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A thin rod (of width zero, but not necessarily uniform) is pivoted freely at one end about the horizontal \(z\) axis, being free to swing in the \(x y\) plane ( \(x\) horizontal, \(y\) vertically down). Its mass is \(m,\) its \(\mathrm{CM}\) is a distance \(a\) from the pivot, and its moment of inertia (about the \(z\) axis) is \(I\). (a) Write down the equation of motion \(\dot{L}_{z}=\Gamma_{z}\) and, assuming the motion is confined to small angles (measured from the downward vertical), find the period of this compound pendulum. ("Compound pendulum" is traditionally used to mean any pendulum whose mass is distributed \(-\) as contrasted with a "simple pendulum," whose mass is concentrated at a single point on a massless arm.) (b) What is the length of the "equivalent" simple pendulum, that is, the simple pendulum with the same period?

Short Answer

Expert verified
(a) Period: \( T = 2\pi \sqrt{\frac{I}{mga}} \); (b) Length of equivalent simple pendulum: \( L_{eq} = \frac{I}{ma} \).

Step by step solution

01

Write Down the Equation of Motion

For a rotating system about z-axis, the angular momentum Lz changes with time due to a torque τz. The equation of motion is given by \( \dot{L}_z = \tau_z \). For a thin rod pivoted at one end and swinging in the xy-plane, the torque arises due to gravity, \( \tau_z = m g a \sin(\theta) \), where θ is the angle the rod makes with the vertical. For small angles, we approximate \( \sin(\theta) \approx \theta \). Thus, the equation becomes \( I \ddot{\theta} = -m g a \theta \).
02

Simplify and Identify Conditions for Simple Harmonic Motion

Rearranging the equation \( I \ddot{\theta} = -m g a \theta \), we have \( \ddot{\theta} + \frac{mga}{I} \theta = 0 \). This differential equation represents simple harmonic motion, where \( \omega^2 = \frac{mga}{I} \).
03

Find the Period of Oscillation

The angular frequency \( \omega \) of the simple harmonic motion is \( \omega = \sqrt{\frac{mga}{I}} \). The period \( T \) of the pendulum is related to the angular frequency by \( T = \frac{2\pi}{\omega} \). Thus, the period of oscillation is \( T = 2\pi \sqrt{\frac{I}{mga}} \).
04

Determine the Length of the Equivalent Simple Pendulum

For a simple pendulum of length \( L_{eq} \), the period is given by \( T = 2 \pi \sqrt{\frac{L_{eq}}{g}} \). Equating the expression for the period of the compound pendulum to that of a simple pendulum, we have \( \sqrt{\frac{I}{mga}} = \sqrt{\frac{L_{eq}}{g}} \). Solving for \( L_{eq} \), we get \( L_{eq} = \frac{I}{ma} \).

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Key Concepts

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

Equation of Motion
In physics, the equation of motion for a pendulum describes how its angular position changes over time. For a compound pendulum, like a rod pivoted to swing, this involves understanding the interplay between angular momentum and torque. The general form of this equation is represented as \( \dot{L}_z = \tau_z \), where \( \dot{L}_z \) is the rate of change of angular momentum around the \( z \)-axis, and \( \tau_z \) is the torque due to gravity.

For a thin rod pivoted at one end, the torque \( \tau_z \) can be expressed based on the gravitational force acting on its center of mass. It's given by \( \tau_z = m g a \sin(\theta) \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, \( a \) is the distance from the pivot to the center of mass, and \( \theta \) is the angular displacement. For small angles, \( \sin(\theta) \approx \theta \), simplifying the equation to \( I \ddot{\theta} = -m g a \theta \), where \( I \) is the moment of inertia.
Moment of Inertia
The moment of inertia is a crucial parameter that measures how much an object resists rotation. It's often referred to as the rotational equivalent of mass in linear motion. For the compound pendulum described here, the moment of inertia \( I \) is about the pivot point, crucial for calculating oscillation periods.

