This document contains Homer Reid's solutions to problems from Goldstein's Classical Mechanics textbook. The solutions covered include:
1) Finding the momentum and kinetic energy of a nucleus that decays by emitting an electron and neutrino.
2) Deriving the escape velocity of Earth from the conservation of energy.
3) Deriving an equation of motion for a vertically launched rocket where mass decreases over time.
4) Deriving equations relating kinetic energy to force and momentum for particles with constant and varying mass.
5) Deriving an expression for the distance of a center of mass from an origin.
6) Deriving constraint equations for a system of two wheels on an axle rolling without slipping
This document contains Homer Reid's solutions to problems from Goldstein's Classical Mechanics textbook. The first problem solved involves a radioactive nucleus decaying and emitting an electron and neutrino. The solution finds the direction and momentum of the recoiling nucleus. The next problems solved include deriving the escape velocity of Earth, the equation of motion for a rocket, relating kinetic energy to momentum for varying mass systems, and several other kinematic and constraint problems.
1. The document discusses various kinematics problems involving motion under uniform acceleration. It provides solutions using graphical, analytical and vector methods.
2. Methods include calculating time taken to cross a river based on velocities and angles, determining average and instantaneous velocities from distance-time graphs, resolving velocities into components, and finding the distance between particles moving with different initial velocities.
3. One problem involves three particles moving in a circle such that they are always at the vertices of an equilateral triangle, and calculates the distance traveled by one particle before they meet.
1. The document provides solutions to 7 miscellaneous physics problems involving kinematics, dynamics, and rotational motion. The problems involve calculating quantities like angular speeds, forces, and times using principles like conservation of energy, angular momentum, and equations of motion. Complex algebraic and calculus solutions are shown.
2. Key steps involve setting up and solving equations derived from applying relevant physics principles to diagrams of the systems. Calculus techniques like differentiation and numerical integration are used.
3. Emphasis is placed on being able to instantly solve equations of the form f(x)=0 that arise, with references provided to resources on techniques like Newton-Raphson iteration.
Serway, raymond a physics for scientists and engineers (6e) solutionsTatiani Andressa
This document contains a chapter outline and sample questions and solutions for a physics and measurement chapter. The chapter outline lists topics like standards of length, mass and time, density and atomic mass, and dimensional analysis. The questions and solutions provide examples of calculations involving converting between units, determining densities, and applying dimensional analysis.
This chapter discusses spin in strong interactions like pion-nucleon and nucleon-nucleon scattering. It introduces the density matrix and reaction matrix to describe mixed spin states. The density matrix is determined by the mean values of spin operators and completely characterizes the spin state. The reaction matrix relates the initial and final density matrices. This allows calculating observables in the final state given the initial state parameters and scattering matrix. Pauli matrices provide a complete spin operator basis for pion-nucleon reactions. The density matrix is expressed in terms of the target or beam polarization vector. Constraints from rotational, parity and time reversal symmetries on the nucleon-nucleon scattering matrix are also discussed.
This document discusses various types of mechanical vibrations including undamped, underdamped, critically damped, and overdamped motion. It provides examples of solving differential equations describing simple harmonic motion under different damping conditions and applying the solutions to problems involving springs, pendulums, and other oscillating systems. Specific cases are worked through to determine characteristics like frequency, period, and damping behavior.
Applications of Differential Equations of First order and First DegreeDheirya Joshi
The document describes how to calculate the time it takes for a population growing at 5% annually to double in size using a differential equation model. It is also solved to be 20loge2 years, or approximately 14 years. A second problem involves calculating the final temperature of liquid in an insulated cylindrical tank over 5 days using a heat transfer model. A third problem uses kinematics equations to find how far a drag racer will travel in 8 seconds if its speed increases by 40 feet per second each second.
The document provides an overview of key physics equations and concepts related to forces and motion, including equations for relative deviation, prefixes, units of area and volume, average speed, velocity, acceleration, momentum, Newton's laws of motion, and impulse. Key variables and their units are defined for each equation. Examples of displacement-time and velocity-time graphs are also included to illustrate the relationships between displacement, velocity, time, and acceleration.
This document contains Homer Reid's solutions to problems from Goldstein's Classical Mechanics textbook. The first problem solved involves a radioactive nucleus decaying and emitting an electron and neutrino. The solution finds the direction and momentum of the recoiling nucleus. The next problems solved include deriving the escape velocity of Earth, the equation of motion for a rocket, relating kinetic energy to momentum for varying mass systems, and several other kinematic and constraint problems.
1. The document discusses various kinematics problems involving motion under uniform acceleration. It provides solutions using graphical, analytical and vector methods.
2. Methods include calculating time taken to cross a river based on velocities and angles, determining average and instantaneous velocities from distance-time graphs, resolving velocities into components, and finding the distance between particles moving with different initial velocities.
3. One problem involves three particles moving in a circle such that they are always at the vertices of an equilateral triangle, and calculates the distance traveled by one particle before they meet.
1. The document provides solutions to 7 miscellaneous physics problems involving kinematics, dynamics, and rotational motion. The problems involve calculating quantities like angular speeds, forces, and times using principles like conservation of energy, angular momentum, and equations of motion. Complex algebraic and calculus solutions are shown.
2. Key steps involve setting up and solving equations derived from applying relevant physics principles to diagrams of the systems. Calculus techniques like differentiation and numerical integration are used.
3. Emphasis is placed on being able to instantly solve equations of the form f(x)=0 that arise, with references provided to resources on techniques like Newton-Raphson iteration.
Serway, raymond a physics for scientists and engineers (6e) solutionsTatiani Andressa
This document contains a chapter outline and sample questions and solutions for a physics and measurement chapter. The chapter outline lists topics like standards of length, mass and time, density and atomic mass, and dimensional analysis. The questions and solutions provide examples of calculations involving converting between units, determining densities, and applying dimensional analysis.
This chapter discusses spin in strong interactions like pion-nucleon and nucleon-nucleon scattering. It introduces the density matrix and reaction matrix to describe mixed spin states. The density matrix is determined by the mean values of spin operators and completely characterizes the spin state. The reaction matrix relates the initial and final density matrices. This allows calculating observables in the final state given the initial state parameters and scattering matrix. Pauli matrices provide a complete spin operator basis for pion-nucleon reactions. The density matrix is expressed in terms of the target or beam polarization vector. Constraints from rotational, parity and time reversal symmetries on the nucleon-nucleon scattering matrix are also discussed.
This document discusses various types of mechanical vibrations including undamped, underdamped, critically damped, and overdamped motion. It provides examples of solving differential equations describing simple harmonic motion under different damping conditions and applying the solutions to problems involving springs, pendulums, and other oscillating systems. Specific cases are worked through to determine characteristics like frequency, period, and damping behavior.
Applications of Differential Equations of First order and First DegreeDheirya Joshi
The document describes how to calculate the time it takes for a population growing at 5% annually to double in size using a differential equation model. It is also solved to be 20loge2 years, or approximately 14 years. A second problem involves calculating the final temperature of liquid in an insulated cylindrical tank over 5 days using a heat transfer model. A third problem uses kinematics equations to find how far a drag racer will travel in 8 seconds if its speed increases by 40 feet per second each second.
The document provides an overview of key physics equations and concepts related to forces and motion, including equations for relative deviation, prefixes, units of area and volume, average speed, velocity, acceleration, momentum, Newton's laws of motion, and impulse. Key variables and their units are defined for each equation. Examples of displacement-time and velocity-time graphs are also included to illustrate the relationships between displacement, velocity, time, and acceleration.
