Classical Mechanics 1 (PHY6001) - UMM AL-QURA UNIVERSITY PDF

Part 1 Survey of the Elementary Principles By Dr. Fatma El-Sayed Classical Mechanics (PHY6001) Chapter 1 Outlines 1. Mechanics of a particle 2. Mechanics of system of particles 3. Constraints 4. D’Alembert’s princi...

Part 1 Survey of the Elementary Principles By Dr. Fatma El-Sayed Classical Mechanics (PHY6001) Chapter 1 Outlines 1. Mechanics of a particle 2. Mechanics of system of particles 3. Constraints 4. D’Alembert’s principle and Lagrange’s equations 5. Velocity –dependent potentials and the dissipation function. 6. Simple Applications of the Lagrangian formulation. 2 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ Let 𝑟Ԧ be the radius vector of a particle from some given origin and 𝑣Ԧ its velocity vector: 𝑑 𝑟Ԧ 𝑣Ԧ = 1 𝑑𝑡 □ The linear momentum 𝑝Ԧ of the particle is defined as the product of the particle mass and its velocity: 𝑝Ԧ = 𝑚𝑣Ԧ 2 □ In consequence of interactions with external objects and fields, the particle may experience forces of various types, e.g., gravitational or electrodynamic; the vector sum of these force exerted on the particle Ԧ is the total force 𝐹. 3 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ The mechanics of the particle is contained in Newton’s second law of motion, which states that there exist frames of reference in which the motion of the particle is described by the differential equation: 𝑑 𝑝Ԧ Ԧ 𝐹= = 𝑝Ԧሶ (3) 𝑑𝑡 □ Or 𝑑 𝐹Ԧ = 𝑚𝑣Ԧ (4) 𝑑𝑡 □ In most instances, the mass of the particle is constant, and Eq. (4) reduces to 𝑑 𝑣Ԧ Ԧ 𝐹=𝑚 = 𝑚𝑎Ԧ 5 𝑑𝑡 □ Where 𝑎Ԧ is the acceleration vector of the particle defined by 𝑑 𝑣Ԧ 𝑑 𝑑 𝑟Ԧ 𝑑 2 𝑟Ԧ 𝑎Ԧ = = = 2 (6) 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 4 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ The equation of motion is a differential equation of second order, assuming 𝐹Ԧ does not depend on higher-order derivatives □ A reference frame in which Eq. (3) is valid is called an inertial or Galilean system. □ Many of the important conclusions of mechanics can be expressed in the form of conservation theorems, which indicate under what conditions various mechanical quantities are constant in time. □ The Conservation theorem for the Linear momentum of a particle: if the total force 𝐹Ԧ is zero, then 𝑝Ԧሶ = 0 and the linear momentum 𝑝Ԧ is conserved. 𝑑 𝑝Ԧ ∵ 𝐹Ԧ = =0 ⇒ ∴ 𝑝Ԧ = constant 𝑑𝑡 5 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ The angular momentum 𝐿 of the particle about point 𝑂 is 𝐿 = 𝑟Ԧ × 𝑝Ԧ 7 □ where 𝑟Ԧ is the radius vector from 𝑂 to the particle. Notice that the order of the factors is important. We now define the moment of the force or torque about 𝑂 as 𝑁 = 𝑟Ԧ × 𝐹Ԧ 8 □ The equation analogous to Eq. (3) for 𝑁 is obtained by forming the cross product of 𝑟Ԧ with Eq. (4): 𝑑 𝑑 𝑑𝐿 𝑁 = 𝑟Ԧ × 𝐹Ԧ = 𝑟Ԧ × 𝑚𝑣Ԧ = 𝑟Ԧ × 𝑚𝑣Ԧ = = 𝐿ሶ (9) 𝑑𝑡 𝑑𝑡 𝑑𝑡 6 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ Note that both 𝑁 and 𝐿 depend on the point 𝑂, about which the moments are taken. □ The conservation theorem for the angular momentum of a particle: if the total torque 𝑁 is zero, then 𝐿ሶ = 0, and the angular momentum 𝐿 is conserved. 𝑑𝐿 ∵𝑁= =0 ⇒ 𝐿 = constant 𝑑𝑡 7 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ Consider the work done by the external force 𝐹Ԧ upon the particle in going from point 1 to point 2, 2 Ԧ 𝑑 𝑠Ԧ 𝑊12 = න 𝐹. 10 1 □ For constant mass, the integral in Eq. (10) reduces to 2 2 𝑑 𝑣Ԧ 𝑚 2𝑑 2 𝑚 2 Ԧ න 𝐹. 𝑑 𝑠Ԧ = 𝑚 න. 𝑣Ԧ 𝑑𝑡 = න 𝑣 𝑑𝑡 = 𝑣2 − 𝑣12 (11) 1 1 𝑑𝑡 2 1 𝑑𝑡 2 □ The scalar quantity 𝑚𝑣 2 /2 is called the kinetic energy of the particle and is denoted by 𝑇, so the work done is equal to the change in the kinetic energy, 𝑊12 = 𝑇2 − 𝑇1 (12) 8 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ If the force field is such that the work 𝑊12 is the same for any physically possible path between points 1 and 2, then the force (and the system) is said to be conservative. □ An alternative description of a conservative system is obtained by imagining the particle being taken from point 1 to point 2 by one possible path and then being returned to point 1 by another path. The independence of 𝑊12 on the particle path implies that the work done around such a closed circuit is zero, i.e. Ԧ 𝑑 𝑠Ԧ = 0 ර 𝐹. 13 □ Physically, a system cannot be conservative if friction or other dissipation forces are present, because 𝐹.