For a rod of non-uniform density, \( I \) wouldn't be as straightforward as for a uniform rod, but it generally combines the rod's mass distribution and geometry. This term \( I \) appears in the simplified equation of motion \( I \ddot{\theta} = -m g a \theta \), linking the pendulum's inertia to its oscillatory behavior. Understanding \( I \) helps us predict how quickly the pendulum swings back and forth, rooting the pendulum's dynamics in its physical characteristics.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic oscillation. It's characterized by the restoring force being directly proportional to the displacement. For the compound pendulum, when reduced to small angle approximations, its motion resembles SHM.

The governing equation \( \ddot{\theta} + \frac{mga}{I} \theta = 0 \) highlights that the compound pendulum behaves like an SHM system. Here, \( \frac{mga}{I} \) represents the square of the angular frequency \( \omega^2 \), dictating how rapidly oscillations occur.
  • The angular frequency is \( \omega = \sqrt{\frac{mga}{I}} \), leading to periodic swings.
  • The period of oscillation is \( T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{mga}} \), defining the time taken for a full swing cycle.
This SHM framework lets us predict oscillations based on physical properties, grounding the behavior of the pendulum in classical physics.
Equivalent Simple Pendulum
An equivalent simple pendulum refers to a point mass suspended on a massless rod, having the same period as a complex oscillating system like the compound pendulum. By equating the oscillation period \( T \) of both pendulums, we can find this "equivalent length".

For the compound pendulum, its period is \( T = 2\pi \sqrt{\frac{I}{mga}} \). A simple pendulum's period is \( T = 2 \pi \sqrt{\frac{L_{eq}}{g}} \). Setting these equal, \( \sqrt{\frac{I}{mga}} = \sqrt{\frac{L_{eq}}{g}} \), solves to \( L_{eq} = \frac{I}{ma} \).
  • Finding this length helps visualize and compare the dynamics of different pendulum types.
  • The equivalent length simplifies analysis by providing a straightforward relation to a simple pendulum's period.
Thus, understanding equivalent simple pendulums bridges complex systems with more intuitive models, enhancing comprehension of periodic motion.

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Most popular questions from this chapter

Consider a rigid plane body or "lamina," such as a flat piece of sheet metal, rotating about a point \(O\) in the body. If we choose axes so that the lamina lies in the \(x y\) plane, which elements of the inertia tensor \(\mathbf{I}\) are automatically zero? Prove that \(I_{z z}=I_{x x}+I_{y y}\).

Find the inertia tensor for a uniform, thin hollow cone, such as an ice-cream cone, of mass \(M,\) height \(h,\) and base radius \(R,\) spinning about its pointed end.

A triangular prism (like a box of Toblerone) of mass \(M,\) whose two ends are equilateral triangles parallel to the \(x y\) plane with side \(2 a,\) is centered on the origin with its axis along the \(z\) axis. Find its moment of inertia for rotation about the \(z\) axis. Without doing any integrals write down and explain its two products of inertia for rotation about the \(z\) axis.

(a) A thin uniform rod of mass \(M\) and length \(L\) lies on the \(x\) axis with one end at the origin. Find its moment of inertia for rotation about the \(z\) axis. [Here the sum of (10.25) must be replaced by an integral of the form \(\int x^{2} \mu d x\) where \(\mu\) is the linear mass density, mass/length.] (b) What if the rod's center is at the origin?

An axially symmetric space station (principal axis \(\mathbf{e}_{3},\) and \(\lambda_{1}=\lambda_{2}\) ) is floating in free space. It has rockets mounted symmetrically on either side that are firing and exert a constant torque \(\Gamma\) about the symmetry axis. Solve Euler's equations exactly for \(\omega\) (relative to the body axis) and describe the motion. At \(t=0\) take \(\omega=\left(\omega_{10}, 0, \omega_{30}\right)\)

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