Forces, Energy and Equations of Motion

Transcription

1 Forces, Energy and Equations of Motion Chapter 1 P. J. Grandinetti Chem Jan. 8, 2019 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

2 Galileo taught us how to measure time. Pendulum s period is approximately independent of swing amplitude Galileo Galilei ( ) P. J. Grandinetti (Chem. 4300) In 1632 he published Dialogue Concerning the Two Chief World Systems where he proposed his law of inertia and introduces inertial frames. Forces, Energy and Equations of Motion Jan. 8, / 50

3 In 1686 Issac Newton publishes Principia which gives his laws of motion, his law of universal gravitation, and his derivation of Kepler s laws of planetary motion. Sir Issac Newton ( ) P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

4 Newton s Laws of Motion - The First Law Newton s First Law is the Law of Inertia. In an inertial reference frame, a particle either remains at rest or continues to move at a constant velocity, v, unless acted upon by a net external force. Definition Inertial frame: any reference frame in which Newton s first law holds, that is, a non-accelerating, non-rotating frame. Can you give examples of inertial and non-inertial frames? P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

5 Newton s Laws of Motion - The Second Law Newton s Second Law For any particle of mass, m, the net force, F, on the particle is always equal to the mass, m, times the particle s acceleration, a, F = m a The 2 nd law is more generally given in terms of the particle s momentum, p = m v, as F = d p dt A force acting on a particle can cause a change in the particle s momentum vector (direction and magnitude). P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

6 Newton s Laws of Motion Consider a system of N particles each of different mass m α. z m 2 m 5 m 3 m 1 y x m 4 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

7 Newton s Laws of Motion Consider a system of N particles each of different mass m α. z Net force on α th particle, F net, is given by α m 1 m 2 m 3 m 5 F net α = F ext α + N β=1 β α f αβ x m 4 y F ext α is external force acting on α th particle f αβ is internal force on α th particle by β th particle. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

8 Newton s Laws of Motion Consider a system of N particles each of different mass m α. Total momentum, p total, of system of particles is p α is momentum of α th particle. p total = N p α α=1 We write Newton s 2nd law as F total = d p total dt = N α=1 d p α dt = N α=1 F net α = N α=1 F ext α + N N α=1 β=1 β α f αβ P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

9 Newton s Laws of Motion - The Third Law Consider a system of N particles each of different mass m α. F total = d p total dt N = N N F α ext + f αβ α=1 α=1 β=1 β α Newton s third law If object 1 exerts force f 21 on object 2, then object 2 always exerts reaction force f 12 on object 1 f 12 = f 21 Double summation term reduces to zero with 3 rd law leaving F total = d p total dt = N 0 N N f αβ = F ext α + α=1 α=1 β=1 β α N α=1 F ext α = F ext total P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

10 Conservation of Linear Momentum Consider a system of N particles each of different mass m α. if d p total dt = 0, then p total = constant When no net external force acts on system of particles, F ext total = 0, then p total is constant. This is the principle of conservation of linear momentum. It is true in classical and quantum mechanics. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

11 Conservation of Linear Momentum Example A 100 kg astronaut was doing repairs outside her space ship when suddenly the onboard computer malfunctions and disconnects her tether line, leaving her out of reach of the airlock hatch door. In an attempt to return to the ship she throws a 3.0 kg wrench away from her at 5.0 m/s relative to the space station. With what speed and in which direction will she begin to move? Assume the astronaut and wrench are initially at rest relative to the space ship so that v wrench = v astro = 0. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

12 Conservation of Linear Momentum Solution Conservation of momentum says total momentum must remain unchanged before and after she throws wrench p total, initial = p total, final = constant If she throws wrench in x direction we have m (astro) v (astro) x,initial + m(wrench) v (wrench) = m (astro) v (astro) + m (wrench) v (wrench) = constant x,initial x,final x,final Taking initial velocities of v wrench = v astro = 0 leaves m (astro) v (astro) x,final + m (wrench) v (wrench) x,final = 0 or v (astro) x,final = m(wrench) v(wrench) m (astro) x,final She moves in opposite direction that wrench is thrown. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

