LECTURES 20 - 29 INTRODUCTION to CLASSICAL MECHANICS Prof

LECTURES 20 - 29 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford HT 2017

1 OUTLINE : INTRODUCTION TO MECHANICS LECTURES 20-29 20.1 The hyperbolic orbit 20.2 Hyperbolic orbit : the distance of closest approach 20.3 Hyperbolic orbit: the angle of deﬂection, φ 20.3.1 Method 1 : using impulse 20.3.2 Method 2 : using hyperbola orbit parameters 20.4 Hyperbolic orbit : Rutherford scattering 21.1 NII for system of particles - translation motion 21.1.1 Kinetic energy and the CM 21.2 NII for system of particles - rotational motion 21.2.1 Angular momentum and the CM 21.3 Introduction to Moment of Inertia 21.3.1 Extend the example : J not parallel to ω 21.3.2 Moment of inertia : mass not distributed in a plane 21.3.3 Generalize for rigid bodies 22.1 Moment of inertia tensor

2 22.1.1 Rotation about a principal axis 22.2 Moment of inertia & energy of rotation 22.3 Calculation of moments of inertia 22.3.1 MoI of a thin rectangular plate 22.3.2 MoI of a thin disk perpendicular to plane of disk 22.3.3 MoI of a solid sphere 23.1 Parallel axis theorem 23.1.1 Example : compound pendulum 23.2 Perpendicular axis theorem 23.2.1 Perpendicular axis theorem : example 23.3 Example 1 : solid ball rolling down slope 23.4 Example 2 : where to hit a ball with a cricket bat 23.5 Example 3 : an aircraft landing 24.1 Lagrangian mechanics : Introduction 24.2 Introductory example : the energy method for the E of M 24.3 Becoming familiar with the jargon 24.3.1 Generalised coordinates 24.3.2 Degrees of Freedom 24.3.3 Constraints

3 24.3.4 Conﬁguration Space 25.1 The Lagrangian : simplest illustration 25.2 A formal derivation of the Lagrange Equation The calculus of variations 25.3 A sanity check 25.4 Fermat’s Principle & Snell’s Law 25.5 Hamilton’s principle (Principle of Stationary Action) 26.1 Conjugate momentum and cyclic coordinates 26.2 Example : rotating bead 26.3 Example : simple pendulum 26.3.1 Dealing with forces of constraint 26.3.2 The Lagrange multiplier method 27.1 The Lagrangian in various coordinate systems 27.2 Example 1: the rotating bead 27.3 Example 2: bead on rotating hoop 28.1 Hamilton mechanics 28.2 The physical signiﬁcance of the Hamiltonian 28.3 Example: re-visit bead on rotating hoop 28.4 Noether’s theorem

29.14 Re-examine the sliding blocks using E-L 29.2 Normal modes of coupled identical springs

29.3 Final example: a rotating coordinate system

5 20.1 The hyperbolic orbit

r0 I Orbit equation : r(θ) = 1+e cos θ x 2 y 2 I Ellipse : e < 1 a + b = 1

I Hyperbola : x 2 y 2 e > 1 a − b = 1 y x b When x&y → ∞, b = a → y = ( a )x 6 20.2 Hyperbolic orbit : the distance of closest approach For example a comet deviated by the gravitational attraction of a

planet. Velocity v = v0 when r → ∞.

I h is known as the impact parameter

I Angular momentum J = m r × v

→ |J| = mvr sin γ = mv0h (as r → ∞) 1 2 → Total energy E = 2 mv0 (again as r → ∞) 7 Distance of closest approach continued

1 2 J2 α I = ˙ + − α = E 2 mr 2mr 2 r , where GmM I At distance of closest approach r = rmin → r˙ = 0 J2 α I E = 2 − r 2mrmin min 2 α J2 → rmin + E rmin − 2mE = 0

I Same form of solution as for the ellipse : 2 α 2EJ 1 2 2 1 2 I r = − [1 − (1 + ) 2 ] (J = (mv h) ; E = mv ) min 2E mα2 0 2 0 | {z } α e → r = − (1 − e) min 2E | {z } = a(e - 1) 0 I Velocity v at distance of closest approach: line to trajectory is a right angle.