Ԧ 𝑑 𝑠Ԧ due to friction is always positive and the integral cannot vanish. 9 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a Particle □ A necessary condition that the work 𝑊12 be independent of the physical path taken by the particle is that 𝐹Ԧ be the gradient of some scalar function of position, 𝑉(𝑟): 𝐹Ԧ = −∇ 𝑉 𝑟 14 □ Where 𝑉 is called the potential, or potential energy. For conservative system, the work done by the force is 𝑊12 = 𝑉1 − 𝑉2 (15) □ Combining Eq. (15) with Eq. (12), we have the result 𝑊12 = 𝑇2 − 𝑇1 = 𝑉1 − 𝑉2 ⟹ ∴ 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 (16) □ The Energy conservation theorem for a particle: if the forces acting on a particle are conservative, then the total energy of the particle is conserved. 10 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ When dealing the systems of many particles, we must distinguish between the external forces acting on the particles due to sources outside the system and internal forces on some particle 𝑖 due to all other particles in the system. □ The equation of motion (Newton’s second law) for the 𝑖 -th particle is 𝑒 ෍ 𝐹Ԧ𝑗𝑖 + 𝐹Ԧ𝑖 = 𝑝Ԧሶ𝑖 (17) 𝑗 𝑒 □ where 𝐹Ԧ𝑖 is the external force and 𝐹Ԧ𝑗𝑖 is the internal force on the i-th particle due to the j-th particle (𝐹Ԧ𝑖𝑖 = 0). 11 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles 𝑒 □ We shall assume that the 𝐹Ԧ𝑖𝑗 (like the 𝐹Ԧ𝑖 ) obey Newton’s third law of motion in its original form: that the forces of two particles exert on each other are equal and opposite. □ By applying the summation over all particles, Eq. (17) becomes 𝑑2 𝑒 𝑒 ෍ 𝑚 𝑟 Ԧ 𝑖 𝑖 = ෍ Ԧ 𝐹𝑗𝑖 + Ԧ 𝐹 = ෍ Ԧ 𝐹 (18) 𝑑𝑡 2 𝑖 𝑖 𝑖 𝑖≠𝑗 𝑖 □ The second term on the right side is the total external force, while the first term vanishes, since the law of action and reaction states that each pair 𝐹Ԧ𝑖𝑗 + 𝐹Ԧ𝑗𝑖 is zero. 12 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ To reduce the left-hand side, we define a vector 𝑅 as the average of the radii vectors of the particles, weighted in proportion to their mass: σ 𝑚𝑖 𝑟Ԧ𝑖 σ 𝑚𝑖 𝑟Ԧ𝑖 𝑅= = 19 σ 𝑚𝑖 𝑀 □ The vector 𝑅 defines a point known as the center of mass, or as the center of gravity, of the system. 13 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ With this definition, Eq. (18) reduces to 𝑑2 𝑑2𝑅 ෍ 𝑚 𝑖 𝑟 Ԧ𝑖 = 𝑀 = ෍ Ԧ 𝑒 = 𝐹Ԧ 𝐹 𝑒 (20) 𝑑𝑡 2 𝑑𝑡 2 𝑖 𝑖 𝑖 □ which states that the center of mass moves as if the total external force were acting on the entire mass of the system concentrated at the center of mass. □ From the definition of the center of mass, Eq. (19), the total linear momentum of the system is the total mass of the system times the velocity of the center of mass. 𝑑𝑟Ԧ𝑖 𝑑 𝑑 𝑑𝑅 𝑃 = ෍ 𝑚𝑖 =෍ 𝑚𝑖 𝑟Ԧ𝑖 = ෍ 𝑚𝑖 𝑟Ԧ𝑖 = 𝑀 (21) 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑖 𝑖 𝑖 14 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ The conservation theorem for the linear momentum of a system of particles: if the total external force is zero, the total linear momentum is conserved. □ We obtain the total angular momentum of the system by forming the cross product 𝑟Ԧ𝑖 × 𝑝Ԧ𝑖 and summing over 𝑖, as follows: 𝑑 𝑒 ෍ 𝑟Ԧ𝑖 × 𝑝Ԧሶ𝑖 = ෍ 𝑟Ԧ𝑖 × 𝑝Ԧ𝑖 = 𝐿ሶ = ෍ 𝑟Ԧ𝑖 × 𝐹Ԧ𝑖 + ෍ 𝑟Ԧ𝑖 × 𝐹Ԧ𝑗𝑖 22 𝑑𝑡 𝑖 𝑖 𝑖 𝑖≠𝑗 □ The last term on the right in Eq. (22) can be considered as a sum of the pairs of the form ∵ 𝑟Ԧ𝑖 × 𝐹Ԧ𝑗𝑖 + 𝑟Ԧ𝑗 × 𝐹Ԧ𝑖𝑗 = 𝑟Ԧ𝑖 × 𝐹Ԧ𝑗𝑖 − 𝑟Ԧ𝑗 × 𝐹Ԧ𝑗𝑖 = 𝑟Ԧ𝑖 − 𝑟Ԧ𝑗 × 𝐹Ԧ𝑗𝑖 ∴ 𝑟Ԧ𝑖 × 𝐹Ԧ𝑗𝑖 + 𝑟Ԧ𝑗 × 𝐹Ԧ𝑖𝑗 = 𝑟Ԧ𝑖𝑗 × 𝐹Ԧ𝑗𝑖 (23) 15 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ If the internal forces between two particles, in addition to being equal and opposite, also lie along the line joining the particles, a condition known as the strong law of action and reaction, then all of these cross products vanish. The sum over pairs is zero under this assumption, and Eq. (22) becomes 𝑑𝐿 𝑒 =𝑁 (24) 𝑑𝑡 □ The time derivative of the total angular momentum is equal to the moment of the external force about the given point. 16 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ The conservation theorem for the total angular momentum: 𝐿 is constant in time if the applied (external) torque is zero. 