1) The document discusses elastic instability in structures, using the example of buckling of bars and columns under compressive loading. Elastic instability occurs when a structure transitions from stable to unstable deformation modes with increasing load.
2) As an introductory example, the document analyzes a rigid bar with a torsional spring, subject to horizontal and vertical forces. It derives an expression for the critical vertical load that causes instability, equal to the torsional spring stiffness divided by the bar length.
3) Looking ahead, the document notes it will apply these concepts of instability and critical load to analyze the buckling of a compressed column, which has a continuous distribution of stiffness rather than
Cluster-cluster aggregation with (complete) collisional fragmentationColm Connaughton
This document summarizes a presentation on cluster-cluster aggregation models with collisional fragmentation. It discusses mean-field theories of aggregation with a source of monomers and collision-induced fragmentation. Stationary solutions to the Smoluchowski equation are presented for both local and nonlocal aggregation kernels. While stationary nonlocal solutions exist, they are dynamically unstable. Simplified models with complete fragmentation and a source/sink of monomers produce exact solutions analogous to previous work. Nonlocality and the instability of stationary states require further study.
Cluster aggregation with complete collisional fragmentationColm Connaughton
The document summarizes research on cluster-cluster aggregation (CCA) models where particles stick together upon contact. It discusses mean-field kinetic equations to model CCA with sources and sinks of particles. For the case of complete fragmentation, it presents an exact solution to the kinetic equations. It finds that nonlocal cascades where larger clusters interact mostly with smaller ones can be unstable, leading to oscillatory behavior over time rather than a stationary state. The document outlines approaches to model the nonlocal case using approximations to the Smoluchowski kinetic equation.
A Fast Algorithm for Solving Scalar Wave Scattering Problem by Billions of Pa...A G
This document proposes a fast algorithm for solving wave scattering problems involving billions of particles using the convolution theorem and fast Fourier transforms (FFTs). The algorithm represents the Green's function as a vector and stores particle positions on a uniform grid, allowing the scattering calculation to be computed as a 3D convolution. This convolution can be rapidly evaluated using FFTs, significantly improving the efficiency over direct matrix-vector multiplication. The algorithm distributes data across multiple machines in a cluster to parallelize the computations.
This document summarizes a paper that considers the Klein-Gordon scalar field in a two-dimensional Rindler spacetime. It writes a two-dimensional action for the Klein-Gordon scalar in this background and obtains the equation of motion. The equation of motion can be solved exactly in imaginary time, with the solution taking an oscillatory form with a frequency equal to an integer number.
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
Understanding lattice Boltzmann boundary conditions through momentsTim Reis
This document provides an overview of the key points that will be covered in a lecture about understanding lattice Boltzmann boundary conditions through moments. The lecture will focus on connecting LB boundary conditions to LB distribution function moments. It will discuss interpreting other boundary condition methods in terms of moments and analyzing boundary condition accuracy, including for slip and no-slip flow conditions. The goal is to provide a clear perspective on LB boundary conditions and how to determine the appropriate conditions for different applications.
The document is a multi-part physics problem involving the analysis of dropping objects through air resistance and into snow.
The first part asks the student to show that the speed of a raindrop falling under gravity and gaining mass over time will eventually become constant, deriving an expression for the terminal speed.
The second part involves calculating the loss of mechanical energy when a sliding block hits a spring and experiences friction slowing its motion.
The third part examines using a ballistic pendulum to measure the speed of a bullet by calculating momentum and energy transfers between the bullet and suspended block.
The final part evaluates whether parachutes would be needed to drop soldiers packed into hay bales, based on observing the sinking
1. This document contains solutions to problems from Chapter 1 of the textbook "The Physics of Vibrations and Waves".
2. It provides detailed calculations and explanations for problems related to simple harmonic motion, including determining restoring forces, stiffness, frequencies, and solving differential equations of motion.
3. Examples include a simple pendulum, a mass on a spring, and vibrations of strings, membranes, and gas columns.
This document defines and provides examples of curves and parametrized curves. It discusses regular and unit-speed curves. The key points are:
i) A parametrized curve is a continuous function from an interval to Rn. Examples of parametrized curves include ellipses, parabolas, and helices.
ii) A regular curve is one where the derivative of the parametrization is never zero. A unit-speed curve has a derivative of constant length 1.
iii) The arc-length of a curve is defined as the integral of the derivative of the parametrization. Any reparametrization of a regular curve is also regular. A curve has a unit-
1. The document defines various physics concepts including units of measurement, kinematics equations, forces and motion, energy, heat, waves, and optics.
2. Prefixes are provided for the standard form of various units including tera, giga, mega, kilo, deci, centi, and milli.
3. Formulas are given for average speed, velocity, acceleration, linear motion, momentum, impulse, work, power, and more.
4. Concepts around pressure, density, buoyancy, heat transfer, the gas laws, refraction, lenses, and telescopes are also summarized.
Nonequilibrium Statistical Mechanics of Cluster-cluster Aggregation Warwick ...Colm Connaughton
This document summarizes a talk on nonequilibrium statistical mechanics of cluster-cluster aggregation. The talk focused on theoretical models of particle clustering, including simple models where particles perform random walks and merge upon contact, and more sophisticated models that track the distribution of cluster sizes over time using the Smoluchowski equation. It discussed self-similar solutions and stationary solutions of the Smoluchowski equation. It also described the gelation transition that can occur when clusters absorb smaller clusters rapidly, violating the assumption of mass conservation and leading to clusters of infinite size.
This document discusses techniques for calculating electric potential, including:
1. Laplace's equation and its solutions in 1D, 2D, and 3D, including boundary conditions.
2. The method of images, which uses fictitious "image" charges to solve problems involving conductors. The classical image problem and induced surface charge on a conductor are examined.
3. Other techniques like multipole expansion, separation of variables, and numerical methods like relaxation are mentioned but not explained in detail.
The one-dimensional wave equation governs vibrations of an elastic string. It is solved by separating variables, yielding solutions of the form F(x)G(t) where F and G satisfy ordinary differential equations. Boundary conditions require F(x) to be sinusoidal, with wavelengths that are integer multiples of the string length. The general solution is a superposition of these sinusoidal modes, with coefficients determined by the initial conditions. Motions of strings with different initial displacements are expressed as solutions to the one-dimensional wave equation.
The document contains 14 problems related to classical mechanics. Problem 1 describes two ladders leaning against two walls, with their intersection point 3 m above the ground, and asks for the distance between the walls. Problem 3 involves a pendulum whose length below a board can be varied, and asks the reader to show various relationships involving the semi-vertical angle, angular speed, and pendulum length are constant. Problem 4 describes a rod initially vertical on a table that is given an angular displacement and falls over, and asks for expressions involving the rod's angular speed, center speed, lower end speed, and the table's normal reaction.
1. The document discusses forces between charged particles and how to calculate those forces using Coulomb's law.
2. It presents examples of calculating the original charges on two spheres after they are connected by a wire, allowing their charges to redistribute evenly.
3. It also examines the forces and charges in a system of three particles in equilibrium.
The document provides an overview of key physics equations and concepts for Form 4 students, including equations for relative deviation, prefixes, units for area and volume, equations for average speed, velocity, acceleration, momentum, Newton's laws of motion, and impulse. Key terms are defined for important concepts like displacement, time, mass, force, and velocity. Formulas are presented for calculations involving these fundamental physics quantities and relationships.
