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Solved Problems in Lagrangian and Hamiltonian Mechanics
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This textbook covers all the standard introductory topics in classical mechanics, including Newton's laws, oscillations, energy, momentum, angular momentum, planetary motion, and special relativity. It also explores more advanced topics, such as normal modes, the Lagrangian method, gyroscopic motion, fictitious forces, 4-vectors, and general relativity. It contains more than 250 problems with detailed solutions so students can easily check their understanding of the topic. There are also over 350 unworked exercises, which are ideal for homework assignments. Password-protected solutions are available to instructors at www.cambridge.org/9780521876223. The vast number of problems alone makes it an ideal supplementary book for all levels of undergraduate physics courses in classical mechanics. The text also includes many additional remarks which discuss issues that are often glossed over in other textbooks, and it is thoroughly illustrated with more than 600 figures to help demonstrate key concepts.
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13.8: More Lagrangian Mechanics Examples
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- Page ID 8445
- Jeremy Tatum
- University of Victoria
Example \(\PageIndex{1}\)
The kinetic energy is
\[ T=\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m_{1}(\dot{x}-\dot{y})^{2}+\frac{1}{2}m_{2}(\dot{x}+\dot{y})^{2}. \label{13.8.1} \]
The potential energy is
\[ V=g[Mx-m_{1}(x-y)-m_{2}(x+y)]+constant. \label{13.8.2} \]
Apply Lagrange’s equation (13.4.13) in turn to the coordinates \( x\) and \( y\):
\[ M\ddot{x}+m_{1}(\ddot{x}+\ddot{y})+m_{2}(\ddot{x}+\ddot{y})=-g(M-m_{1}-m_{2}). \label{13.8.3} \]
\[ -m_{1}(\ddot{x}+\ddot{y})+m_{2}(\ddot{x}+\ddot{y})=-g(m_{1}-m_{2}) \label{13.8.4} \]
These two equations can be solved at one’s leisure for \( \ddot{x}\) and \( \ddot{y}\).
Example \(\PageIndex{2}\)
A torus of mass \( M\) and radius \( a\) rolls without slipping on a horizontal plane. A pearl of mass \( m\) slides smoothly around inside the torus. Describe the motion.
I have marked in the several velocity vectors. The torus is rolling at angular speed \( \dot{\phi}\) Consequently the linear speed of the centre of mass of the hoop is \( a\dot{\phi}\) and the pearl also shares this velocity. In addition, the pearl is sliding relative to the torus at an angular speed \( \dot{\theta}\) and consequently has a component to its velocity of \( a\dot{\theta}\) tangential to the torus. We are now ready to start.
The kinetic energy of the torus is the sum of its translational and rotational kinetic energies:
\( \frac{1}{2}M(a\dot{\phi})+\frac{1}{2}(Ma^{2})\dot{\phi}^{2}=Ma^{2}\dot{\phi}^{2}\)
The kinetic energy of the pearl is
\( \frac{1}{2}ma^{2}(\dot{\theta}^{2}+\dot{\phi}^{2}-2\dot{\theta}\dot{\phi}\cos\theta)\)
\[ T=Ma^{2}\dot{\phi}^{2}+\frac{1}{2}ma^{2}(\dot{\theta}^{2}+\dot{\phi}^{2}-2\dot{\theta}\dot{\phi}\cos\theta). \label{13.8.5} \]
\[ V=constant-mga\cos\theta. \label{13.8.6} \]
The lagrangian equation in \( \theta\) becomes
\[ a(\ddot{\theta}-\ddot{\phi}\cos\theta)+g\sin\theta=0. \label{13.8.7} \]
The lagrangian equation in \( \phi\) becomes
\[ (2M+m)\ddot{\phi}=m(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta) \label{13.8.8} \]
These, then, are two differential equations in the two variables. The lagrangian part of the analysis is over; we now have to see if we can do anything with these equations.