𝑑𝐿 ∵𝑁= =0 ⇒ ∴ 𝐿 = constant 𝑑𝑡 □ According to Eq. (21), the total linear momentum of the system is the same as if the entire mass were concentrated at the center of mass and moving with it. □ The analogous theorem for angular momentum is more complicated. With the origin 𝑂 as reference point, the total angular momentum of the system is 𝐿 = ෍ 𝑟Ԧ𝑖 × 𝑝Ԧ𝑖 𝑖 17 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ Let 𝑅 be the radius vector from 𝑂 to the center of mass and let 𝑟Ԧ𝑖′ be the radius vector from the center of mass to the i-th particle. Then 𝑟Ԧ𝑖 = 𝑟Ԧ𝑖′ + 𝑅 □ and 𝑣Ԧ𝑖 = 𝑣Ԧ𝑖′ + 𝑣Ԧ □ where 𝑣Ԧ𝑖′ is the velocity of the 𝑖-th particle relative to the center of mass of the system, and 𝑣Ԧ is the velocity of the center of mass relative to 𝑂. 18 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ The total angular momentum becomes 𝐿 = ෍ 𝑟Ԧ𝑖 × 𝑝Ԧ𝑖 = ෍ 𝑟Ԧ𝑖′ + 𝑅 × 𝑚𝑖 ( 𝑣Ԧ𝑖′ + 𝑣) Ԧ 𝑖 𝑖 𝑑 = ෍ 𝑟Ԧ𝑖′ × 𝑚𝑖 𝑣Ԧ𝑖′ + ෍ 𝑅 × 𝑚𝑖 𝑣Ԧ + ෍ 𝑚𝑖 𝑟Ԧ𝑖′ × 𝑣Ԧ + 𝑅 × ෍ 𝑚𝑖 𝑟Ԧ𝑖′ 𝑑𝑡 𝑖 𝑖 𝑖 𝑖 □ From the definition of the center of mass, ∵ ෍ 𝑚𝑖 𝑟Ԧ𝑖′ = ෍ 𝑚𝑖 𝑟Ԧ𝑖 − 𝑅 = ෍ 𝑚𝑖 𝑟Ԧ𝑖 − ෍ 𝑚𝑖 𝑅 = 0 𝑖 𝑖 𝑖 𝑖 □ So, the last two terms vanish and the total angular momentum about 𝑂 is ∴ 𝐿 = ෍ 𝑟Ԧ𝑖′ × 𝑚𝑖 𝑣Ԧ𝑖′ + ෍ 𝑅 × 𝑚𝑖 𝑣Ԧ = 𝑅 × 𝑀𝑣Ԧ + ෍ 𝑟Ԧ𝑖′ × 𝑝Ԧ𝑖′ 𝑖 𝑖 𝑖 19 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles 𝐿 = 𝑅 × 𝑀𝑣Ԧ + ෍ 𝑟Ԧ𝑖′ × 𝑝Ԧ𝑖′ (25) 𝑖 □ Eq. (25) says that the total angular momentum about a point 𝑂 is the angular momentum of motion concentrated at the center of mass, plus the angular momentum of motion about the center of mass. □ Finally, let us consider the energy equation, as in the case of a single particle, we calculate the work done by all forces in moving the system from an initial configuration 1 to a final configuration 2: 20 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles 2 2 2 𝑒 𝑊12 = ෍ න 𝐹Ԧ𝑖. 𝑑 𝑠Ԧ𝑖 = ෍ න 𝐹Ԧ𝑖. 𝑑 𝑠Ԧ𝑖 + ෍ න 𝐹Ԧ𝑗𝑖. 𝑑𝑠Ԧ𝑖 26 𝑖 1 𝑖 1 𝑖≠𝑗 1 □ The equations of motion can be used to reduce the integrals to 2 2 2 1 ෍ න 𝐹Ԧ𝑖. 𝑑𝑠Ԧ = ෍ න 𝑚𝑖 𝑣Ԧሶ𝑖. 𝑣Ԧ𝑖 𝑑𝑡 = ෍ න 𝑑 𝑚𝑖 𝑣𝑖2 1 1 1 2 𝑖 𝑖 𝑖 □ The work done can still be written as the difference of the final and initial kinetic energies: 𝑊12 = 𝑇2 − 𝑇1 □ where, the total kinetic energy of the system is 1 𝑇 = ෍ 𝑚𝑖 𝑣𝑖2 2 21 𝑖 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ Making use of the transformations to center-of-mass coordinates, we may also write 𝑇 as 1 𝑇 = ෍ 𝑚𝑖 𝑣Ԧ + 𝑣Ԧ𝑖′. 𝑣Ԧ + 𝑣Ԧ𝑖′ 2 𝑖 1 1 1 𝑑 = ෍ 𝑚𝑖 𝑣 + ෍ 𝑚𝑖 𝑣𝑖′2 + 2 2 𝑣. Ԧ ෍ 𝑚𝑖 𝑟Ԧ𝑖′ 2 2 2 𝑑𝑡 𝑖 𝑖 𝑖 1 1 = 𝑀𝑣 + ෍ 𝑚𝑖 𝑣𝑖′2 2 2 2 𝑖 □ The kinetic energy, like the angular momentum, also consists of two parts: the kinetic energy obtained if all the mass were concentrated at the center of mass, plus the kinetic energy of motion about the center of mass. 22 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ Now, consider the right-hand side of Eq. (26). In the special case that the external forces are derivable in terms of the gradient of a potential, the first term can be written as 2 2 2 𝑒 ෍ න 𝐹Ԧ𝑖. 𝑑 𝑠Ԧ𝑖 = − ෍ න ∇𝑖 𝑉𝑖. 𝑑 𝑠Ԧ𝑖 = − ෍ 𝑉𝑖 ቚ 1 1 1 𝑖 𝑖 𝑖 □ Where the subscript 𝑖 on the del operator indicates that the derivatives are with respect to the components of 𝑟Ԧ𝑖. □ If the internal forces are also conservative, then the mutual forces between the 𝑖-th and 𝑗-th particles, 𝐹Ԧ𝑖𝑗 and 𝐹Ԧ𝑗𝑖 , can be obtained from a potential function 𝑉𝑖𝑗. 23 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ To satisfy the strong law of action and reaction, 𝑉𝑖𝑗 can be a function only of the distance between the particles: 𝑉𝑖𝑗 = 𝑉𝑖𝑗 𝑟Ԧ𝑖 − 𝑟Ԧ𝑗 □ The two forces are then automatically equal and opposite, 𝐹Ԧ𝑗𝑖 = −∇𝑖 𝑉𝑖𝑗 = +∇𝑗 𝑉𝑖𝑗 = −𝐹Ԧ𝑖𝑗 □ and lie along the line joining the two particles, ∇ 𝑉𝑖𝑗 𝑟Ԧ𝑖 − 𝑟Ԧ𝑗 = 𝑟Ԧ𝑖 − 𝑟Ԧ𝑗 𝑓 □ where 𝑓 is some scalar function. 24 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ If 𝑉𝑖𝑗 were also a function of the difference of some other pair of vectors associated with the particles, such as their velocities or their intrinsic “spin” angular momenta, then the forces would still be equal and opposite, but would not necessarily lie along the direction between the particles. □ When the forces are all conservative, the second term in Eq. (26) can be rewritten as a sum over pairs of particles, the terms for each pair being of the form 2 2 ෍ න 𝐹Ԧ𝑗𝑖. 𝑑 𝑠Ԧ𝑖 = − ෍ න ∇𝑖 𝑉𝑖𝑗. 𝑑 𝑠Ԧ𝑖 + ∇𝑗 𝑉𝑖𝑗. 𝑑𝑠Ԧ𝑗 𝑖≠𝑗 1 𝑖≠𝑗 1 25 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ If the difference vector 𝑟Ԧ𝑖 − 𝑟Ԧ𝑗 is denoted by 𝑟Ԧ𝑖𝑗 , and if ∇𝑖𝑗 stands for the gradient with respect to 𝑟Ԧ𝑖𝑗 , then ∇𝑖 𝑉𝑖𝑗 = ∇𝑖𝑗 𝑉𝑖𝑗 = −∇𝑗 𝑉𝑖𝑗 and 𝑑 𝑠Ԧ𝑖 − 𝑑 𝑠Ԧ𝑗 = 𝑑 𝑟Ԧ𝑖 − 𝑑 𝑟Ԧ𝑗 = 𝑑 𝑟Ԧ𝑖𝑗 □ So that the term for the 𝑖𝑗 pair has the form 2 2 2 න 𝐹Ԧ𝑗𝑖. 𝑑 𝑠Ԧ𝑖 = − න ∇𝑖 𝑉𝑖𝑗. 𝑑 𝑠Ԧ𝑖 + ∇𝑗 𝑉𝑖𝑗. 𝑑 𝑠Ԧ𝑗 = − න ∇𝑖𝑗 𝑉𝑖𝑗. 𝑑 𝑟Ԧ𝑖𝑗 1 1 1 □ The total work arising from the internal forces then reduces to 2 1 1 2 − ෍ න ∇𝑖𝑗 𝑉𝑖𝑗. 