This document provides a summary of Abhimanyu Roy's qualifications and experience. It lists his areas of expertise as profit center management, business administration, customer service management, and others. It then details his 17 years of experience in management roles for hotels and casinos in India and Europe, highlighting his responsibilities and achievements in increasing revenues, reducing costs, and improving operations. It concludes with his educational background of multiple MBAs and skills in areas like food safety, wine tasting, and training.
Save money for booking Paris international airport transport in advance with Paris Go Airport. Enjoy world-class services for Paris Orly transport with the experienced service providercompany.
Introductie presentatie voor Hogeschool Arnhem Nijmegen voor student referrals en het beter benutten van de netwerken van werknemers en/of studenten. Hoe social sharing met The Cure extra mogelijkheden biedt om nieuwe studenten te bereiken en te verleiden met de juiste content, zonder moeite.
Our President, Phil Hill, came to Tampa Bay in the Fall of 2001, fell in love with the warm waters, friendly people and picturesque palm trees and by March of 2004 he had moved down to Tampa and opened Xerographic Digital Solutions Inc. He also decided to start his family here and is currently expanding it along with XDS/Florida Copiers—“it’s a great place to raise kids”.
1) The document discusses elastic instability in structures, using the example of buckling of bars and columns under compressive loading. Elastic instability occurs when a structure transitions from stable to unstable deformation modes with increasing load.
2) As an introductory example, the document analyzes a rigid bar with a torsional spring, subject to horizontal and vertical forces. It derives an expression for the critical vertical load that causes instability, equal to the torsional spring stiffness divided by the bar length.
3) Looking ahead, the document notes it will apply these concepts of instability and critical load to analyze the buckling of a compressed column, which has a continuous distribution of stiffness rather than
Cluster-cluster aggregation with (complete) collisional fragmentationColm Connaughton
This document summarizes a presentation on cluster-cluster aggregation models with collisional fragmentation. It discusses mean-field theories of aggregation with a source of monomers and collision-induced fragmentation. Stationary solutions to the Smoluchowski equation are presented for both local and nonlocal aggregation kernels. While stationary nonlocal solutions exist, they are dynamically unstable. Simplified models with complete fragmentation and a source/sink of monomers produce exact solutions analogous to previous work. Nonlocality and the instability of stationary states require further study.
Cluster aggregation with complete collisional fragmentationColm Connaughton
The document summarizes research on cluster-cluster aggregation (CCA) models where particles stick together upon contact. It discusses mean-field kinetic equations to model CCA with sources and sinks of particles. For the case of complete fragmentation, it presents an exact solution to the kinetic equations. It finds that nonlocal cascades where larger clusters interact mostly with smaller ones can be unstable, leading to oscillatory behavior over time rather than a stationary state. The document outlines approaches to model the nonlocal case using approximations to the Smoluchowski kinetic equation.
A Fast Algorithm for Solving Scalar Wave Scattering Problem by Billions of Pa...A G
This document proposes a fast algorithm for solving wave scattering problems involving billions of particles using the convolution theorem and fast Fourier transforms (FFTs). The algorithm represents the Green's function as a vector and stores particle positions on a uniform grid, allowing the scattering calculation to be computed as a 3D convolution. This convolution can be rapidly evaluated using FFTs, significantly improving the efficiency over direct matrix-vector multiplication. The algorithm distributes data across multiple machines in a cluster to parallelize the computations.
This document summarizes a paper that considers the Klein-Gordon scalar field in a two-dimensional Rindler spacetime. It writes a two-dimensional action for the Klein-Gordon scalar in this background and obtains the equation of motion. The equation of motion can be solved exactly in imaginary time, with the solution taking an oscillatory form with a frequency equal to an integer number.
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
Understanding lattice Boltzmann boundary conditions through momentsTim Reis
This document provides an overview of the key points that will be covered in a lecture about understanding lattice Boltzmann boundary conditions through moments. The lecture will focus on connecting LB boundary conditions to LB distribution function moments. It will discuss interpreting other boundary condition methods in terms of moments and analyzing boundary condition accuracy, including for slip and no-slip flow conditions. The goal is to provide a clear perspective on LB boundary conditions and how to determine the appropriate conditions for different applications.
The document is a multi-part physics problem involving the analysis of dropping objects through air resistance and into snow.
The first part asks the student to show that the speed of a raindrop falling under gravity and gaining mass over time will eventually become constant, deriving an expression for the terminal speed.
The second part involves calculating the loss of mechanical energy when a sliding block hits a spring and experiences friction slowing its motion.
The third part examines using a ballistic pendulum to measure the speed of a bullet by calculating momentum and energy transfers between the bullet and suspended block.
The final part evaluates whether parachutes would be needed to drop soldiers packed into hay bales, based on observing the sinking
1. This document contains solutions to problems from Chapter 1 of the textbook "The Physics of Vibrations and Waves".
2. It provides detailed calculations and explanations for problems related to simple harmonic motion, including determining restoring forces, stiffness, frequencies, and solving differential equations of motion.
3. Examples include a simple pendulum, a mass on a spring, and vibrations of strings, membranes, and gas columns.
This document defines and provides examples of curves and parametrized curves. It discusses regular and unit-speed curves. The key points are:
i) A parametrized curve is a continuous function from an interval to Rn. Examples of parametrized curves include ellipses, parabolas, and helices.
ii) A regular curve is one where the derivative of the parametrization is never zero. A unit-speed curve has a derivative of constant length 1.
iii) The arc-length of a curve is defined as the integral of the derivative of the parametrization. Any reparametrization of a regular curve is also regular. A curve has a unit-
1. The document defines various physics concepts including units of measurement, kinematics equations, forces and motion, energy, heat, waves, and optics.
2. Prefixes are provided for the standard form of various units including tera, giga, mega, kilo, deci, centi, and milli.
3. Formulas are given for average speed, velocity, acceleration, linear motion, momentum, impulse, work, power, and more.
4. Concepts around pressure, density, buoyancy, heat transfer, the gas laws, refraction, lenses, and telescopes are also summarized.
Nonequilibrium Statistical Mechanics of Cluster-cluster Aggregation Warwick ...Colm Connaughton
This document summarizes a talk on nonequilibrium statistical mechanics of cluster-cluster aggregation. The talk focused on theoretical models of particle clustering, including simple models where particles perform random walks and merge upon contact, and more sophisticated models that track the distribution of cluster sizes over time using the Smoluchowski equation. It discussed self-similar solutions and stationary solutions of the Smoluchowski equation. It also described the gelation transition that can occur when clusters absorb smaller clusters rapidly, violating the assumption of mass conservation and leading to clusters of infinite size.
This document discusses techniques for calculating electric potential, including:
1. Laplace's equation and its solutions in 1D, 2D, and 3D, including boundary conditions.
2. The method of images, which uses fictitious "image" charges to solve problems involving conductors. The classical image problem and induced surface charge on a conductor are examined.
3. Other techniques like multipole expansion, separation of variables, and numerical methods like relaxation are mentioned but not explained in detail.
The one-dimensional wave equation governs vibrations of an elastic string. It is solved by separating variables, yielding solutions of the form F(x)G(t) where F and G satisfy ordinary differential equations. Boundary conditions require F(x) to be sinusoidal, with wavelengths that are integer multiples of the string length. The general solution is a superposition of these sinusoidal modes, with coefficients determined by the initial conditions. Motions of strings with different initial displacements are expressed as solutions to the one-dimensional wave equation.
