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Using the usual angle \(\phi\) as generalized coordinate, write down the Lagrangian for a simple pendulum of length \(l\) suspended from the ceiling of an elevator that is accelerating upward with constant acceleration \(a\). (Be careful when writing \(T\); it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that \(g\) has been replaced by \(g+a\). In particular, the angular frequency of small oscillations is \(\sqrt{(g+a) / l}\).

Short Answer

Expert verified
The Lagrangian is \(\frac{1}{2} m (l\dot{\phi})^2 + m(g+a) l \cos(\phi)\); equation of motion: \(\ddot{\phi} + \frac{g+a}{l} \sin(\phi) = 0\).

Step by step solution

01

Understand the Physical Setup

A simple pendulum consists of a mass (bob) attached to a string of length \(l\), suspended from a ceiling inside an elevator that is moving upwards with constant acceleration \(a\). The generalized coordinate \(\phi\) represents the angular displacement of the pendulum from vertical.
02

Set Up the Coordinate System

Assume the origin at the pivot point of the pendulum. The position of the bob in the elevator frame is described by \(x = l\sin(\phi)\) and \(y = l\cos(\phi)\).
03

Calculate the Bob's Velocity Components

Compute the velocities \(\dot{x} = l\cos(\phi)\ \dot{\phi}\) and \(\dot{y} = -l\sin(\phi)\ \dot{\phi}\). The velocity squared is then given by \(v^2 = \dot{x}^2 + \dot{y}^2 = (l\dot{\phi})^2\).
04

Write Down the Kinetic Energy (T)

The kinetic energy \(T\) is \( \frac{1}{2} m v^2 = \frac{1}{2} m (l\dot{\phi})^2 \).
05

Write Down the Potential Energy (V)

The potential energy comes from gravity, acting effectively as \(g + a\) because of the upward acceleration: \(V = -m(g+a) y = -m(g+a) l \cos(\phi)\).
06

Form the Lagrangian

The Lagrangian \(\mathcal{L}\) is the difference between kinetic and potential energy: \(\mathcal{L} = T - V = \frac{1}{2} m (l\dot{\phi})^2 + m(g+a) l \cos(\phi)\).
07

Derive the Euler-Lagrange Equation

Apply the Euler-Lagrange equation: \(\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}} - \frac{\partial \mathcal{L}}{\partial \phi} = 0\). Calculate \(\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = ml^2\dot{\phi}\) and derive \(\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = ml^2\ddot{\phi}\). Calculate \(\frac{\partial \mathcal{L}}{\partial \phi} = -m(g+a)l\sin(\phi)\).
08

Find the Equation of Motion

Putting these results into the Euler-Lagrange equation gives \(ml^2\ddot{\phi} + m(g+a)l\sin(\phi) = 0\), which simplifies to \(\ddot{\phi} + \frac{g+a}{l} \sin(\phi) = 0\).
09

Linearize for Small Oscillations

For small oscillations, approximate \(\sin(\phi) \approx \phi\), yielding \(\ddot{\phi} + \frac{g+a}{l} \phi = 0\).
10

Identify Angular Frequency

The angular frequency of small oscillations is \(\omega = \sqrt{\frac{g+a}{l}}\).

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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 is a fundamental concept in physics, representing a mass, often referred to as a "bob," attached to a string or rod that swings back and forth under the influence of gravity. For our exercise, the pendulum is inside an elevator which is an interesting twist as the elevator is accelerating upwards.
This setup changes how we calculate forces acting on the pendulum. Typically, a simple pendulum's motion is influenced by gravity alone. However, here the upward acceleration of the elevator adds an extra force, similar to increasing gravity.
The angular displacement, which is how far the pendulum deviates from its resting vertical position, is represented by the angle \( \phi \). This angle is crucial for analyzing the pendulum's motion in terms of Lagrangian mechanics.
Euler-Lagrange Equation
The Euler-Lagrange equation is a foundational principle in Lagrangian mechanics used to find the equations of motion for a system. It’s derived from the Lagrangian, a function representing the difference between a system's kinetic and potential energy. In the context of the pendulum within an accelerating elevator, this equation helps us understand how the motion changes due to additional forces.
The Lagrangian \( \mathcal{L} \) is expressed as \( \mathcal{L} = T - V\), where \( T \) is the kinetic energy and \( V \) is the potential energy. To find the pendulum's equation of motion, we apply the Euler-Lagrange equation: \[ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}} - \frac{\partial \mathcal{L}}{\partial \phi} = 0 \]
This equation yields a differential equation that governs how \( \phi \) changes over time, which is vital to understanding how the pendulum behaves dynamically.
Equations of Motion
The equations of motion explain how the position of objects evolves over time under the influence of different forces. For the simple pendulum in our scenario, the motion is influenced by both gravity and the elevator's upward acceleration.
Through the Euler-Lagrange approach described previously, we derived the equation \[ \ddot{\phi} + \frac{g+a}{l} \sin(\phi) = 0 \]
This second-order differential equation describes how the angular position \( \phi \) changes. For small oscillations, we simplify this by approximating \( \sin(\phi) \approx \phi \), resulting in a linear differential equation: \[ \ddot{\phi} + \frac{g+a}{l} \phi = 0 \]
This linearization is crucial as it allows us to delve into simple harmonic motion, a staple in classical mechanics, providing insights into how systems oscillate around a stable point.
Angular Frequency
Angular frequency is a concept used to describe how fast an object rotates or oscillates. For small oscillations, the simple pendulum in our accelerating elevator can be analyzed using the formula for angular frequency.
From the simplified equation of motion, the angular frequency \( \omega \) is determined by the expression \[ \omega = \sqrt{\frac{g+a}{l}} \]
This formula highlights how the effective gravitational force—adjusted by both gravity and the elevator's acceleration—affects the rate of oscillation. Understanding angular frequency is valuable for predicting how fast the pendulum completes its back-and-forth swings, which can be directly correlated to periodic behavior seen throughout physics.
It also shows how adjustments in either gravitational forces or pendulum length \( l \) alter the pendulum's responsive timing to those forces.

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Most popular questions from this chapter

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

Consider a mass \(m\) moving in two dimensions with potential energy \(U(x, y)=\frac{1}{2} k r^{2},\) where \(r^{2}=x^{2}+y^{2} .\) Write down the Lagrangian, using coordinates \(x\) and \(y,\) and find the two Lagrange equations of motion. Describe their solutions. [This is the potential energy of an ion in an "ion trap," which can be used to study the properties of individual atomic ions.]

Lagrange's equations in the form discussed in this chapter hold only if the forces (at least the nonconstraint forces) are derivable from a potential energy. To get an idea how they can be modified to include forces like friction, consider the following: A single particle in one dimension is subject to various conservative forces (net conservative force \(=F=-\partial U / \partial x)\) and a nonconservative force (let's call it \(F_{\text {fric }}\) ). Define the Lagrangian as \(\mathcal{L}=T-U\) and show that the appropriate modification is $$\frac{\partial \mathcal{L}}{\partial x}+F_{\mathrm{fric}}=\frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$

Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

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