𝑑 𝑟Ԧ𝑖𝑗 = − ෍ 𝑉𝑖𝑗 ቚ 2 1 2 1 𝑖≠𝑗 𝑖≠𝑗 26 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ The factor ½ appears in the last Equation because in summing over both 𝑖 and 𝑗 each member of a given pair is included twice, first in the 𝑖 summation and then in the 𝑗 summation. □ If the external and internal forces are both derivable from potentials, it is possible to define a total potential energy 𝑉 of the system, 1 𝑉 = ෍ 𝑉𝑖 + ෍ 𝑉𝑖𝑗 (27) 2 𝑖 𝑖≠𝑗 □ such that the total energy 𝑇 + 𝑉 is conserved, the analog of the conservation theorem, Eq. 16, for a single particle. 27 Classical Mechanics (PHY6001) Chapter 1 Mechanics of a System of Particles □ The second term on the right side in Eq.(27) will be called the internal potential energy of the system. It need not be zero and more important, it may vary as the system changes with time. Only for the class of systems known as rigid bodies will the internal potential always be constant. □ A rigid body can be defined as a system of particles in which the distances 𝑟𝑖𝑗 are fixed and cannot vary with time. The vectors 𝑑 𝑟Ԧ𝑖𝑗 can only be perpendicular to the corresponding 𝑟Ԧ𝑖𝑗 and therefore to the 𝐹Ԧ𝑖𝑗. □ Therefore, in a rigid body the internal forces do not work, and the internal potential must remain constant. 28 Classical Mechanics (PHY6001) Chapter 1 Constraints □ All problems in mechanics have been reduced to solving the set of differential equations: 𝑒 𝑚𝑖 𝑟Ԧ𝑖ሷ = 𝐹Ԧ𝑖 + ෍ 𝐹Ԧ𝑗𝑖 𝑗 □ It’s necessary to consider the constraints that limit the motion of the system. □ We have already met one type of system involving constraints, namely, rigid bodies, where the constraints on the motions of the particles keep the distances 𝑟𝑖𝑗 unchanged. 29 Classical Mechanics (PHY6001) Chapter 1 Constraints □ Other examples of constrained systems are □ The beads of an abacus are constrained to one-dimensional motion by the supporting wires. □ Gas molecules within a container are constrained by the walls of the vessel to move only inside the container. □ A particle placed on the surface of a solid sphere is subject to the constraint that it can move only on the surface or in the region exterior to the sphere. 30 Classical Mechanics (PHY6001) Chapter 1 Constraints □ Constraints may be classified in various ways: □ (1) If the conditions of constraint can be expressed as equations connecting the coordinates of the particles (and possibly the time) having the form 𝑓 𝑟Ԧ1 , 𝑟Ԧ2 , 𝑟Ԧ3 , … , 𝑡 = 0, the constraints are said to be holonomic. □ The simplest example of holonomic constraints is the rigid body, where the constraints are expressed by 2 2 2 2 𝑟Ԧ𝑖𝑗 = 𝑐𝑖𝑗 ⟹ 𝑟Ԧ𝑖 − 𝑟Ԧ𝑗 − 𝑐𝑖𝑗 =0 □ (2) If the constraints not expressible in this manner are called nonholonomic. For example, the walls of a gas container and the constraint involved in the example of a particle on the surface of a sphere is also nonholonomic, for it can expressed as 𝑟 2 − 𝑎2 ≥ 0. 31 Classical Mechanics (PHY6001) Chapter 1 Constraints □ Constraints are further classified according to whether the equations of constraint contain the time as an explicit variable (rheonomous) or are not explicitly dependent on time (scleronomous). □ A bead sliding on a rigid curved wire fixed in space is obviously subject to a scleronomous constraint; if the wire is moving in some prescribed fashion, the constraint is rheonomous. □ Note that if the wire moves, say, as a reaction to the bead’s motion, then the time dependence of the constraint enters in the equation of the constraint only through the coordinates of the curved wire (which are now part of the system coordinates), the overall constraint is then scleronomous. 32 Classical Mechanics (PHY6001) Chapter 1 Constraints □ Difficulties with constraints: □ (1) The equations of motion are not all independent, because the coordinates are no longer all independent. □ (2) The forces of constraint (e.g., the force that the wire exerts on the bead or the wall on the gas particle) are not known beforehand and must be obtained from solution we seek. □ In the case of holonomic constraints, the first difficulty is solved by the introduction of generalized coordinates. A system of 𝑁 particles, free from constraints, has 3𝑁 independent coordinates or degree of freedom. If there exist holonomic constraints, we want to eliminate the constraints 𝑘 from 3𝑁 coordinates and the independent coordinates are 3𝑁 − 𝑘 and the system is said to have 3𝑁 − 𝑘 degree of freedom. 