The document contains 14 problems related to classical mechanics. Problem 1 describes two ladders leaning against two walls, with their intersection point 3 m above the ground, and asks for the distance between the walls. Problem 3 involves a pendulum whose length below a board can be varied, and asks the reader to show various relationships involving the semi-vertical angle, angular speed, and pendulum length are constant. Problem 4 describes a rod initially vertical on a table that is given an angular displacement and falls over, and asks for expressions involving the rod's angular speed, center speed, lower end speed, and the table's normal reaction.
1. The document discusses forces between charged particles and how to calculate those forces using Coulomb's law.
2. It presents examples of calculating the original charges on two spheres after they are connected by a wire, allowing their charges to redistribute evenly.
3. It also examines the forces and charges in a system of three particles in equilibrium.
The document provides an overview of key physics equations and concepts for Form 4 students, including equations for relative deviation, prefixes, units for area and volume, equations for average speed, velocity, acceleration, momentum, Newton's laws of motion, and impulse. Key terms are defined for important concepts like displacement, time, mass, force, and velocity. Formulas are presented for calculations involving these fundamental physics quantities and relationships.
This document provides a summary of Abhimanyu Roy's qualifications and experience. It lists his areas of expertise as profit center management, business administration, customer service management, and others. It then details his 17 years of experience in management roles for hotels and casinos in India and Europe, highlighting his responsibilities and achievements in increasing revenues, reducing costs, and improving operations. It concludes with his educational background of multiple MBAs and skills in areas like food safety, wine tasting, and training.
Save money for booking Paris international airport transport in advance with Paris Go Airport. Enjoy world-class services for Paris Orly transport with the experienced service providercompany.
Introductie presentatie voor Hogeschool Arnhem Nijmegen voor student referrals en het beter benutten van de netwerken van werknemers en/of studenten. Hoe social sharing met The Cure extra mogelijkheden biedt om nieuwe studenten te bereiken en te verleiden met de juiste content, zonder moeite.
Our President, Phil Hill, came to Tampa Bay in the Fall of 2001, fell in love with the warm waters, friendly people and picturesque palm trees and by March of 2004 he had moved down to Tampa and opened Xerographic Digital Solutions Inc. He also decided to start his family here and is currently expanding it along with XDS/Florida Copiers—“it’s a great place to raise kids”.
El documento presenta conceptos clave sobre el desarrollo sostenible según varios autores y organizaciones. El desarrollo sostenible se refiere a satisfacer las necesidades actuales sin comprometer las necesidades de las futuras generaciones, cuidando y usando los recursos naturales de forma responsable. Varios autores enfatizan la importancia de proteger los recursos naturales, usarlos a un ritmo normal y no causar un declive no razonable para mantener los sistemas naturales.
3D Print Amman | مقدمة للطباعة ثلاثية الابعادMohammad Misto
مقدمة عن الطباعة ثلاثية الابعاد اعدها لكم محمد مستو وتحتوي على مقدمة عن الطباعة ثلاثية الابعاد و ما هي انواع الطباعة و اشهر استخداماتها بالاضافة للمواد المستخدمة في عملية الطباعة
3D Printing
Principales agresiones humanas a la naturaleza.Alfredo García
Tema ecológico sobre las agresiones que realiza el hombre a la naturaleza, se pone especial énfasis en las agresiones atmosféricas, pero también se tratan el resto.
This document discusses simple and compound pendulums. A simple pendulum consists of a mass attached to a string that swings back and forth. Its period depends only on its length and gravity. A compound pendulum is a rigid object that pivots, like a door. Its period depends on its length of gyration, moment of inertia, mass, and gravity. For small angles, a compound pendulum behaves similarly to a simple pendulum with an effective length. Both types of pendulums exhibit simple harmonic motion that can be modeled by the same equation.
The document is a PowerPoint presentation about the simple pendulum and periodic motion. It defines a simple pendulum as a mass attached to a string that is free to swing back and forth. The motion of the pendulum is periodic, meaning it repeats at regular time intervals. The time taken to complete one full swing from side to side is called the periodic time or period of the pendulum.
1) The document discusses the problem of a particle sliding off a moving hemisphere, with the goal of finding the angle at which the particle's horizontal velocity is maximum.
2) Conservation of momentum and energy equations are used to derive an expression for the particle's horizontal velocity in terms of the angle, which reaches a maximum at a certain angle.
3) Setting the derivative of this expression to zero results in a cubic equation that determines the maximum angle, which can be solved numerically for different ratios of particle to hemisphere mass.
1) The document discusses the problem of a particle sliding off a moving hemisphere, using conservation of momentum and energy equations to derive an expression for the particle's horizontal velocity vx as a function of the angle θ.
2) Setting the derivative of vx equal to zero yields a cubic equation that determines the angle θ at which vx is maximized, corresponding to the particle losing contact with the hemisphere.
3) For the special case where the particle and hemisphere masses are equal (ratio r = 1), the cubic equation can be solved to find θ ≈ 42.9 degrees.
This document summarizes solutions to three theoretical questions:
1) Describes how to use measurements of gravitational redshift to determine the mass and radius of a star.
2) Explains Snell's law and how it can be used to determine the path of light rays through a medium with a linear change in refractive index.
3) Analyzes the motion of a floating cylindrical buoy, determining equations for its vertical and rotational oscillations and relating the periods.
The document discusses motion under central forces. It shows that the equations of motion for a two-body central force problem can be reduced to a one-body problem. For planetary motion with gravitational force, the integrals of motion lead to Kepler's laws, with orbits taking the form of ellipses, parabolas or hyperbolas depending on the value of the eccentricity parameter ε. An eccentricity of 0 gives a circular orbit, 1 gives a parabolic orbit, and above 1 gives a hyperbolic orbit.
1. The document provides solutions to physics problems involving charged spheres and cubes, rotating wheels, friction on motorcycles, chasing foxes and rabbits, growing raindrops, and springs wrapped around poles.
2. It derives equations and ratios for potentials, field strengths, minimum distances, meeting points, accelerations, temperatures, spring constants, and more using principles of electrostatics, kinematics, dynamics, and thermodynamics.
3. The solutions are detailed and rigorous, employing techniques like Gauss' law, dimensional analysis, conservation of energy, and modeling circular and conical motions.
Lagrangian formulation provides an alternative but equivalent way to derive equations of motion compared to Newtonian mechanics.
The document provides examples of deriving equations of motion for simple harmonic oscillators, Atwood's machine, and a spring pendulum using the Lagrangian formulation. It also shows the equivalence between Lagrange's equations and Newton's second law.
Specifically, it demonstrates that for a conservative system using generalized coordinates, Lagrange's equations reduce to F=ma, where the generalized forces are equal to the negative gradient of the potential energy.
- The document discusses the 4/3 problem as it relates to the gravitational field of a uniform massive ball moving at constant velocity.
- It derives expressions for the gravitational field potentials both inside and outside the moving ball using the superposition principle and Lorentz transformations.
- Calculations show that the effective mass of the gravitational field found from the field energy does not equal the effective mass found from the field momentum, with a ratio of approximately 4/3, demonstrating that the 4/3 problem exists for gravitational fields as it does for electromagnetic fields.
