Chapter 14: 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
Step by step solution
Define the Problem
Calculate the Period for Pendulum A
Calculate the Moment of Inertia for Pendulum B
Calculate the Period for Pendulum B
Compare the Periods of Both Pendulums
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Simple Pendulum
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:
- \( T = 2\pi \sqrt{\frac{L}{g}} \)
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
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:
- \( T = 2\pi \sqrt{\frac{I}{mgd}} \)
This setup increases the complexity but gives insights into how mass distribution affects oscillation in physical objects.
Moment of Inertia
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 \( \frac{mL^2}{6} \) for its portion of the mass.
- The moment of inertia for the bob, concentrated at a distance \( L \), is \( \frac{mL^2}{2} \).
- \( I = \frac{2mL^2}{3} \)