0 0 v0h → J = mv rmin = mv0h → v = rmin 8 20.3 Hyperbolic orbit: the angle of deﬂection, φ

9 20.3.1 Method 1 : using impulse

I Directly from the diagram : ∆vx = 2v0 cos θ∞ (1)

I By symmetry, integrated change in vy = 0 : ∆vy = 0 2 ˙ R +∞ R +∞ mr θ I Change in ∆p : m∆v = F dt = F ( ) dt x x −∞ x −∞ x J → m∆v = ( m ) R +θ∞ F r 2dθ | {z } x J −θ∞ x = 1 α α I F = −( )ˆr → = − θ But r 2 Fx r 2 cos mα R θ∞ → m∆vx = −2( J ) 0 cos θdθ 2α → ∆vx = −( J ) sin θ∞ (2) 2α I From (1) & (2) → −( J ) sin θ∞ = 2v0 cos θ∞ Jv0 I tan θ∞ = − α → θ∞ + β = π ; φ + 2β = π φ π φ π φ I θ∞ = 2 + 2 ; tan θ∞ = tan( 2 + 2 ) = − cot 2 2 2 φ J v0 mhv0 hv0 cot 2 = α = α = GM 10 20.3.2 Method 2 : using hyperbola orbit parameters

r0 I Orbit equation : r(θ) = 1+e cos θ 1 I r → ∞ , cos θ∞ = − e φ π φ 1 I θ∞ + β = π ; φ + 2β = π → 2 = θ∞ − 2 → sin 2 = e

I From before : 2 α 2EJ 1 r = − [1 − (1 + ) 2 ] min 2E mα2 | {z } e

φ 2 1 2EJ2 1 I = [ − ] 2 = [ ] 2 cot 2 e 1 mα2 1 2 I E = 2 mv0 ; J = mv0h

φ mv 2h v 2h I 0 0 cot 2 = α = GM as before

11 20.4 Hyperbolic orbit : Rutherford scattering A particle of charge +q deviated by the Coulomb repulsion of a nucleus, charge +Q. Analogous hyperbolic motion as previously: α = GMm → α = − Qq . As before, velocity v = v when r → ∞. 4π0 0 Here we take the other branch of the hyperbola.

I Analogous equations : 1 2 J2 Qq I E = mr˙ + 2 + 2 2mr 4π0 r 2 φ (4π0) mv0 h I cot 2 = qQ qQ 2 1 2EJ 2 I r = [1 + (1 + 2 ) ] min 2E(4π0) mα 12 21.1 NII for system of particles - translation motion

Reminder from MT lectures: d2 ext int I (r ) = F + F Force on particle i: mi dt2 i i i N 2 N N X d X ext X int PN ext I m (r ) = F + F = F i dt2 i i i i i i i i | {z } | {z } | {z } all masses external forces internal forces = zero

PN mi ri I rCM = i M PN where M = i mi PN mi r˙ i I vCM =r ˙ CM = i M PN → PCM = i mi r˙i = M vCM ext PN mi ¨ri Fi I aCM = ¨rCM = i M = M

13 21.1.1 Kinetic energy and the CM

1 PN 2 0 I Lab kinetic energy : T = 2 i mi vi ; vi = vi + vCM 0 where vi is velocity of particle i in the CM

1 P 02 1 P 2 P 0 I T = 2 i mi vi + 2 i mi vCM + i mi vi · vCM

P 0 P 0 i mi vi I But m v · v = · M v i i i CM M CM | {z } = 0

0 1 2 I T = T + 2 MvCM

Same expression as was derived in MT

14 21.2 NII for system of particles - rotational motion

I Angular momentum of particle i about O : J = r × p i i i dJi I Torque of i about O: τ = = r × p˙ +r ˙ × p = r × F i dt i i i i i i | {z } = 0 I Total ang. mom. of system J = PN J = PN r × p i i i i i

I Internal forces: Pint int int pair τ (i,j) = ri × Fij + rj × Fji int = (ri − rj ) × Fij int = 0 since (ri − rj ) parallel to Fij Hence total torque PN ext dJ τ = i (ri × Fi ) = dt If external torque = 0, J is const.

15 21.2.1 Angular momentum and the CM

0 0 I Lab to CM : ri = ri + rCM ; vi = vi + vCM 0 0 where ri , vi are position & velocity of particle i wrt the CM

I Total ang. mom. of system PN PN 0 0 J = i Ji = i mi (ri + rCM ) × (vi + vCM )

P 0 0 P 0 P 0 P = i mi (ri × vi ) + i mi (ri × vCM ) + i mi (rCM × vi ) + i mi (rCM × vCM ) P 0 X 0 P 0 X 0 But i mi (ri × vCM ) = [ mi ri ] ×vCM ; i mi (rCM × vi ) = rCM × [ mi vi ] i i | {z } | {z } = 0 in CM = 0 in CM 0 J = J + rCM × M vCM I Hence |{z} | {z } J wrt CM J of CM translation

16 What we have learned so far

I Newton’s laws relate to rotating systems in the same way that the laws relate to transitional motion.

I For any system of particles, the rate of change of internal angular momentum about an origin is equal to the total torque of the external forces about the origin.

I The total angular momentum about an origin is the sum of the total angular momentum about the CM plus the angular momentum of the translation of the CM.