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Newton™s Laws; Moment of a Vector; Gravitation; Finite Rotations; Trajectory of a Projectile with Air Resistance; The Simple Pendulum; The Linear Harmonic Oscillator; The Damped Harmonic Oscillator
1. The document analyzes the forces acting on a rope as it falls over a support. It derives an expression for the force N from the support as a function of time t until the free end of the rope reaches a distance of 2L.
2. It examines the motion of a mass attached to a string winding around a thin pole, and derives an expression for the angle θ at which the string becomes fully unwound using conservation of energy and the horizontal component of tension.
3. It considers electrical resistance in a circuit composed of squares connected at the corners. By simplifying the circuit in stages, it derives an expression for the desired resistance factor x between opposite corners of the original circuit.
1. The document analyzes the forces acting on a rope as it falls over a support. It derives an expression for the force N from the support as a function of time t until the free end of the rope falls a distance of 2L.
2. It examines the motion of a mass attached to a string winding around a thin pole, and derives an expression for the angle θ that the string will make with the horizontal when fully unwound.
3. It considers electrical resistance in a circuit made of squares connected at the corners. It derives an expression for the factor x that determines the desired resistance between opposite corners.
Gravitation, one of the fundamental forces governing the universe, holds profound significance in the realm of physics. It elucidates the captivating dance of celestial bodies, the gentle sway of tides, and the stability of planetary orbits. At the core of gravitation lies Sir Isaac Newton's revolutionary insights, encapsulated in his Law of Universal Gravitation. This law elegantly describes how every particle in the cosmos attracts every other particle with a force directly proportional to their masses and inversely proportional to the square of the distance between them. Through the study of gravitation, Class 11 physics students embark on a journey to comprehend the intricate interplay of masses, distances, and forces that shape the cosmos. From unraveling the mysteries of planetary motion to delving into the concept of escape velocity, the exploration of gravitation unveils the underlying principles governing the harmony of celestial bodies. As students delve deeper, they encounter Kepler's Laws, which offer profound insights into the motion of planets around the Sun, and delve into gravitational potential, energy, and fields, paving the way for a comprehensive understanding of gravitational phenomena. Thus, in the vast expanse of the universe, gravitation stands as a cornerstone of physical understanding, beckoning curious minds to unravel its mysteries and explore the cosmos. Gravitation is the force of attraction that exists between any two masses.
For more information, visit-www.vavaclasses.com
I am Kennedy G. I am a Single Variable Calculus Assignment Solver at mathhomeworksolver.com. I hold a Master's in Mathematics, from Indiana, USA. I have been helping students with their assignments for the past 15 years. I solved assignments related to Single Variable Calculus.
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BIO: Sostenitrice del software libero e dei formati standard e aperti. È stata un membro attivo dei progetti Fedora e openSUSE e ha co-fondato l'Associazione LibreItalia dove è stata coinvolta in diversi eventi, migrazioni e formazione relativi a LibreOffice. In precedenza ha lavorato a migrazioni e corsi di formazione su LibreOffice per diverse amministrazioni pubbliche e privati. Da gennaio 2020 lavora in SUSE come Software Release Engineer per Uyuni e SUSE Manager e quando non segue la sua passione per i computer e per Geeko coltiva la sua curiosità per l'astronomia (da cui deriva il suo nickname deneb_alpha).
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My slides at Nordic Testing Days 6.6.2024
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Climate Impact of Software Testing at Nordic Testing Days
Gold1
1. Solutions to Problems in Goldstein,
Classical Mechanics, Second Edition
Homer Reid
August 22, 2000
Chapter 1
Problem 1.1
A nucleus, originally at rest, decays radioactively by emitting an electron of mo-
mentum 1.73 MeV/c, and at right angles to the direction of the electron a neutrino
with momentum 1.00 MeV/c. ( The MeV (million electron volt) is a unit of energy,
used in modern physics, equal to 1.60 x 10−6
erg. Correspondingly, MeV/c is a
unit of linear momentum equal to 5.34 x 10−17
gm-cm/sec.) In what direction does
the nucleus recoil? What is its momentum in MeV/c? If the mass of the residual
nucleus is 3.90 x 10−22
gm, what is its kinetic energy, in electron volts?
Place the nucleus at the origin, and suppose the electron is emitted in the
positive y direction, and the neutrino in the positive x direction. Then the
resultant of the electron and neutrino momenta has magnitude
|pe+ν| = (1.73)2 + 12 = 2 MeV/c,
and its direction makes an angle
θ = tan−1 1.73
1
= 60◦
with the x axis. The nucleus must acquire a momentum of equal magnitude
and directed in the opposite direction. The kinetic energy of the nucleus is
T =
p2
2m
=
4 MeV2
c−2
2 · 3.9 · 10−22 gm
·
1.78 · 10−27
gm
1 MeV c−2
= 9.1 ev
This is much smaller than the nucleus rest energy of several hundred GeV, so
the non-relativistic approximation is justified.
1
2. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 2
Problem 1.2
The escape velocity of a particle on the earth is the minimum velocity required
at the surface of the earth in order that the particle can escape from the earth’s
gravitational field. Neglecting the resistance of the atmosphere, the system is con-
servative. From the conservation theorem for potential plus kinetic energy show
that the escape velocity for the earth, ignoring the presence of the moon, is 6.95
mi/sec.
If the particle starts at the earth’s surface with the escape velocity, it will
just manage to break free of the earth’s field and have nothing left. Thus after
it has escaped the earth’s field it will have no kinetic energy left, and also no
potential energy since it’s out of the earth’s field, so its total energy will be zero.
Since the particle’s total energy must be constant, it must also have zero total
energy at the surface of the earth. This means that the kinetic energy it has at
the surface of the earth must exactly cancel the gravitational potential energy
it has there:
1
2
mv2
e − G
mMR
RR
= 0
so
v =
2GMR
RR
=
2 · (6.67 · 1011
m3
kg−3
s−2
) · (5.98 · 1024
kg)
6.38 · 106 m
1/2
= 11.2 km/s ·
1 m
1.61 km
= 6.95 mi/s.
3. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 3
Problem 1.3
Rockets are propelled by the momentum reaction of the exhaust gases expelled from
the tail. Since these gases arise from the reaction of the fuels carried in the rocket
the mass of the rocket is not constant, but decreases as the fuel is expended. Show
that the equation of motion for a rocket projected vertically upward in a uniform
gravitational field, neglecting atmospheric resistance, is
m
dv
dt
= −v
dm
dt
− mg,
where m is the mass of the rocket and v is the velocity of the escaping gases relative
to the rocket. Integrate this equation to obtain v as a function of m, assuming a
constant time rate of loss of mass. Show, for a rocket starting initially from rest,
with v equal to 6800 ft/sec and a mass loss per second equal to 1/60th of the initial
mass, that in order to reach the escape velocity the ratio of the weight of the fuel
to the weight of the empty rocket must be almost 300!
Suppose that, at time t, the rocket has mass m(t) and velocity v(t). The
total external force on the rocket is then F = gm(t), with g = 32.1 ft/s2
, pointed
downwards, so that the total change in momentum between t and t + dt is
Fdt = −gm(t)dt. (1)
At time t, the rocket has momentum
p(t) = m(t)v(t). (2)
On the other hand, during the time interval dt the rocket releases a mass
∆m of gas at a velocity v with respect to the rocket. In so doing, the rocket’s
velocity increases by an amount dv. The total momentum at time t + dt is the
sum of the momenta of the rocket and gas:
p(t + dt) = pr + pg = [m(t) − ∆m][v(t) + dv] + ∆m[v(t) + v ] (3)
Subtracting (2) from (3) and equating the difference with (1), we have (to
first order in differential quantities)
−gm(t)dt = m(t)dv + v ∆m
or
dv
dt
= −g −
v
m(t)
∆m
dt
4. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 4
which we may write as
dv
dt
= −g −
v
m(t)
γ (4)
where
γ =
∆m
dt
=
1
60
m0s−1
.