13 Center of Mass Baton toss in strobe light P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

14 Center of Mass Consider a system of N particles each of different mass m α. Equation of motion for α th particle is given by m α a α = F net α or m α d 2 r α dt 2 Summing over all N particles gives 0 N d 2 r m α N N N α dt = F ext α=1 2 α + α=1 f αβ α=1 β=1 β α Define total force acting on system of particles as N F = F ext α to get i=1 = F ext α + N β=1 β α f αβ 3 rd law makes double sum zero N α=1 m α d 2 r α dt 2 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50 = F

15 Center of Mass Consider a system of N particles each of different mass m α. Defining the total mass as M = N m α α=1 the center of mass as R = 1 M N α=1 N m α r α α=1 m α d 2 r α dt 2 = F we obtain M d2 R dt 2 Center of mass obeys same equation of motion as single particle of total mass M acted on by force, F. Particles thrown into the air execute a complicated motion while center of mass follows a simple parabolic path like a single particle. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50 = F

16 Baton toss in strobe light P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

17 Center of mass of water Example Find center of mass of water molecule. Take OH bond length as d = Å, H-O-H angle as Ω = 105, and mass of oxygen and hydrogen as 16.0 u and 1.0 u. (1 u = kg) z With mass of atom concentrated at nucleus we safely approximate each atom as point mass. Place oxygen atom at origin and hydrogen symmetrically in x-z plane above and below x axis. 105 x P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

18 Center of mass of water Example Find center of mass of water molecule. Take OH bond length as d = Å, H-O-H angle as Ω = 105, and mass of oxygen and hydrogen as 16.0 u and 1.0 u. (1 u = kg) z x coordinate of each hydrogen is d cos Ω 2 z coordinate is ±d sin Ω 2 for H above and below x axis, respectively. 105 x # atom mass x y z 1 H 1 u Å Å 2 H 1 u Å Å 3 O 16 u P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

19 Center of mass of water Example Find center of mass of water molecule. Take OH bond length as d = Å, H-O-H angle as Ω = 105, and mass of oxygen and hydrogen as 16.0 u and 1.0 u. Substituting values into formulas # atom mass x y z 1 H 1 u Å Å 2 H 1 u Å Å 3 O 16 u X = m H x 1 + m H x 2 + m O x 3 m H + m H + m O = (1.0 u)(0.5826å) + (1.0 u)(0.5826å) 1.0 u u u = Å Z = m Hz 1 + m H z 2 + m O z 3 m H + m H + m O = (1.0 u)(0.7592å) + (1.0 u)( Å) 1.0 u u u = 0 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

20 Center of mass of water Example Center of mass of water is along x axis and shifted slightly to the of oxygen position by Å (see small black dot). z 105 x P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

21 Work, Power, and Kinetic Energy P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

22 Work and Power Work Work done by external force on particle moving it from r 1 to r 2 is given by path integral w = r 2 r 1 path F ext d r The SI unit for work is the joule (1 J = 1 kg m 2 /s 2 ). Power Work done per unit time is called power P = dw dt = F ext d r dt = F ext v The SI unit for power is the watt (1 W = 1 J/s = 1 kg m 2 /s 3 ). P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

23 Work-Energy Theorem P = dw dt Substituting Newton s 2 nd law, F ext = m a = m d v dt dw dt = F ext d r dt = F ext v = m d v dt v = mv dv x dv y dv z x + mv dt y + mv dt z dt Note time dependence of v(t) and rewrite the above expression as dw = d ( 1 dt dt 2 mv2 x (t) mv2 y (t) + 1 ) 2 mv2 z (t) = d ( 1 (t)) dt 2 mv2 Work done on the particle is transformed into kinetic energy Work-Energy Theorem Work done by net force acting on particle equals change in kinetic energy of particle. w(t 1 t 2 ) = ΔK = 1 2 mv2 (t 2 ) 1 2 mv2 (t 1 ) P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

24 Conservative Forces and Potential Energy Conservative Force Conservative force acting on particle depends only particle s position, r, and not on particle s velocity, v, time, t, or any other variable. When conservative force acts on particle traveling in closed loop the net work done on particle is zero, that is, w = F d r = 0 Potential Energy Since work done by conservative force is same for all paths between points 1 and 2, we can write r 2 w(1 2) = F d r = [ V( r 2 ) V( r 1 ) ] r 1 and define V( r) as a quantity called the potential energy. V( r) is a scalar field: a physical quantity that can have a different value at each point in space and time. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

