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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.

alt

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 \(\PageIndex{3}\)

alt

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 \(\PageIndex{4}\)

alt

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

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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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