This is a differential equation for the function v(t) giving the velocity of the
rocket as a function of time. We would now like to recast this as a differential
equation for the function v(m) giving the rocket’s velocity as a function of its
mass. To do this, we first observe that since the rocket is releasing the mass
∆m every dt seconds, the time derivative of the rocket’s mass is
dm
dt
= −
∆m
dt
= −γ.
We then have
dv
dt
=
dv
dm
dm
dt
= −γ
dv
dm
.
Substituting into (4), we obtain
−γ
dv
dm
= −g −
v
m
γ
or
dv =
g
γ
dm + v
dm
m
.
Integrating, with the condition that v(m0) = 0,
v(m) =
g
γ
(m − m0) + v ln
m
m0
.
Now, γ=(1/60)m0 s−1
, while v =-6800 ft/s. Then
v(m) = 1930 ft/s ·
m
m0
− 1 + 6800 ft/s · ln
m0
m
For m0 m we can neglect the first term in the parentheses of the first term,
giving
v(m) = −1930 ft/s + 6800 ft/s · ln
m0
m
.
The escape velocity is v = 6.95 mi/s = 36.7 · 103
ft/s. Plugging this into the
equation above and working backwards, we find that escape velocity is achieved
when m0/m=293.
Thanks to Brian Hart for pointing out an inconsistency in my original choice
of notation for this problem.
5. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 5
Problem 1.4
Show that for a single particle with constant mass the equation of motion implies
the following differential equation for the kinetic energy:
dT
dt
= F · v,
while if the mass varies with time the corresponding equation is
d(mT )
dt
= F · p.
We have
F = ˙p (5)
If m is constant,
F = m ˙v
Dotting v into both sides,
F · v = mv · ˙v =
1
2
m
d
dt
|v|2
=
dT
dt
(6)
On the other hand, if m is not constant, instead of v we dot p into (5):
F · p = p · ˙p
= mv ·
d(mv)
dt
= mv · v
dm
dt
+ m
dv
dt
=
1
2
v2 d
dt
m2
+
1
2
m2 d
dt
(v2
)
=
1
2
d
dt
(m2
v2
) =
d(mT )
dt
.
6. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 6
Problem 1.5
Prove that the magnitude R of the position vector for the center of mass from an
arbitrary origin is given by the equation
M2
R2
= M
i
mir2
i −
1
2 ij
mimjr2
ij.
We have
Rx =
1
M i
mixi
so
R2
x =
1
M2
i
m2
i x2
i +
i=j
mimjxixj
and similarly
R2
y =
1
M2
i
m2
i y2
i +
i=j
mimjyiyj
R2
z =
1
M2
i
m2
i z2
i +
i=j
mimjzizj
.
Adding,
R2
=
1
M2
i
m2
i r2
i +
i=j
mimj(ri · rj)
. (7)
On the other hand,
r2
ij = r2
i + r2
j − 2ri · rj
and, in particular, r2
ii = 0, so
i,j
mimjr2
ij =
i=j
[mimjr2
i + mimjr2
j − 2mimj(ri · rj)]
= 2
i=j
mimjr2
i − 2
i=j
mimj(ri · rj). (8)
Next,
M
i
mir2
i =
j
mj
i
mir2
i =
i
m2
i r2
i +
i=j
mimjr2
i . (9)
7. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 7
θ
r, φ
r , φ
(x, y)
Figure 1: My conception of the situation of Problem 1.8
Subtracting half of (8) from (9), we have
M mir2
i −
1
2
ijmimjr2
ij =
i
m2
i r2
i +
i=j
mimj(ri · rj)
and comparing this with (7) we see that we are done.
Problem 1.8
Two wheels of radius a are mounted on the ends of a common axle of length b such
that the wheels rotate independently. The whole combination rolls without slipping
on a plane. Show that there are two nonholonomic equations of constraint,
cos θ dx + sin θ dy = 0
sin θ dx − cos θ dy = a(dφ + dφ )
(where θ, φ, and φ have meanings similar to the problem of a single vertical disc,
and (x, y) are the coordinates of a point on the axle midway between the two wheels)
and one holonomic equation of constraint,
θ = C −
a
b
(φ − φ )
where C is a constant.
My conception of the situation is illustrated in Figure 1. θ is the angle
between the x axis and the axis of the two wheels. φ and φ are the rotation
angles of the two wheels, and r and r are the locations of their centers. The
center of the wheel axis is the point just between r and r :
(x, y) =
1
2
(rx + rx, ry + ry).
8. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 8
If the φ wheel rotates through an angle dφ, the vector displacement of its center
will have magnitude adφ and direction determined by θ. For example, if θ = 0
then the wheel axis is parallel to the x axis, in which case rolling the φ wheel
clockwise will cause it to move in the negative y direction. In general, referring
to the Figure, we have
dr = a dφ[sin θˆi − cos θˆj] (10)
dr = a dφ [sin θˆi − cos θˆj] (11)
Adding these componentwise we have1
dx =
a
2
[dφ + dφ ] sin θ
dy = −
a
2
[dφ + dφ ] cos θ
Multiplying these by sin θ or − cos θ and adding or subtracting, we obtain
sin θ dx − cos θ dy = a[dφ + dφ ]
cos θ dx + sin θ dy = 0.
Next, consider the vector r12 = r − r connecting the centers of the two wheels.
The definition of θ is such that its tangent must just be the ratio of the y and
x components of this vector:
tan θ =
y12
x12
→ sec2
θ dθ = −
y12
x2
12
dx12 +
1
x12
dy12.
Subtracting (11) from (10),
sec2
θdθ = a[dφ − dφ ] −
y12
x2
12
sin θ −
1
x12
cos θ
Again substituting for y12/x12 in the first term in parentheses,
sec2
θdθ = −a[dφ − dφ ]
1
x12
(tan θ sin θ + cos θ)
or
dθ = −a[dφ − dφ ]
1
x12
(sin2
θ cos θ + cos3
θ)
= −a[dφ − dφ ]
1
x12
cos θ. (12)
1Thanks to Javier Garcia for pointing out a factor-of-two error in the original version of
these equations.
9. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 9
However, considering the definition of θ, we clearly have
cos θ =
x12
(x2
12 + y2
12)1/2
=
x12
b
because the magnitude of the distance between r1 and r2 is constrained to be b
by the rigid axis. Then (12) becomes
dθ = −
a
b
[dφ − dφ ]
with immediate solution
θ = C −
a
b
[φ − φ ].
with C a constant of integration.
Problem 1.9
A particle moves in the x − y plane under the constraint that its velocity vector is
always directed towards a point on the x axis whose abscissa is some given function
of time f(t). Show that for f(t) differentiable, but otherwise arbitrary, the constraint
is nonholonomic.