33 Classical Mechanics (PHY6001) Chapter 1 Constraints □ If the constraint is nonholonomic, the equations expressing the constraint cannot be used to eliminate the dependent coordinates. □ An example of a nonholonomic constraint is that of an object rolling on a rough surface without slipping. The coordinate used to describe the system will generally involve angular coordinates to specify the orientation of the body, plus a set of coordinates describing the location of the point of contact on the surface. The constraint of “rolling” connects these two sets of coordinates; they are not independent. A change in the position of the point of contact means a change in its orientation. 34 Classical Mechanics (PHY6001) Chapter 1 Constraints □ Examples of generalized coordinates: □ (1) Two angles expressing position on the sphere that a particle is constrained to move on. □ (2) Two angles for a double pendulum moving in a plane. □ (3) Amplitudes in a Fourier expansion of 𝑟Ԧ𝑗. □ (4) Quantities with dimensions of energy or angular momentum. 35 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ A virtual (infinitesimal) displacement of a system refers to a change in the configuration of the system as the result of any arbitrary infinitesimal change of the coordinates 𝛿 𝑟Ԧ𝑖 , consistent with the forces and constraints imposed on the system at the given instant 𝑡. □ The displacement is called virtual to distinguish it from an actual displacement of the system occurring in a time interval 𝑑𝑡 , during which the forces and constraints may be changing. □ Suppose the system is in equilibrium; the total force on each particle vanishes, 𝐹Ԧ𝑖 = 0. Then, the dot product 𝐹Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 vanishes, which is the virtual work of the force 𝐹Ԧ𝑖 in the displacement 𝛿 𝑟Ԧ𝑖. The sum of these vanishing products over all particles must be zero. 36 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations ෍ 𝐹Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 𝑖 𝑎 □ Decompose 𝐹Ԧ𝑖 into the applied force, 𝐹Ԧ𝑖 and the force of constraint 𝑓Ԧ𝑖 , 𝑎 𝐹Ԧ𝑖 = 𝐹Ԧ𝑖 + 𝑓Ԧ𝑖 (28) □ So, 𝑎 ෍ 𝐹Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 = ෍ 𝐹Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 + ෍ 𝑓Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 (29) 𝑖 𝑖 𝑖 □ With the restriction to systems for which the net virtual work of the forces of constraint is zero. The condition for equilibrium of a system that the virtual work of the applied forces vanishes is 𝑎 ෍ 𝐹Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 30 𝑖 37 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ Eq. (30) is called the principle of virtual work. Note that, the coefficients of 𝛿 𝑟Ԧ𝑖 can no longer be set equal to zero; i.e., in 𝑎 general, 𝐹Ԧ𝑖 ≠ 0, since the 𝛿 𝑟Ԧ𝑖 are not completely independent but are connected by the constraints. □ To equate the coefficients to zero, we must transform the principle into a form involving the virtual displacements of the 𝑞𝑖 , which are independent. □ Eq. (30) satisfies our needs in that it does not contain the 𝑓Ԧ𝑖 , but it deals only with statics; we want a condition involving the general motion of the system. 38 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ The D’Alembert’s principle is developed by D’Alembert and thought of first by Bernoulli. The equation of motion, 𝑑𝑝Ԧ𝑖 𝐹Ԧ𝑖 = = 𝑝Ԧሶ𝑖 or 𝐹Ԧ𝑖 − 𝑝Ԧሶ𝑖 = 0 𝑑𝑡 □ which states that the particles in the system will be in equilibrium under a force equal to the actual force plus a “reversed effective force” (−𝑝Ԧሶ𝑖 ), and ෍ 𝐹Ԧ𝑖 − 𝑝Ԧሶ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 𝑖 □ and making the same resolution into applied forces and forces of constraint, there results 39 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations 𝑎 ෍ 𝐹Ԧ𝑖 − 𝑝Ԧሶ𝑖. 𝛿 𝑟Ԧ𝑖 + ෍ 𝑓Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 𝑖 𝑖 □ with regarding with the systems for which the virtual work of the forces of constraint vanishes and therefore obtain 𝑎 ෍ 𝐹Ԧ𝑖 − 𝑝Ԧሶ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 𝑖 □ which is often called D’Alembert’s principle. We have achieved our aim, in that the forces of constraint no longer appear, and the superscript (𝑎) can now be dropped without ambiguity. ෍ 𝐹Ԧ𝑖 − 𝑝Ԧሶ𝑖. 