It is easy to eliminate \( \ddot{\phi}\) and hence get a single differential equation in \( \theta\) :.
\[ (2M+m\sin^{2}\theta)a\ddot{\theta}+ma\sin\theta\cos\theta\dot{\theta}^{2}+(2M+m)g\sin\theta=0. \label{13.8.9} \]
If you are good at differential equations, you might be able to do something with this, and get \( \theta\) as a function of the time. In the meantime, I think I can get the “first space integral” (see Chapter 6) – i.e. \( \dot{\theta}\) as a function of \( \theta\). Thus, the total energy is constant:
\[ Ma^{2}\dot{\phi}^{2}+\frac{1}{2}ma^{2}(\dot{\theta}^{2}+\dot{\phi}^{2}-2\dot{\theta}\dot{\phi}\cos\theta)-mga\cos\theta=E. \label{13.8.10} \]
Equation \( \ref{13.8.8}\) can easily be integrated once with respect to time, since \( \ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta=\frac{d}{dt}(\dot{\theta}\cos\theta)\) as would have been apparent during the derivation of Equation \( \ref{13.8.8}\). With the condition that the kinetic energy was initially zero, integration of Equation \( \ref{13.8.8}\) gives
\[ (2M+m)\dot{\phi}=m\dot{\theta}\cos\theta. \label{13.8.11} \]
Now we can easily eliminate \( \dot{\phi}\) between Equations \( \ref{13.8.10}\) and \( \ref{13.8.11}\), to obtain a single equation relating \( \dot{\theta}\) and \( \theta\) :
\[ b\dot{\theta}^{2}(1+c\sin^{2}\theta)-d\cos\theta-1=0, \label{13.8.12} \]
\[ b=\frac{Mma^{2}}{(2M+m)E},\quad c=\frac{m}{2M},\quad d=\frac{mga}{E}=-\sec\alpha. \label{13.8.13 a,b,c} \]
Example \(\PageIndex{3}\)
As in example ii, we have a torus of radius \( a\) and mass \( M\), and a pearl of mass \( m\) which can slide freely and without friction around the torus. This time, however, the torus is not rolling along the table, but is spinning about a vertical axis at an angular speed \( \dot{\phi}\). The pearl has a velocity component \( a\dot{\theta}\) because it is sliding around the torus, and a component \( a\sin\theta\dot{\phi}\) because the torus is spinning. The resultant speed is the orthogonal sum of these. The kinetic energy of the system is the sum of the translational kinetic energy of the pearl and the rotational kinetic energy of the torus:
\[ T=\frac{1}{2}ma^{2}(\dot{\theta}^{2}+\sin^{2}\theta\dot{\phi}^{2})+\frac{1}{2}(\frac{1}{2}Ma^{2})\dot{\phi}^{2}. \label{13.8.14} \]
If we refer potential energy to the centre of the torus:
\[ V=mga\cos\theta. \label{13.8.15} \]
The lagrangian equations with respect to the two variables are:
\( \theta\):
\[ a(\ddot{\theta}-\sin\theta\cos\theta\dot{\phi}^{2})-g\sin\theta=0. \label{13.8.16} \]
\[ m\sin^{2}\theta\dot{\phi}+\frac{1}{2}M\dot{\phi}=constant. \label{13.8.17} \]
The constant is equal to whatever the initial value of the left hand side was. E.g., maybe the initial values of \( \theta\) and \( \dot{\phi}\) were \( \alpha\) and \( \omega\). This finishes the lagrangian part of the analysis. The rest is up to you. For example, it would be easy to eliminate \( \dot{\phi}\) between these two equations to obtain a differential equation between \( \theta\) and the time. If you then write \( \ddot{\theta}\) as \( \dot{\theta}\frac{d\dot{\theta}}{d\theta}\) in the usual way, I think it wouldn’t be too difficult to obtain the first space integral and hence get \( \dot{\theta}\) as a function of \( \theta\) . I haven’t tried it, but I’m sure it’ll work.
Example \(\PageIndex{4}\)
Figure XIII.10 shows a pendulum. The mass at the end is \( m\). It is at the end not of the usual inflexible string, but of an elastic spring obeying Hooke’s law, of force constant \( k\). The spring is sufficiently stiff at right angles to its length that it remains straight during the motion, and all the motion is restricted to a plane. The unstretched natural length of the spring is \( l\), and, as shown, its extension is \( r\). The spring itself is “light” in the sense that it does not contribute to the kinetic or potential energies. (You can give the spring a finite mass if you want to make the problem more difficult.) The kinetic and potential energies are
\[ T=\frac{1}{2}m(\dot{r}^{2}+(l+r)^{2}\dot{\theta}^{2}) \label{13.8.18} \]
\[ V=constant-mg(l+r)\cos\theta+\frac{1}{2}kr^{2}. \label{13.8.19} \]
Apply Lagrange’s equation in turn to \( r\) and to \( \theta\) and see where it leads you.
Example \(\PageIndex{5}\)
Another example suitable for lagrangian methods is given as problem number 11 in Appendix A of these notes.
