \sqrt{\frac{L}{g}} \) and \( T_B = 2\pi \sqrt{\frac{3L}{2g}} \). Since \( \frac{3}{2} > 1 \), \( T_B \gt T_A \). This means that pendulum B takes longer for a swing.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simple Pendulum
A simple pendulum consists of a weight, or bob, attached to the end of a string or rod. The other end is fixed, allowing the bob to swing back and forth. The defining characteristic of a simple pendulum is that all of the mass is concentrated at a single point. This point mass, in the context of the textbook problem, makes up pendulum A. The formula for the period (i.e., the time for one complete oscillation) of a simple pendulum is:
  • \( T = 2\pi \sqrt{\frac{L}{g}} \)
  • where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
This formula only depends on the length of the pendulum and the gravitational pull, making calculations straightforward. The simplicity of this setup allows us to predict when and how the pendulum will swing under small angles and ideal conditions.
Physical Pendulum
In contrast to a simple pendulum, a physical pendulum involves a rigid body swinging around a pivot point. Here, the mass is distributed over a length or shape, rather than being a point mass. In the textbook problem, pendulum B features mass distributed between a ball and a bar.The period of a physical pendulum is given by:
  • \( T = 2\pi \sqrt{\frac{I}{mgh}} \)
  • where \( I \) is the moment of inertia, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the distance from the pivot to the center of mass.
The complexity arises in calculating the moment of inertia and identifying the right center of mass. This makes understanding and solving physical pendulum problems more involved than with simple pendulums. However, this also provides insight into how shape and mass distribution affect motion.
Moment of Inertia
Moment of inertia is a critical concept for understanding physical pendulums. It describes how the mass of an object is distributed in relation to an axis, impacting how easily the object can rotate around that axis. For pendulum B in the textbook problem, the moment of inertia consists of two parts:
  • The ball at the end, which contributes \( \frac{m}{2}L^2 \)
  • The bar, contributing \( \frac{1}{3} \left(\frac{m}{2}\right)L^2 \)
Adding these gives:
  • \( I = \frac{mL^2}{3} + \frac{mL^2}{6} = \frac{mL^2}{2} \)
The moment of inertia is fundamentally crucial in determining the period of a physical pendulum as it affects the swing efficiency and speed. It shows that mass distribution can significantly impact rotational movement.
Center of Mass
The center of mass is the point in a body or system where its whole mass can be considered to be concentrated for purposes of analysis. In pendulum problems, the center of mass influences the period of oscillation, especially for physical pendulums. For pendulum B in the exercise, the center of mass is calculated using:
  • Contribution of the ball (closer to the pivot) and the remaining bar.
  • The calculated position \( h = \frac{2}{3}L \) means the center is two-thirds the length of the pendulum from the pivot point.
This impacts the period calculation, as it affects the distance used in the pendulum's formula for period. Grasping the center of mass location helps visualize how the pendulum balances and distributes its weight during motion. It is a pivotal point that shifts dynamics from simply swinging to understanding rotatory balance.

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Most popular questions from this chapter

On the planet Newtonia, a simple pendulum having a bob with mass 1.25 \(\mathrm{kg}\) and a length of 185.0 \(\mathrm{cm}\) takes 1.42 \(\mathrm{s}\) , when released from rest, to swing through an angle of \(12.5^{\circ},\) where it again has zero speed. The circumference of Newtonia is measured to be \(51,400 \mathrm{km}\) . What is the mass of the planet Newtonia?

The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 \(\mathrm{Hz}\) . At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s}\) , respectively (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and accelerationcomponents of the point as a function of time.

Jerk. A guitar string vibrates at a frequency of 440 \(\mathrm{Hz}\) . A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are added to the original mass, and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

A \(1.50-\mathrm{kg}\) ball and a \(2.00-\mathrm{kg}\) ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The gine connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.
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