𝛿 𝑟Ԧ𝑖 = 0 (31) 𝑖 40 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ Now we want to transform the principle into an expression involving virtual displacements of the generalized coordinates, which are then independent of each other (for holonomic constraints), so that the coefficients of the 𝛿𝑞𝑖 can be set separately equal to zero. 𝑟Ԧ𝑖 = 𝑟Ԧ𝑖 𝑞1 , 𝑞2 , … , 𝑞𝑛 , 𝑡 𝑑𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖 𝑑𝑞𝑘 𝜕𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖 𝑣Ԧ𝑖 = =෍ + =෍ 𝑞ሶ + 𝑑𝑡 𝜕𝑞𝑘 𝑑𝑡 𝜕𝑡 𝜕𝑞𝑘 𝑘 𝜕𝑡 𝑘 𝑘 □ The arbitrary virtual displacement 𝛿 𝑟Ԧ𝑖 is 𝜕𝑟Ԧ𝑖 𝛿 𝑟Ԧ𝑖 = ෍ 𝛿𝑞 𝜕𝑞𝑗 𝑗 41 𝑗 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ The virtual work of the 𝐹Ԧ𝑖 , in terms of the generalized coordinates, becomes 𝜕𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖 ෍ 𝐹Ԧ𝑖. 𝛿 𝑟Ԧ𝑖 = ෍ 𝐹Ԧ𝑖. Ԧ 𝛿𝑞𝑗 = ෍ 𝐹𝑖. 𝛿𝑞𝑗 = ෍ 𝑄𝑗 𝛿𝑞𝑗 𝜕𝑞𝑗 𝜕𝑞𝑗 𝑖 𝑖,𝑗 𝑖,𝑗 𝑗 □ where 𝑄𝑗 are called the components of the generalized force, and 𝜕𝑟Ԧ𝑖 Ԧ 𝑄𝑗 = ෍ 𝐹𝑖. (32) 𝜕𝑞𝑗 𝑖,𝑗 □ where 𝑄𝑗 𝛿𝑞𝑗 must always have the dimensions of work, for example, 𝑄𝑗 might be a torque and 𝑑𝑞𝑗 a differential angle 𝑑𝜃, which makes 𝑁𝑗 𝑑𝜃𝑗 a differential of work. 42 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ The other term involved in Eq. (31), D’Alembert’s principle, which 𝜕𝑟Ԧ𝑖 Ԧ Ԧ Ԧ ෍ 𝑝ሶ𝑖. 𝛿 𝑟Ԧ𝑖 = ෍ 𝑚𝑖 𝑟𝑖ሷ. 𝛿 𝑟Ԧ𝑖 = ෍ 𝑚𝑖 𝑟𝑖ሷ. 𝛿𝑞𝑗 𝜕𝑞𝑗 𝑖 𝑖 𝑖 □ Now, consider the relation 𝜕𝑟Ԧ𝑖 𝑑 𝜕𝑟Ԧ𝑖 𝑑 𝜕𝑟Ԧ𝑖 ෍ 𝑚𝑖 𝑟Ԧ𝑖ሷ. =෍ Ԧ 𝑚𝑖 𝑟𝑖ሶ. Ԧ − 𝑚𝑖 𝑟𝑖ሶ. (33) 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞𝑗 𝑖 𝑖 □ In the last term of Eq. (33), we can interchange the differentiation with respect to 𝑡 and 𝑞𝑗 , for 𝑑 𝜕𝑟Ԧ𝑖 𝜕 𝑑𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖ሶ 𝜕𝑣Ԧ𝑖 𝜕𝑣Ԧ𝑖 𝜕𝑟Ԧ𝑖 /𝜕𝑡 𝜕𝑟Ԧ𝑖 = = = and = = 𝑑𝑡 𝜕𝑞𝑗 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞𝑗 𝜕𝑞𝑗 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 /𝜕𝑡 𝜕𝑞𝑗 43 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ Substitution of these changes in Eq. (33) leads to the result that 𝜕𝑟Ԧ𝑖 𝑑 𝜕𝑟Ԧ𝑖 𝑑 𝜕𝑟Ԧ𝑖 ෍ 𝑚𝑖 𝑟Ԧ𝑖ሷ. =෍ 𝑚𝑖 𝑟Ԧ𝑖ሶ. − 𝑚𝑖 𝑟Ԧ𝑖ሶ. 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞𝑗 𝑖 𝑖 𝑑 𝜕𝑣Ԧ𝑖 𝜕𝑣Ԧ𝑖 =෍ 𝑚𝑖 𝑣Ԧ𝑖. − 𝑚𝑖 𝑣Ԧ𝑖. 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 𝑖 So, 𝑑 𝜕 1 𝜕 1 ෍ ෍ 𝑚𝑖 𝑣𝑖2 − ෍ 𝑚𝑖 𝑣𝑖2 − 𝑄𝑗 𝛿𝑞𝑗 𝑑𝑡 𝜕𝑞ሶ 𝑗 2 𝜕𝑞𝑗 2 𝑗 𝑖 𝑖 𝑑 𝜕𝑇 𝜕𝑇 =෍ − − 𝑄𝑗 𝛿𝑞𝑗 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 □ So, D’Alembert’s principle becomes 𝑑 𝜕𝑇 𝜕𝑇 ෍ − − 𝑄𝑗 𝛿𝑞𝑗 = 0 (34) 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 44 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ The variables 𝑞𝑗 can be any set of coordinates used to describe the motion of the system. If, however, the constraints are holonomic, then it is possible to find sets of independent coordinates 𝑞𝑗 that contain the constraint conditions implicitly in the transformation equations 𝑟Ԧ = 𝑟Ԧ 𝑞1 , 𝑞2 , …. , 𝑞3𝑁−𝑘 , 𝑡 and 𝑟Ԧ𝑁 = 𝑟Ԧ𝑁 (𝑞1 , 𝑞2 , … , 𝑞3𝑁−𝑘 , 𝑡) □ Any virtual displacement 𝛿𝑞𝑗 is then independent of 𝛿𝑞𝑘 , and therefore the only way for Eq. (34) to hold is for the individual coefficients to vanish: 𝑑 𝜕𝑇 𝜕𝑇 − = 𝑄𝑗 (35) 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 □ There are 𝑛 such equations in all. 45 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ When the forces are derivable from a scalar potential function 𝑉, 𝐹Ԧ𝑖 = −∇𝑖 𝑉 □ Then, the generalized forces can be written as 𝜕𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖 𝜕𝑉 𝜕𝑟Ԧ𝑖 𝜕𝑉 𝑄𝑗 = ෍ 𝐹Ԧ𝑖. = − ෍ ∇𝑖 𝑉. = −෍. =− 𝜕𝑞𝑗 𝜕𝑞𝑗 𝜕𝑟Ԧ𝑖 𝜕𝑞𝑗 𝜕𝑞𝑗 𝑖 𝑖 𝑖 □ So, Eq. (35) can be written as 𝑑 𝜕𝑇 𝜕𝑇 𝑑 𝜕𝑇 𝜕𝑇 𝜕𝑉 − = 𝑄𝑗 ⇒ − =− 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 𝜕𝑞𝑗 𝑑 𝜕𝑇 𝜕 ∴ − 𝑇−𝑉 =0 36 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 46 Classical Mechanics (PHY6001) Chapter 1 D’Alembert’s Principle and Lagrange’s Equations □ The potential 𝑉 does not depend on the generalized velocities, so 𝜕𝑇 𝜕(𝑇 − 𝑉) = 𝜕𝑞ሶ 𝑗 𝜕𝑞ሶ 𝑗 □ With the definition of a new function, the Lagrangian 𝐿, as 𝐿 =𝑇−𝑉 □ Eq. (36) become 𝑑 𝜕𝐿 𝜕𝐿 = (37) 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 □ These equations referred to as “Lagrange’s equations”. 47 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ Lagrang’s equations can be put in the form, Eq. (37), even if there is no potential function, 𝑉 , in the usual sense, providing the generalized forces are obtained from a function 𝑈 𝑞𝑗 , 𝑞ሶ 𝑗 by the following equation: 𝜕𝑈 𝑑 𝜕𝑈 𝑄𝑗 = − + 38 𝜕𝑞𝑗 𝑑𝑡 𝜕𝑞ሶ 𝑗 □ The Lagrangian function is given by 𝐿 =𝑇−𝑈 39 □ Here, 𝑈 may be called a “generalized potential”, or “velocity- dependent potential”. It applies to one very important type of force field, namely, the electromagnetic forces on moving charges. 