Lagrangian methods are particularly applicable to vibrating systems, and examples of these will be discussed in Chapter 17. These chapters are being written in more or less random order as the spirit moves me, rather than in logical order, so that vibrating systems appear after the unlikely sequence of relativity and hydrostatics.
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- Foreword Synoptic Tables.
- Chapter 1 : The Lagrangian formulation (1 1 problems)
- Chapter 2 : Lagrangian systems (14 problems)
- Chapter 3 : The Hamilton's principle (15 problems)
- Chapter 4 : The Hamiltonian formalism (17 problems)
- Chapter 5 : The Hamilton-Jacobi formalism (1 1 problems)
- Chapter 6 : Integrable systems (18 problems)
- Chapter 7 : Quasi-integrable systems (9 problems)
- Chapter 8 : From order to chaos (12 problems). Bibliography.
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Solved Problems in Lagrangian and Hamiltonian Mechanics
- Claude Gignoux 0 ,
- Bernard Silvestre-Brac 1
Lab. Physique Subatomique et Cosmologie CNRS-IN2P3, Université Joseph Fourier, Grenoble, France
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The chosen problems often originate from real physical situations
They are worked out in detail so that the reader may perfectly integrate all the notions presented in the summary of the course
37k Accesses
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Table of contents (8 chapters)
Front matter, the lagrangian formulation.
Claude Gignoux, Bernard Silvestre-Brac
Lagrangian Systems
Hamilton’s principle, hamiltonian formalism, hamilton–jacobi formalism, integrable systems, quasi-integrable systems, from order to chaos, back matter.
- Hamilton-Jacobi
- Hamiltonian mechanics
- Quasi-integrable systems
From the reviews:
Book Title : Solved Problems in Lagrangian and Hamiltonian Mechanics
Authors : Claude Gignoux, Bernard Silvestre-Brac
DOI : https://doi.org/10.1007/978-90-481-2393-3
Publisher : Springer Dordrecht
eBook Packages : Physics and Astronomy , Physics and Astronomy (R0)
Copyright Information : Springer Science+Business Media B.V. 2009
Hardcover ISBN : 978-90-481-2392-6 Published: 17 July 2009
Softcover ISBN : 978-94-007-9176-3 Published: 16 November 2014
eBook ISBN : 978-90-481-2393-3 Published: 14 July 2009
Edition Number : 1
Number of Pages : XVIII, 464
Additional Information : Original French edition published by Grenoble Sciences, 2004
Topics : Classical Mechanics , Complex Systems , Classical Electrodynamics , Statistical Physics and Dynamical Systems
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The Lagrangian Method Copyright 2007 by David Morin, [email protected](draft version) In this chapter, we're going to learn about a whole new way of looking at things. Consider the system of a mass on the end of a spring. We can analyze this, of course, by usingF=ma to write downmx˜ =¡kx.
Lagrangian Problems 1. Cube on Top of a Cylinder Consider the gure below which shows a cube of mass m with a side length of 2b sitting on top of a xed rubber horizontal cylinder of radius r. The cube cannot slip on the cylinder, but it can rock from side to side.
The vast number of problems alone makes it an ideal supplementary book for all levels of undergraduate physics courses in classical mechanics. The text also includes many additional remarks which discuss issues that are often glossed over in other textbooks, and it is thoroughly illustrated with more than 600 figures to help demonstrate key ...
Lagrangian Mechanics 6.1 Generalized Coordinates set of generalized coordinates q1, . . . , qn completely describes the positions of all particles in a mechanical system. In a system with df degrees of freedom and k constraints, n = df −k independent generalized coordinates are needed to completely specify all the positions.
2.1 Lagrangian for unconstrained systems For a collection of particles with conservative forces described by a potential, we have in inertial cartesian coordinates m¨x i=F i: The left hand side of this equation is determined by the kinetic energy func- tion as the time derivative of the momentump i=@T=@x_
3.7.1 Physical interpretation of the Lagrange multipliers 84 3.8 The invariance of the Lagrange equations 86 3.9 Problems 88 II HAMILTONIAN MECHANICS 91 4 Hamilton's equations 93 4.1 The Legendre transformation 93 4.1.1 Application to thermodynamics 95 4.2 Application to the Lagrangian. The Hamiltonian 97 4.3 Hamilton's canonical equations 98
n = mα xα,i xα,i = + y2 + z2 ) . α=1 n ∑ mα ( x2 α=1 (4.3) 4.2 Introduction Although Newton's equation F = p correctly describes the motion of a particle (or a system of particles), it is often the case that a problem will be too complicated to solve using this formalism.