25 Conservative Forces and the Work-Energy Theorem When work-energy theorem involves conservative force it becomes w(1 2) = K 2 K 1 = V( r 2 ) + V( r 1 ) Total mechanical energy, E, in moving particle between r 1 and r 2 by conservative force is E = K 1 + V( r 1 ) = K 2 + V( r 2 ) Total mechanical energy is constant when forces acting on particle are conservative. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

26 Conservative Forces and Potential Energy When all forces acting on particle are conservative we calculate force vector from gradient of scalar potential energy, F( r) = V( r) is gradient operator = e x x + e y y + e z z When applied to potential energy function we obtain gradient vector of V(x, y, z): V(x, y, z) = e x V(x, y, z) x + e y V(x, y, z) y + e z V(x, y, z) z V(x, y, z) is vector that points in direction of greatest increase of potential energy function. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

27 Conservative Forces and Potential Energy Potential Energy Surface in One Dimension F = dv(x) dx E = K+V(x) P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

28 Equations of Motion in terms of Potential Energy For particle of mass m acted upon by conservative forces Newton s 2 nd law becomes three coupled second-order partial differential equations: m d2 x V(x, y, z) + = 0 dt2 x m d2 y V(x, y, z) + = 0 dt2 y m d2 z V(x, y, z) + = 0 dt2 z Equations are coupled because potential energy function, V(x, y, z), is present in every equation. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

29 Classification of Differential Equations Definition Order of differential equation is order of highest derivative in equation. Definition Ordinary differential equation, is linear if F is linear function of variables f, dy dx,..., d n f dx n. General linear ordinary differential equation of order n has form ( F t, f, df ) dx,..., dn f dx n = 0, a 0 (t) dn f dx n + a 1(t) dn 1 f dx n a n(t)f = g(t) Similar definition applies to partial differential equations. In this course we focus primarily on physical systems described by linear differential equations. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

30 Order and Linearity of Differential Equations Example Determine differential equation order and whether or not it s a linear differential equation. t 2 d2 y dt + t dy + 2y = sin t 2 dt 2nd order, linear dy dt + ty2 = 0 1st order, non-linear (1 + y 2 ) d2 y dt + t dy 2 dt + y = et 2nd order, non-linear d 3 y dt + t dy 3 dt + (cos2 t)y = t 3 3rd order, linear d 4 y dt + d3 y 4 dt + d2 y 3 dt + dy 2 dt + y = 1 4th order, linear d 2 y + sin(t + y) = sin t dt2 2nd order, non-linear P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

31 Equations of Motion in terms of Potential Energy z For N particles we get 3N coupled partial differential equations of motion: m 1 m 2 m 3 m 5 y m i d 2 x i dt 2 + V(x 1,, x 3N ) x i = 0 Note change in notation: x 1 x x 2 y x 3 z m 1 m m 2 m m 3 m } for particle 1 x m 4 x 4 x x 5 y x 6 z m 4 m m 5 m m 6 m } for particle 2 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

32 Equations of Motion in terms of Kinetic and Potential Energy Since kinetic energy is K(ẋ 1,, ẋ 3N ) = 3N i=1 1 2 m iẋ 2 dx i, where ẋ = dt Take partial derivative of both sides with respect to ẋ i K(ẋ 1,, ẋ 3N ) = m ẋ i ẋ i i Then take partial derivative with respect to time ( ) d K(ẋ1,, ẋ 3N ) dẋ = m i d 2 x dt ẋ i = m i dt i dt 2 and m i d 2 x i dt 2 + V(x 1,, x 3N ) x i = 0 becomes d dt ( K(ẋ1,, ẋ 3N ) ẋ ) + V(x 1,, x 3N ) x i = 0 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

33 Equations of Motion in terms of Kinetic and Potential Energy For N particles the 3N coupled partial differential equations of motion can be written ( ) d K(ẋ1,, ẋ 3N ) + V(x 1,, x 3N ) = 0 dt ẋ x i Equations of motion can be derived from scalar functions for kinetic and potential energy. No force vectors are needed. Keep in mind that general concept of energy, and particularly those of kinetic and potential energies, did not appear until well over 100 years after Newton death. This kind of reformulation of Newtonian Mechanics in terms of energy functions eventually led to a more versatile approaches called Lagrangian Mechanics and Hamiltonian Mechanics (more later). P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