48 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ Example: consider an electric charge, 𝑞, of mass 𝑚 moving at a velocity, 𝑣, Ԧ in an otherwise charge-free region containing both an electric field, 𝐸, and a magnetic field, 𝐵, which may depend upon time and position. The charge experiences a force, called the Lorentz force, given by 𝐹Ԧ = 𝑞 𝐸 + 𝑣Ԧ × 𝐵 (40) □ Both 𝐸 𝑡, 𝑥, 𝑦, 𝑧 and 𝐵 𝑡, 𝑥, 𝑦, 𝑧 are continuous functions of time and position derivable from a scalar potential 𝜙(𝑡, 𝑥, 𝑦, 𝑧) and a Ԧ 𝑥, 𝑦, 𝑧) by vector potential 𝐴(𝑡, 49 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function 𝜕𝐴Ԧ 𝐸 = −∇𝜙 − and 𝐵 = ∇ × 𝐴Ԧ (41) 𝜕𝑡 □ The force on the charge can be derived from the following velocity-dependent potential energy Ԧ 𝑣Ԧ = 𝑞𝜙 − 𝑞𝐴. 𝑈 = 𝑞 𝜙 − 𝐴. Ԧ 𝑣Ԧ (42) □ So, the Lagrangian, 𝐿 = 𝑇 − 𝑈, is 1 1 Ԧ Ԧ 𝑣Ԧ 𝐿 = 𝑚𝑣 − 𝑞𝜙 − 𝑞𝐴. 𝑣Ԧ = 𝑚𝑣 2 − 𝑞𝜙 + 𝑞𝐴. 2 (43) 2 2 □ Considering just the 𝑥-component of Lagrange’s equations gives 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 𝑑𝐴𝑥 𝑚𝑥ሷ = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 −𝑞 + 44 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑑𝑡 50 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ Considering just the 𝑥-component of Lagrange’s equations gives 𝜕𝐿 𝜕𝐿 = = 𝑚𝑣 + 𝑞𝐴𝑥 = 𝑚𝑥ሶ + 𝑞𝐴𝑥 𝜕𝑣 𝜕𝑥ሶ 𝜕𝐿 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 −𝑞 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑑 𝜕𝐿 𝜕𝐿 ∵ = 𝑑𝑡 𝜕𝑥ሶ 𝜕𝑥 𝑑 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 ∴ 𝑚𝑥ሶ + 𝑞𝐴𝑥 = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 −𝑞 𝑑𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑑𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 ∴ 𝑚𝑥ሷ + 𝑞 = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 −𝑞 𝑑𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 □ So, 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 𝑑𝐴𝑥 ∴ 𝑚𝑥ሷ = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 −𝑞 + 44 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑑𝑡 51 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ The total time derivative of 𝐴𝑥 is related to the particle time derivative through 𝑑𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 = + 𝑣. Ԧ ∇𝐴𝑥 = + 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 45 𝑑𝑡 𝜕𝑡 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 □ Equation (41) gives 𝜕𝐴Ԧ ∵ 𝐵 = ∇ × 𝐴Ԧ and 𝐸 = −∇𝜙 − 𝜕𝑡 𝜕𝐴𝑦 𝜕𝐴𝑥 𝜕𝐴𝑧 𝜕𝐴𝑥 ∴ 𝑣Ԧ × 𝐵 = 𝑣𝑦 𝐵𝑧 − 𝑣𝑧 𝐵𝑦 = 𝑣𝑦 − + 𝑣𝑧 − 𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑧 𝜕𝜙 𝜕𝐴𝑥 ∴ 𝐸𝑥 = − − 𝜕𝑥 𝜕𝑡 52 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ The equation of motion in the 𝑥-direction becomes 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 𝑑𝐴𝑥 𝑚𝑥ሷ = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 −𝑞 + 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑑𝑡 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝜙 𝜕𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 = 𝑞 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 − − − 𝑣𝑥 − 𝑣𝑦 − 𝑣𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧 𝜕𝐴𝑥 𝜕𝐴𝑥 𝜕𝐴𝑥 = 𝑞 ൭ 𝑣𝑥 + 𝑣𝑦 + 𝑣𝑧 − 𝑣𝑥 − 𝑣𝑦 − 𝑣𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝜙 𝜕𝐴𝑥 + − − ൱ 𝜕𝑥 𝜕𝑡 𝜕𝐴𝑦 𝜕𝐴𝑥 𝜕𝐴𝑧 𝜕𝐴𝑥 𝜕𝜙 𝜕𝐴𝑥 = 𝑞 𝑣𝑦 − + 𝑣𝑧 − + − − 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑡 ∴ 𝑚𝑥ሷ = 𝑞 𝐸𝑥 + 𝑣Ԧ × 𝐵 (46) 𝑥 53 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ Note that if not all the forces acting on the system are derivable from a potential, then Lagrange’s equations can be written in the form 𝑑 𝜕𝐿 𝜕𝐿 − = 𝑄𝑗 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 □ where 𝐿 contains the potential of the conservative forces as before, and 𝑄𝑗 represents the forces not arising from a potential. Such a situation often occurs when frictional forces are present. 54 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ If frequently happens that the frictional force is proportional to the velocity of the particle, so that its 𝑥-component has the form 𝐹𝑓𝑥 = −𝑘𝑥 𝑣𝑥 □ Frictional forces of this type may be derived in terms of a function ℱ, known as Rayleigh’s dissipation function, which is 1 2 2 2 ℱ = ෍ 𝑘𝑥 𝑣𝑖𝑥 + 𝑘𝑦 𝑣𝑖𝑦 + 𝑘𝑧 𝑣𝑖𝑧 (47) 2 𝑖 □ where the summation is over the particles of the system. From this definition it is clear that 𝜕ℱ 𝐹𝑓𝑥 = − or 𝐹Ԧ𝑓 = −∇𝑣 ℱ (48) 𝜕𝑣𝑥 55 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ The physical interpretation to the dissipation function: □ The work done by the system against the friction is 𝑑𝑊𝑓 = −𝐹Ԧ𝑓. 