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Chapter 13: Problem 51

Show that the expression for the period of a physical penduhm reduces to that of a simple pendulum if the physical pendulum consists of a particle with mass \(m\) on the end of a massless string of length \(L .\)

Short Answer

Expert verified
The period is \( T = 2\pi \sqrt{\frac{L}{g}} \), matching that of a simple pendulum.

Step by step solution

01

Understanding the Physical Pendulum

The period of a physical pendulum is given by the formula \( T = 2\pi \sqrt{\frac{I}{mgd}} \), where \( I \) is the moment of inertia of the pendulum about its pivot point, \( m \) is the mass of the pendulum, \( g \) is the acceleration due to gravity, and \( d \) is the distance from the pivot to the center of mass.
02

Moment of Inertia for a Mass on a String

For a particle of mass \( m \) attached to a massless string of length \( L \), the moment of inertia \( I \) about the pivot (the point where the string is attached) is \( I = mL^2 \).
03

Determine Distance to Center of Mass

Since the particle is at the end of the string, the distance \( d \) from the pivot to the center of mass is simply the length of the string \( L \).
04

Substitute Values into Period Formula

Substitute \( I = mL^2 \) and \( d = L \) into the period formula: \[ T = 2\pi \sqrt{\frac{mL^2}{mgL}} \]
05

Simplify the Expression

Simplify the expression inside the square root:\[ T = 2\pi \sqrt{\frac{mL^2}{mgL}} = 2\pi \sqrt{\frac{L}{g}} \]This matches the period of a simple pendulum with length \( L \).

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Key Concepts

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

Physical Pendulum
A physical pendulum, unlike its simpler counterpart, considers the full distribution of mass around its pivot point. This means it can be any rigid body that swings back and forth about a fixed horizontal axis. One common example is a swinging solid rod. In physics, understanding its motion requires knowing how its entire mass is spread out, which affects its oscillation period.
For a physical pendulum, each piece of the body has its own role to play, as they contribute differently depending on their distance from the pivot. This complexity secures the need for considering the moment of inertia when calculating its period.
Simple Pendulum
The simple pendulum is an idealized version, consisting merely of a point mass at the end of a massless string. Its movement is governed primarily by its length and the gravitational pull it experiences. This simplicity allows us to deduce its period using a straightforward formula.
The simple pendulum oscillates back and forth, and its motion is predominantly influenced by gravity and the length of the string. Its uncomplicated nature makes it a staple example in introductory physics, illustrating fundamental principles of harmonic motion.
Period Formula
The formula connecting the swinging time of a pendulum to its physical properties is central in its study. For a physical pendulum, the period's expression is derived by considering both mass placement and gravity's effect. It is presented as:
  • \[ T = 2\pi \sqrt{\frac{I}{mgd}} \]

Where:
  • \(T\) is the period of oscillation.
  • \(I\) is the moment of inertia about the pivot.
  • \(m\) is the mass of the pendulum.
  • \(g\) is the acceleration due to gravity.
  • \(d\) is the distance from the pivot to the center of mass.
For a simple pendulum, this complexity reduces to a form where the moment of inertia is simplified, given that all the mass is concentrated at one point, and the distance \(d\) becomes simply the string's length, \(L\).
Moment of Inertia
Moment of inertia determines how difficult it is to start a body's rotation and how it affects the period of a pendulum. It is crucial in any rotational dynamics analysis. For a simple mass \(m\) attached to a string of length \(L\), the moment of inertia is calculated as follows:
  • \[ I = mL^2 \]

This formula tells us how mass distributed over the length impacts the pendulum's swing. A larger value means a slower start but potentially more sustained motion due to greater rotational inertia.
Acceleration due to Gravity
Gravity is the invisible force pulling objects towards earth. Its acceleration, denoted by \(g\), is a cornerstone in pendulum physics. It's the driving force behind a pendulum's return to equilibrium. Every pendulum experiment essentially measures this gravity, which on average, is about \(9.8 \text{m/s}^2\) at Earth's surface.
The pendulum swings fastest at the lowest point of its path, and slows as it reaches the peak due to this consistent gravitational attraction. This tug causes the pendulum to perpetuate a regular motion cycle, relying on gravity to consistently dictate the speed and period, or time it takes to complete one swing.

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Most popular questions from this chapter

A 0.500 \(\mathrm{kg}\) mass on a spring has velocity as a function of time given by \(v_{x}(t)=(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) . What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

A \(0.0200-\mathrm{kg}\) bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{c})\) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;(\mathrm{d})\) the speed of the bolt when \(x=-0.180 \mathrm{m} .\)

An object is undergoing SHM with period 0.300 s and amplitude 6.00 \(\mathrm{cm} .\) At \(t=0\) the object is instantaneously at rest at \(x=6.00 \mathrm{cm} .\) Calculate the time it takes the object to go from \(x=6.00 \mathrm{cm}\) to \(x=-1.50 \mathrm{cm} .\)

A harmonic oscillator consists of a \(0.500-\mathrm{kg}\) mass attached to an ideal spring with force constant 140 \(\mathrm{N} / \mathrm{m}\) . Find (a) the period; (b) the frequency; (c) the angular frequency of the oscillations.

A large bell is hung from a wooden beam so it can swing back and forth with negligible friction. The center of mass of the bell is 0.60 \(\mathrm{m}\) below the pivot, the bell has mass 34.0 \(\mathrm{kg}\) , and the moment of inertia of the bell about an axis at the pivot is 18.0 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The clapper is a small, 1.8-kg mass attached to one end of a slender rod that has length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell so it 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 for the clapper?
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