The particle’s position is (x(t), y(t)), while the position of the moving point
is (f(t), 0). Then the vector d from the particle to the point has components
dx = x(t) − f(t) dy = y(t). (13)
The particle’s velocity v has components
vx =
dx
dt
vy =
dy
dt
(14)
and for the vectors in (13) and (?? to be in the same direction, we require
vy
vx
=
dy
dx
or
dy/dt
dx/dt
=
dy
dx
=
y(t)
x(t) − f(t)
so
dy
y
=
dx
x − f(t)
(15)
For example, if f(t) = αt, then we may integrate to find
ln y(t) = ln[x(t) − α(t)] + C
or
y(t) = C · [x(t) − αt]
which is a holonomic constraint. But for general f(t) the right side of (15) is
not integrable, so the constraint is nonholonomic.
10. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 10
θ
φ
l
a
Figure 2: My conception of the situation of Problem 1.10
Problem 1.10
Two points of mass m are joined by a rigid weightless rod of length l, the center of
which is constrained to move on a circle of radius a. Set up the kinetic energy in
generalized coordinates.
My conception of this one is shown in Figure 2. θ is the angle representing
how far around the circle the center of the rod has moved. φ is the angle the
rod makes with the x axis.
The position of the center of the rod is (x, y) = (a cos θ, a sin θ). The
positions of the masses relative to the center of the rod are (xrel, yrel) =
±(1/2)(l cos φ, l sin φ). Then the absolute positions of the masses are
(x, y) = (a cos θ ±
l
2
cos φ, a sin θ ±
l
2
sin φ)
and their velocities are
(vx, vy) = (−a sin θ ˙θ
l
2
sin φ ˙φ, a cos θ ˙θ ±
l
2
cos φ ˙φ).
The magnitudes of these are
|v| = a2 ˙θ2
+
l2
4
˙φ2
± al ˙θ ˙φ(sin θ sin φ + cos θ cos φ)
= a2 ˙θ2
+
l2
4
˙φ2
± al ˙θ ˙φ cos(θ − φ)
When we add the kinetic energies of the two masses, the third term cancels,
and we have
T =
1
2
mv2
= m(a2 ˙θ2
+
l2
4
˙φ2
).
11. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 11
Problem 1.13
A particle moves in a plane under the influence of a force, acting toward a center
of force, whose magnitude is
F =
1
r2
1 −
˙r2
− 2¨rr
c2
,
where r is the distance of the particle to the center of force. Find the generalized
potential that will result in such a force, and from that the Lagrangian for the
motion in a plane. (The expression for F represents the force between two charges
in Weber’s electrodynamics).
If we take
U(r) =
1
r
1 +
v2
c2
=
1
r
+
( ˙r)2
c2r
then
∂U
∂r
= −
1
r2
−
˙r2
c2r2
and
d
dt
∂U
∂ ˙r
=
d
dt
2 ˙r
c2r
=
2¨r
c2r
−
2( ˙r)2
c2r2
so
Qr = −
∂U
∂r
+
d
dt
∂U
∂ ˙r
=
1
r2
1 +
2r¨r − ( ˙r)2
c2
The Lagrangian for motion in a plane is
L = T − V =
1
2
m ˙r2
+
1
2
m ˙r2 ˙θ2
−
1
r2
1 +
2r¨r − ( ˙r)2
c2
.
Problem 1.14
If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange’s
equations, show by direct substitution that
L = L +
dF(q1, . . . , qn, t)
dt
also satisfies Lagrange’s equations, where F is any arbitrary, but differentiable,
function of its arguments.
We have
∂L
∂qi
=
∂L
∂qi
+
∂
∂qi
dF
dt
(16)
12. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 12
and
∂L
∂ ˙qi
=
∂L
∂ ˙qi
+
∂
∂ ˙qi
dF
dt
. (17)
For the function F we may write
dF
dt
=
i
∂F
∂qi
˙qi +
∂F
∂t
and from this we may read off
∂
∂ ˙qi
dF
dt
=
∂F
∂qi
.
Then taking the time derivative of (17) gives
d
dt
∂L
∂ ˙qi
=
d
dt
∂L
∂ ˙qi
+
d
dt
∂F
∂qi
so we have
∂L
∂qi
−
d
dt
∂L
∂ ˙qi
=
∂L
∂qi
−
d
dt
∂L
∂ ˙qi
+
∂
∂qi
dF
dt
−
d
dt
∂F
∂qi
.
The first two terms on the RHS cancel because L satisfies the Euler-Lagrange
equations, while the second two terms cancel because F is differentiable. Hence
L satisfies the Euler-Lagrange equations.
Problem 1.16
A Lagrangian for a particular physical system can be written as
L =
m
2
(a ˙x2
+ 2b ˙x ˙y + c ˙y2
) −
K
2
(ax2
+ 2bxy + cy2
),
where a, b, and c are arbitrary constants but subject to the condition that b2
−ac =
0. What are the equations of motion? Examine particularly the two cases a = 0 = c
and b = 0, c = −a. What is the physical system described by the above Lagrangian?
Show that the usual Lagrangian for this system as defined by Eq. (1-56) is related
to L by a point transformation (cf. Exercise 15 above). What is the significance of
the condition on the value of b2
− ac?
Clearly we have
∂L
∂x
= −Kax − Kby
∂L
∂ ˙x
= ma ˙x + mb ˙y
so the Euler-Lagrange equation for x is
∂L
∂x
=
d
dt
∂L
∂ ˙x
→ m(a¨x + b¨y) = −K(ax + by).
13. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 13
Similarly, for y we obtain
m(b¨y + c¨y) = −K(bx + cy).
These are the equations of motion for a particle of mass m undergoing simple
harmonic motion in two dimensions, as if bound by two springs of spring con-
stant K. Normally we would express the Lagrangian in unravelled form, by
transforming to new coordinates u1 and u2 with
u1 = ax + by u2 = bx + cy.
The condition b2
− ac = 0 is the condition that the coordinate transformation
not be degenerate, i.e. that there are actually two distinct dimensions in which
the particle experiences a restoring force. If b2
= ac then we have just a one-
dimensional problem.
Problem 1.17
Obtain the Lagrangian equations of motion for a spherical pendulum, i.e. a mass
point suspended by a rigid weightless rod.
Let m and L be the mass of the particle and the length of the rod. Since the
particle is constrained to move on the surface of a sphere of radius L, we may
parameterize its position by the angles θ and ϕ, in terms of which the particle’s
position and velocity are
x = L (sin θ cos ϕi + sin θ sin ϕj + cos θk)
v = L (cos θ cos ϕ ˙θ − sin θ sin ϕ ˙ϕ)i + (cos θ sin ϕ ˙θ + sin θ cos ϕ ˙ϕ)j − (sin θ ˙θk) .
so the kinetic energy is
T =
1
2
mv2
=
mL2
2
˙θ2
+
mL2
2
sin2
θ ˙ϕ2
.
On the other hand, the gravitational potential energy depends only on θ :
V = −mgL cosθ
where we take the potential at the height of the fulcrum as the zero of potential.
Then the Lagrangian is
L = T − V =
1
2
mL2 ˙θ2
+
1
2
mL2
sin2
θ ˙ϕ2
+ mgL cosθ
and the equations of motion are
¨ϕ = 0
¨θ = −
g
L
− ˙ϕ2
cos θ sin θ.
14. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 14
If there is no motion in the azimuthal direction, ˙ϕ = 0 and the system is the
elementary one-dimensional mechanical pendulum with frequency ω0 = g/L.