𝑑 𝑟Ԧ = −𝐹Ԧ𝑓. 𝑣𝑑𝑡 Ԧ = 𝑘𝑥 𝑣𝑥2 + 𝑘𝑦 𝑣𝑦2 + 𝑘𝑧 𝑣𝑧2 𝑑𝑡 = 2ℱ 𝑑𝑡 □ Hence, 2ℱ is the rate of energy dissipation due to friction. The component of the generalized force resulting from the force of friction is given by 𝜕𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖 𝜕𝑟Ԧ𝑖ሶ 𝜕ℱ Ԧ 𝑄𝑗 = ෍ 𝐹𝑓𝑖. = − ෍ ∇𝑣 ℱ. = − ෍ ∇𝑣 ℱ. =− (49) 𝜕𝑞𝑗 𝜕𝑞𝑗 𝜕𝑞ሶ 𝑗 𝜕𝑞ሶ 𝑗 𝑖 56 Classical Mechanics (PHY6001) Chapter 1 Velocity-Dependent Potentials and The Dissipation Function □ The Lagrange equations with dissipation become 𝑑 𝜕𝐿 𝜕𝐿 𝜕ℱ − = 𝑄𝑗 and 𝑄𝑗 = − 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 𝜕𝑞ሶ 𝑗 𝑑 𝜕𝐿 𝜕𝐿 𝜕ℱ ∴ − =− 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 𝜕𝑞ሶ 𝑗 𝑑 𝜕𝐿 𝜕𝐿 𝜕ℱ ∴ − + =0 (50) 𝑑𝑡 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 𝜕𝑞ሶ 𝑗 □ So, that two scalar functions, 𝐿 and ℱ, must be specified to obtain the equations of motion. 57 Classical Mechanics (PHY6001) Chapter 1 Simple Applications of the Lagrangian Formulation Example 1: Motion of one particle (using Cartesian coordinates): The kinetic energy, in Cartesian coordinates, is 1 1 𝑇 = 𝑚𝑣 = 𝑚 𝑥ሶ 2 + 𝑦ሶ 2 + 𝑧ሶ 2 2 2 2 𝜕𝑇 𝜕𝑇 𝜕𝑇 = 𝑚𝑥,ሶ = 𝑚𝑦,ሶ = 𝑚𝑧ሶ 𝜕𝑥ሶ 𝜕𝑦ሶ 𝜕𝑧ሶ 𝜕𝑇 𝜕𝑇 𝜕𝑇 = = =0 𝜕𝑥 𝜕𝑦 𝜕𝑧 □ The equations of motion are: 𝑑 𝑑 𝑑 𝑚𝑥ሶ = 𝐹𝑥 , 𝑚𝑦ሶ = 𝐹𝑦 , 𝑚𝑧ሶ = 𝐹𝑧 𝑑𝑡 𝑑𝑡 𝑑𝑡 58 Classical Mechanics (PHY6001) Chapter 1 Simple Applications of the Lagrangian Formulation Example 2: Motion of one particle (using plane polar coordinates): the kinetic energy in terms of 𝑟ሶ and 𝜃ሶ 1 𝑇 = 𝑚 𝑟ሶ 2 + 𝑟 2 𝜃ሶ 2 2 𝜕𝑇 𝜕𝑇 𝜕𝑇 𝜕𝑇 = 𝑚𝑟,ሶ = 𝑚𝑟 2 𝜃,ሶ = 𝑚𝑟𝜃ሶ 2 , =0 𝜕𝑟ሶ 𝜕𝜃ሶ 𝜕𝑟 𝜕𝜃 □ The components of the generalized force are 𝜕𝑟Ԧ 𝜕𝑟Ԧ Ԧ 𝑄𝑟 = 𝐹. Ԧ 𝑟Ƹ = 𝐹𝑟 , = 𝐹. Ԧ 𝑄𝜃 = 𝐹. Ԧ 𝑟𝜃෠ = 𝑟𝐹𝜃 = 𝐹. 𝜕𝑟 𝜕𝜃 □ There are two generalized coordinates, and therefore two Lagrange equations. 59 Classical Mechanics (PHY6001) Chapter 1 Simple Applications of the Lagrangian Formulation □ The equation of motion is: 𝑑 𝜕𝑇 = 𝑚𝑟,ሷ and 𝑚𝑟ሷ − 𝑚𝑟𝜃ሶ 2 = 𝐹𝑟 𝑑𝑡 𝜕𝑟ሶ □ The second term being the centripetal acceleration term. For the θ equation, we have the derivatives 𝜕𝑇 𝜕𝑇 𝑑 = 𝑚𝑟 2 𝜃,ሶ = 0, 𝑚𝑟 2 𝜃ሶ = 𝑚𝑟 2 𝜃ሷ + 2𝑚𝑟𝑟ሶ 𝜃ሶ 𝜕𝜃ሶ 𝜕𝜃 𝑑𝑡 □ and the equation becomes 𝑑 𝑚𝑟 2 𝜃ሶ = 𝑚𝑟 2 𝜃ሷ + 2𝑚𝑟𝑟ሶ 𝜃ሶ = 𝑟𝐹𝜃 𝑑𝑡 □ Note that the left side of the equation is the time derivative of the angular momentum, and the right side is exactly the applied torque. 60 Classical Mechanics (PHY6001) Chapter 1 Simple Applications of the Lagrangian Formulation Example 3: Atwood’s machine □ An example of a conservative system with holonomic, scleronomous (the pulley is assumed frictionless and massless). There is only one independent coordinate 𝑥, the position of the other weight is determined by the constraint that the length of the rope between them is 𝑙. □ The potential energy and the kinetic energy are 𝑉 = −𝑀1 𝑔𝑥 − 𝑀2 𝑔 𝑙 − 𝑥 , 1 𝑇 = 𝑀1 + 𝑀2 𝑥ሶ 2 2 61 Classical Mechanics (PHY6001) Chapter 1 Simple Applications of the Lagrangian Formulation □ The Lagranigan has the form 1 𝐿 = 𝑇 − 𝑉 = 𝑀1 + 𝑀2 𝑥ሶ 2 + 𝑀1 𝑔𝑥 + 𝑀2 𝑔 𝑙 − 𝑥 2 □ The partial derivatives of 𝐿 with respect to 𝑥 and 𝑥ሶ are 𝜕𝐿 𝜕𝐿 = 𝑀1 + 𝑀2 𝑥,ሶ and = 𝑀1 𝑔 − 𝑀2 𝑔 = 𝑀1 − 𝑀2 𝑔 𝜕𝑥ሶ 𝜕𝑥 □ So, 𝑑 𝜕𝐿 𝜕𝐿 𝑑 ∵ = ⇒ ∴ 𝑀1 + 𝑀2 𝑥ሶ = 𝑔 𝑀1 − 𝑀2 𝑑𝑡 𝜕𝑥ሶ 𝜕𝑥 𝑑𝑡 𝑀1 − 𝑀2 𝑀1 + 𝑀2 𝑥ሷ = 𝑀1 − 𝑀2 𝑔 or 𝑥ሷ = 𝑔 𝑀1 + 𝑀2 62 Classical Mechanics (PHY6001) Chapter 1 Problems □ (1) Prove that the magnitude 𝑅 of the position vector for the center of mass from an arbitrary origin is given by 1 𝑀 𝑅 = 𝑀 ෍ 𝑚𝑖 𝑟𝑖 − ෍ 𝑚𝑖 𝑚𝑗 𝑟𝑖𝑗2 2 2 2 2 𝑖 𝑖≠𝑗 □ (2) Show that Lagrange’s equations can also be written as 𝜕𝑇ሶ 𝜕𝑇 −2 = 𝑄𝑗 𝜕𝑞ሶ 𝑗 𝜕𝑞𝑗 □ These are sometimes known as the Nielsen form of the Lagrange equations. □ (3) Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the plane of the disk. Derive Lagrange’s equations and find the generalized force. 63 Classical Mechanics (PHY6001) Chapter 1 Problems □ (4) A particle of mass 𝑚 moves in one dimension such that it has the Lagrangian 𝑚2 𝑥ሶ 4 𝐿= − 𝑚𝑥ሶ 2 𝑉 𝑥 − 𝑉 2 (𝑥) 12 □ Where 𝑉(𝑥) is some differentiable function of 𝑥. Find the equation of motion for 𝑥(𝑡) and describe the physical nature of the system based on this system. □ (5) Obtain the Lagrange equations of motion for spherical pendulum, a mass point suspended by a rigid weightless rod, where the kinetic and potential energies are 1 2 2 𝑇 = 𝑚𝑙 𝜃ሶ + sin2 𝜃 𝜙ሶ 2 and 𝑉 = 𝑚𝑔𝑙 cos 𝜃 2 64