34 height Equations of Motion in terms of Kinetic and Potential Energy Example Neglecting air resistance, determine trajectory of a ball initially thrown straight up in the air with an initial velocity v 0. ( ) d K(ż) + V(z) = 0 dt ż z Kinetic energy is K = 1 2 mv2 z = 1 2 mż2 Potential energy is V = mg 0 z, where g 0 = m/s 2, is magnitude of acceleration due to force of gravity. ( ) ( ) K(ż) d K(ż) = mż, = m z, and V(z) = mg ż dt ẋ z 0 Differential equation of motion is m z + mg 0 = 0 Solution is z(t) = v 0 t 1 2 g 0t 2 (see Problem 1-17) time P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

35 Central Forces Definition A central force is everywhere directed towards a fixed mass or force center, and has the form z F( r) = f ( r) e r e r is unit vector in spherical coordinate system. Spherical coordinate unit vectors are given by e r = sin θ cos φ ê x + sin θ sin φ e y + cos θ e z, e θ = cos θ cos φ ê x + cos θ sin φ e y sin θ e z, e φ = sin φ ê x + cos φ e y, x y P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

36 Conservative Central Forces If central force is conservative then it must be spherically symmetric. Non-conservative central force: Conservative central force: F( r) = f ( r) e r F( r) = f (r) e r 3D problems with conservative central forces can often be transformed into 1D problems. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

37 Fundamental Forces of Nature Gravitational force : a conservative central force Gravitational force on a mass M due to mass m at distance r is F = GMm r M r m r M r m 3, Newton s law of universal gravitation G is gravitational constant, G = m 3 /(kg s 2 ). r M r m is inter-particle vector. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

38 Fundamental Forces of Nature Electric force : a conservative central force Electric force on charge Q due to charge q which is at rest a distance r away is also conservative and central force F = Qq r Q r q 4πε 0 r Q r q, 3 Coulomb s law where ε 0 is the electric constant, ε 0 = s 4 A 2 /(m 3 kg). P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

39 Fundamental Forces of Nature Magnetic force The magnetic force between two closed loops of wire carrying currents I 1 and I 2 is given by the path integral F = μ 0 4π I 1d l 1 I 2 d l 2 r 1 r 2 r 1 r 2 3. μ 0 is the magnetic constant μ 0 = m kg/(s 2 A 2 ) P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

40 Fields P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

41 Michael Faraday ( ) P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

42 Electric field What is force that acts on charge Q due to some fixed arrangement of N charges, q α? We can calculate superposition of forces of each q α on Q to obtain F = F α = Q 4πε α 0 α q α r r α r r α = Q ( r) 3 where electric field at r due to some fixed arrangement of N charges, q α, is given by ( r) = 1 4πε 0 α q α r r α r r α 3 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

43 Electric field lines around single positive charge + Field lines appear closer together when field is stronger and further apart when field is weaker. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

44 Electric field lines around two opposite sign charges A physical dipole + Field lines always point away from positive charges and towards negative charges. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

45 Electric field lines around two identical charges + + Electric field is tangent to field line at any given point in space. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

46 Web Apps 2D Electrostatic Fields 3D Electrostatic Fields P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

47 Magnetic field Biot-Savart law Force exerted on wire carrying current I in magnetic field arising from all surrounding currents given by F = Idl B( r) where B( r) = μ 0 4π I kd r r l k k k r r k 3 P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

48 Magnetic field from a current loop Right hand rule for direction of the magnetic field vector If you point your right hand thumb along direction of current then your fingers will curl in direction of magnetic field. In 1820 Ampére suggested that magnetism in matter arises from a multitude of ring currents circulating at the atomic and molecular scale. P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

49 Web Apps 3D Magnetostatic Fields P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50

50 Lorentz force law Lorentz force law For charge Q in motion with velocity v through region containing both electric and magnetic field in given inertial frame the force on charge is ( ) F = Q + v B Force from magnetic field changes direction of charge s velocity but not its speed. Magnetic fields do not change particle s kinetic energy and thus do no work on particle. In 1905 Albert Einstein published On the electrodynamics of moving bodies, showing that magnetism is a relativistic effect. Electric and magnetic fields transform into each other when moving between different inertial frames. Video Links: Electrons in Magnetic Fields How Special Relativity Makes Magnets Work P. J. Grandinetti (Chem. 4300) Forces, Energy and Equations of Motion Jan. 8, / 50