But any finite velocity of azimuthal spinning gives rise to an additional effect
which we may think of as mitigating the downward force of gravity, yielding an
effective gravitational acceleration
g = g − L ˙ϕ2
cos θ.
The mitigating effect is largest near the trough of the pendulum, vanishes as
the particle passes through the vertical height of the fulcrum, and becomes
an enhancing effect in the upper hemisphere. For small oscillations about the
trough, cos θ ≈ 1 and the pendulum frequency is reduced to ω = ω2
0 − ˙ϕ2.
Thanks to Tomasz Szymanski for pointing out an error in an earlier version
of this solution.
Problem 1.18
A particle of mass m moves in one dimension such that it has the Lagrangian
L =
m2
˙x4
12
+ m ˙x2
V (x) − V 2
(x),
where V is some differentiable function of x. Find the equation of motion for x(t)
and describe the physical nature of the system on the basis of this equation.
We have
∂L
∂x
= m ˙x2 dV
dx
− 2V (x)
dV
dx
∂L
∂ ˙x
=
m2
˙x3
3
+ 2m ˙xV (x)
d
dt
∂L
∂ ˙x
= m2
( ˙x)2
¨x + 2m¨xV (x) + 2m ˙x
d
dt
V (x)
In the last equation we can use
d
dt
V (x) = ˙x
dV
dx
.
Then the Euler-Lagrange equation is
d
dt
∂L
∂ ˙x
−
∂L
∂x
= 0 → m2
( ˙x)2
¨x + 2m¨xV (x) + m ˙x2 dV
dx
+ 2V (x)
dV
dx
or
m¨x +
dV
dx
m ˙x2
+ 2V (x) = 0.
15. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 15
If we identify F = −dV/dx and T = m ˙x2
/2, we may write this as
(F − m¨x)(T + V ) = 0
So, this is saying that, at all times, either the difference between F and ma is
zero, or the sum of kinetic and potential energy is zero.
Problem 1.19
Two mass points of mass m1 and m2 are connected by a string passing through
a hole in a smooth table so that m1 rests on the table and m2 hangs suspended.
Assuming m2 moves only in a vertical line, what are the generalized coordinates
for the system? Write down the Lagrange equations for the system and, if possible,
discuss the physical significance any of them might have. Reduce the problem to a
single second-order differential equation and obtain a first integral of the equation.
What is its physical significance? (Consider the motion only so long as neither m1
nor m2 passes through the hole).
Let d be the height of m2 above its lowest possible position, so that d =
0 when the string is fully extended beneath the table and m1 is just about
to fall through the hole. Also, let θ be the angular coordinate of m1 on the
table. Then the kinetic energy of m2 is just m2
˙d2
/2, while the kinetic energy
of m1 is m1
˙d2
/2 + m1d2 ˙θ2
/2, and the potential energy of the system is just the
gravitational potential energy of m2, U = m2gd. Then the Lagrangian is
L =
1
2
(m1 + m2) ˙d2
+
1
2
m1d2 ˙θ2
− m2gd
and the Euler-Lagrange equations are
d
dt
(m1d2 ˙θ) = 0
(m1 + m2) ¨d = −m2g + m1d ˙θ2
From the first equation we can identify a first integral, m1d2 ˙θ = l where l is a
constant. With this we can substitute for ˙θ in the second equation:
(m1 + m2) ¨d = −m2g +
l2
m1d3
Because the sign of the two terms on the RHS is different, this is saying that, if
l is big enough (if m1 is spinning fast enough), the centrifugal force of m1 can
balance the downward pull of m2, and the system can be in equilibrium.
16. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 16
Problem 1.20
Obtain the Lagrangian and equations of motion for the double pendulum illustrated
in Fig. 1-4, where the lengths of the pendula are l1 and l2 with corresponding masses
m1 and m2.
Taking the origin at the fulcrum of the first pendulum, we can write down
the coordinates of the first mass point:
x1 = l1 sin θ1
y1 = −l1 cos θ1
The coordinates of the second mass point are defined relative to the coordi-
nates of the first mass point by exactly analogous expressions, so relative to the
coordinate origin we have
x2 = x1 + l2 sin θ2
y2 = y1 − l2 cos θ2
Differentiating and doing a little algebra we find
˙x2
1 + ˙y2
1 = l2
1
˙θ2
1
˙x2
2 + ˙y2
2 = l2
1
˙θ2
1 + l2
2
˙θ2
2 − 2l1l2
˙θ1
˙θ2 cos(θ1 − θ2)
The Lagrangian is
L =
1
2
(m1+m2)l2
1
˙θ2
1+
1
2
m2l2
2
˙θ2
2−m2l1l2
˙θ1
˙θ2 cos(θ1−θ2)+(m1+m2)gl2 cos θ1+m2gl2 cos θ2
with equations of motion
d
dt
(m1 + m2)l2
1
˙θ1 − m2l1l2
˙θ2 cos(θ1 − θ2) = −(m1 + m2)gl2 sin θ1
and
d
dt
l2
˙θ2 − l1
˙θ1 cos(θ1 − θ2) = −g sin θ2.
If ˙θ1 = 0, so that the fulcrum for the second pendulum is stationary, then the
second of these equations reduces to the equation we derived in Problem 1.17.
17. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 17
Problem 1.21
The electromagnetic field is invariant under a gauge transformation of the scalar
and vector potential given by
A → A + Ψ(r, t),
Φ → Φ −
1
c
∂Ψ
∂t
,
where Ψ is arbitrary (but differentiable). What effect does this gauge transformation
have on the Lagrangian of a particle moving in the electromagnetic field? Is the
motion affected?
The Lagrangian for a particle in an electromagnetic field is
L = T − qΦ(x(t)) +
q
c
A(x(t)) · v(t)
If we make the suggested gauge transformation, this becomes
→ T − q Φ(x(t)) −
1
c
∂Ψ
∂t x=x(t)
+
q
c
[A(x(t)) · v(t) + v · Ψ(x(t))]
= T − qΦ(x(t)) +
q
c
A(x(t)) · v(t) +
q
c
∂Ψ
∂t
+ v · Ψ(x(t))
= T − qΦ(x(t)) +
q
c
A(x(t)) · v(t) +
q
c
d
dt
Ψ(x(t))
= L +
q
c
d
dt
Ψ(x(t)).
So the transformed Lagrangian equals the original Lagrangian plus a total time
derivative. But we proved in Problem 1.15 that adding the total time derivative
of any function to the Lagrangian does not affect the equations of motion, so
the motion of the particle is unaffected by the gauge transformation.
Problem 1.22
Obtain the equation of motion for a particle falling vertically under the influence
of gravity when frictional forces obtainable from a dissipation function 1
2 kv2
are
present. Integrate the equation to obtain the velocity as a function of time and
show that the maximum possible velocity for fall from rest is v = mg/k.
The Lagrangian for the particle is
L =
1
2
m ˙z2
− mgz
18. Homer Reid’s Solutions to Goldstein Problems: Chapter 1 18
and the dissipation function is k ˙z2
/2, so the equation of motion is
d
dt
∂L
∂ ˙z
−
∂L
∂z
+
∂F
∂ ˙z
→ m¨z = mg − k ˙z.
This says that the acceleration goes to zero when mg = k ˙z, or ˙z = mg/k, so
the velocity can never rise above this terminal value (unless the initial value of
the velocity is greater than the terminal velocity, in which case the particle will
slow down to the terminal velocity and then stay there).