These two equations can be solved at one's leisure for x¨ x ¨ and y¨ y ¨. Example 13.8.2 13.8. 2. A torus of mass M M and radius a a rolls without slipping on a horizontal plane. A pearl of mass m m slides smoothly around inside the torus. Describe the motion. I have marked in the several velocity vectors.
13.1 Introduction The usual way of using newtonian mechanics to solve a problem in dynamics is first of all to draw a large, clear diagram of the system, using a ruler and a compass. Then mark in the forces on the various parts of the system with red arrows and the accelerations of the various parts with green arrows.
The Solved Problems in Lagrangian and Hamiltonian Mechanics Reading Committee included the following members: Robert ARVIEU, Professor at the Joseph Fourier University, Grenoble, France Jacques MEYER, Professor at the Nuclear Physics Institute, Claude Bernard University, Lyon, France with the contribution of: Myriam REFFAY and Bertrand RUPH
Here are some sample problems. I will assign similar problems for the next problem set. Example 1 In Figure 1 we show a box of mass m sliding down a ramp of mass M. The ramp moves without friction on the horizontal plane and is located by coordinate x1.
However, it will go through a practical step by step process such that a person who understands the theory and examples presented in this handout will be able to solve olympiad physics problems through the usage of lagrangian formalism.
Contributor Silvestre-Brac, Bernard. Contents/Summary Bibliography Includes bibliographical references (pages 457-460) and index. Contents Foreword Synoptic Tables. Chapter 1 : The Lagrangian formulation (1 1 problems) Chapter 2 : Lagrangian systems (14 problems) Chapter 3 : The Hamilton's principle (15 problems)
2. Write down the equation of motion for the position(s) of the masses. Solve the equations of motion and show that. (m1 x1(t) = m2)gt2 : (7.1) 2(m1 + m2 + 2mP) Determine the potential energy V , the kinetic energy Tlin of the masses and the rotational energy Trot of the pulley as functions of time t.
1 Strategies for solving problems 1 1.1 General strategies 1 1.2 Units, dimensional analysis 4 1.3 Approximations, limiting cases 7 1.4 Solving differential equations numerically 11 1.5 Problems 14 1.6 Exercises 15 1.7 Solutions 18 2 Statics 22 2.1 Balancing forces 22 2.2 Balancing torques 27 2.3 Problems 30 2.4 Exercises 35 2.5 Solutions 39 3 ...
beyond that as well. The scheme is Lagrangian and Hamiltonian mechanics. Its original prescription rested on two principles. First that we should try to express the state of the mechanical system using the minimum representa-tion possible and which re ects the fact that the physics of the problem is coordinate-invariant.
Lagrangian mechanics is extremely efficient: There are only as many equations to solve as there are physically significant variables. Lagrange did not introduce new physical principles to mechanics. The physical concepts are due to Newton and Galileo.
4 Lagrangian Mechanics 62 4.1 Example: Free Particle in Polar Coordinates 63 4.2 Lagrange s Equations 66 4.3 Hamilton s Principle 77 Exercises 80 ... powerful tools for setting up and solving problems, which then expand the reach of the student s powers and lead to applications in quantum theory and statistical mechanics. However, on the dull ...
Solved Problems in Lagrangian and Hamiltonian Mechanics Home Textbook Authors: Claude Gignoux, Bernard Silvestre-Brac The chosen problems often originate from real physical situations They are worked out in detail so that the reader may perfectly integrate all the notions presented in the summary of the course 37k Accesses 23 Citations 4 Altmetric
Solved Problems in Lagrangian and Hamiltonian Mechanics. C. Gignoux, B. Silvestre-Brac. Published 15 July 2009. Physics. TLDR. This book discusses the Hamiltonian formalism, the Lagrangian formulation, the Hamilton's principle, and the Hamilton-Jacobi formalism as well as other topics. Expand.
1. Give the expression of the kinetic energy in terms of the generalized coordinate. 2. Give the expression of the generalized force. 3. Write down the corresponding Lagrange equation. 4. This simple question illustrates the difference between virtual work and real work. We are interested in the reaction force of the hoop on the bead.
The truth is that the Lagrangian formulation of mechanics makes most problems simpler to solve. That's because F = ma is a PAIN—for all but the most basic setups. You have to deal with vectors and "fictitious forces" and jump through crazy hoops if you want to solve a problem using anything but x, y, z coordinates. The Lagrangian approach does away with all that nonsense.
It's easier to figure out tough problems faster using Chegg Study. Unlike static PDF Solved Problems in Lagrangian and Hamiltonian Mechanics solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn.