/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 A large, 34.0-kg bell is hung fr... [FREE SOLUTION] | 91Ó°ÊÓ

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A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is 0.60 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg \(\cdot\) m\(^2\). The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently\(-\)that is, for the period of oscillation for the bell to equal that of the clapper?

Short Answer

Expert verified
Clapper rod length should be approximately 0.882 m.

Step by step solution

01

Determine the Period of the Bell

The period of oscillation for the bell can be found using the formula for the pendulum: \( T = 2\pi \sqrt{\frac{I}{mgd}} \), where \( I = 18.0 \text{ kg} \cdot \text{m}^2 \) is the moment of inertia, \( m = 34.0 \text{ kg} \) is the mass, \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity, and \( d = 0.60 \text{ m} \) is the distance from the pivot to the center of mass. Substituting these values gives the period of the bell: \( T = 2\pi \sqrt{\frac{18.0}{34.0 \times 9.8 \times 0.60}} \).
02

Calculate the Period of the Bell

Solve the equation from Step 1: \( T = 2\pi \sqrt{\frac{18.0}{34.0 \times 9.8 \times 0.60}} = 2\pi \sqrt{\frac{18.0}{199.92}} \approx 2\pi \sqrt{0.09} = 2\pi \times 0.3 = 0.6\pi \). Thus, the period of the bell \( T \approx 1.884 \text{ s} \).
03

Determine the Period of the Clapper

For the clapper, the period of oscillation \( T' \) is given by \( T' = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the rod. To ring silently, \( T' \) must equal \( T \).
04

Set Equality of Periods and Solve for L

Equate the periods of the bell and the clapper: \( T = T' \Rightarrow 2\pi \sqrt{\frac{L}{g}} = 0.6\pi \). Dividing both sides by \( 2\pi \) leads to \( \sqrt{\frac{L}{9.8}} = 0.3 \). Squaring both sides gives \( \frac{L}{9.8} = 0.09 \). Solving for \( L \) results in \( L = 0.882 \text{ m} \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in physics, particularly relevant when dealing with rotational dynamics. It is often considered the rotational equivalent of mass for linear motion. The moment of inertia, denoted by \( I \), essentially describes how much a rotating object resists changes in its rotational motion. This resistance depends not only on the object's mass but also on how the mass is distributed relative to the axis of rotation. In the case of a bell, the moment of inertia is calculated taking into account both the mass and its distance from the pivot point. This concept is crucial when analyzing objects like pendulums (or bells in this case) and their oscillations around a pivot point. For the given bell, with a moment of inertia \( I = 18.0 \, \text{kg} \, \cdot \, \text{m}^2 \), it influences how quick and smooth the oscillations are. Understanding this might help clarify why the bell’s movement will be unlike a simple pendulum, where its mass would merely influence linearly.
Oscillation Period
The oscillation period \( T \) of an object refers to the time it takes to complete one full cycle of motion from its starting position and back again. For pendulum-like structures, this is an essential way to understand their behavior over time. For the bell, the period, \( T \), can be found using the formula \( T = 2\pi \sqrt{\frac{I}{mgd}} \), where \( I \) is the moment of inertia, \( m \) the mass of the bell, \( g \) the acceleration due to gravity, and \( d \) the distance from the pivot to the center of mass.
  • For the bell: The period is calculated to be approximately 1.884 seconds.
  • For the clapper: The desired period should be the same as the bell to achieve synchronization and 'silent ringing.'
Ensuring both the bell and the clapper have matching periods aligns their movements and avoids unnecessary additional sound waves that might otherwise cause dissonance.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a type of periodic motion where the force restoring the object to its equilibrium position is directly proportional to its displacement and acts in the opposite direction. This fundamental principle in mechanics is what allows pendulums, springs, and other oscillatory systems to function predictably.In the scenario of the bell and the clapper, each behaves like a simple pendulum—an ideal example of simple harmonic motion. For the clapper, we use the equation \( T' = 2\pi \sqrt{\frac{L}{g}} \) to describe its period of oscillation, where \( L \) is the length of the clapper rod.
  • SHM is characterized by its sinusoidal nature—which can be graphically represented as a wave—with its amplitude, frequency, and period defining the object's motion.
  • The period of a pendulum in SHM is independent of its amplitude, provided the amplitude is small.
By using SHM principles, we can ensure that both the bell and its clapper oscillate with the same period, thus allowing them to "ring silently" with synchronized motion.

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Most popular questions from this chapter

An 85.0-kg mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 m long. He holds one end of the rope, and the other end is tied higher up on a rock face. Since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. At the lowest point of his swing, he plans to let go and drop a short distance to the ground. (a) How long after he begins his swing will the climber first reach his lowest point? (b) If he missed the first chance to drop off, how long after first beginning his swing will the climber reach his lowest point for the second time?

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