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Chapter 7: Problem 22

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equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that \(g\) has been replaced by \(g+a\).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鈥檚 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鈥攁djusted by both gravity and the elevator's acceleration鈥攁ffects 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

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

Consider a mass \(m\) moving in a frictionless plane that slopes at an angle \(\alpha\) with the horizontal. Write down the Lagrangian in terms of coordinates \(x,\) measured horizontally across the slope, and \(y\), measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected.

A smooth wire is bent into the shape of a helix, with cylindrical polar coordinates \(\rho=R\) and \(z=\lambda \phi,\) where \(R\) and \(\lambda\) are constants and the \(z\) axis is vertically up (and gravity vertically down). Using \(z\) as your generalized coordinate, write down the Lagrangian for a bead of mass \(m\) threaded on the wire. Find the Lagrange equation and hence the bead's vertical acceleration \(\ddot{z}\). In the limit that \(R \rightarrow 0\), what is \(\ddot{z} ?\) Does this make sense?

A mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]
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