17 21.3 Introduction to Moment of Inertia

I Take the simplest example of 2 particles rotating in circular motion about a common axis of rotation with angular velocity ω = ω zˆ

I Deﬁnition of ω for circular motion : r˙ = ω × r

I Total angular momentum of the system of particles about O

J = r1 × (m1v1) + r2 × (m2v2)

I v1 = ω × r1

v2 = ω × r2

I Since ri ⊥ vi 2 2 J = (m1 r1 + m2 r2 ) ω

I J = Iω (J is parallel to ω) 2 2 I Moment of Inertia I = m1 r1 + m2 r2 P 2 Or more generally I = i [mi ri ] 18 21.3.1 Extend the example : J not parallel to ω Now consider the same system but with rotation tilted wrt rotation axis by an angle φ. Again ω = ω zˆ

I Total angular momentum of the particles about O :

J = r1 × (m1v1) + r2 × (m2v2) 0 I NB. J now points along the z axis

I J vector is ⊥ to line of m1 and m2 and deﬁnes the principal axis of the MoI mass distribution (see later)

d1 I Since v1 = d1 ω ; r1 = sin φ

d2 v2 = d2 ω ; r2 = sin φ 2 2 (m1 d1 +m2 d2 ) ω I Then |J| = sin φ → |J| sin φ ωˆ = Iz ω . Hence Jz = Iz ω where Iz is calculated about the zˆ (i.e. ωˆ) axis. 19 21.3.2 Moment of inertia : mass not distributed in a plane Now take a system of particles rotating in circular motion about a common axis of rotation, all with angular velocity ω (where vi = ω × ri ).

I Total angular momentum of the system of particles about O (which is on the axis of rotation) PN PN J = i ri × (mi vi ) = i mi ri vi uˆi (not necessarily parallel to ω axis)

I As before, resolve the angular momentum about the axis of rotation PN Jz zˆ = i mi ri vi sin φi ωˆ

I vi = ω × ri ; vi = ωri sin φi Also sin φ = di ; v = ω d i ri i i PN 2 n Jz ωˆ = i mi di ω → Jz = Iz ω (Iz : MoI about rot axis) 20 21.3.3 Generalize for rigid bodies

A rigid body may be considered as a collection of inﬁnitesimal point particles whose relative distance does not change during motion. P R I i mi → dm, where dm = ρdV and ρ is the volume density

PN 2 R 2 I Iz = i mi di → V d ρ dV

I This integral gives the moment of inertia about axis of rotation (z axis)

21 22.1 Moment of inertia tensor

I Consider bodies rotating around a common axis ω, no axis of symmetry, J not necessarily parallel to ω, origin O lies on the ω (z) axis.

I Deﬁnition of angular velocity : r˙ = ω × r P P I J = r × p = m r × r˙ i i i i i i i P = i mi ri × (ω × ri )

I Use the vector identity a × (b × c) = (a.c) b − (a.b) c P 2 P J = i mi ri ω − i mi (ri · ω) ri

[Note for circular motion in a plane where P ri is ⊥ to ω , i mi (ri · ω) = 0 , and hence in this case J is k to ω.]

22 Moment of inertia tensor continued

P 2 P I From before : J = i mi ri ω − i mi (ri · ω) ri

I Can express in terms of components P 2 2 2 (Jx , Jy , Jz ) = i mi (xi + yi + zi )(ωx , ωy , ωz )− P i mi (xi ωx + yi ωy + zi ωz )(xi , yi , zi )

I Construct a matrix equation :

   P 2 2 P P    Jx i (yi + zi )mi − i (xi yi )mi − i (xi zi )mi ωx P P 2 2 P  Jy  =  − i (xi yi )mi i (xi + zi )mi − i (yi zi )mi   ωy  J P P P 2 2 ω z − i (xi zi )mi − i (yi zi )mi i (xi + yi )mi z If mass is continuous (rigid body) :

   R 2 2 R R    Jx (y + z ) dm − xy dm − xz dm ωx R R 2 2 R  Jy  =  − xy dm (x + z ) dm − yz dm   ωy  R R R 2 2 Jz − xz dm − yz dm (x + y ) dm ωz

I Hence we can write : J = eIω

I eI is the Moment of Inertia tensor of the system 23 22.1.1 Rotation about a principal axis

I In general J = eI ω , where eI is the Moment of Inertia Tensor       Jx Ixx Ixy Ixz ωx  Jy  =  Iyx Iyy Iyz   ωy  Jz Izx Izy Izz ωz

I Whenever possible, one aligns the axes of the coordinate system in such a way that the mass of the body evenly distributes around the axes: we choose axes of symmetry.       Jx Ixx 0 0 ωx  Jy  =  0 Iyy 0   ωy  Jz 0 0 Izz ωz

The diagonal terms are called the principal axes of the moment of inertia.

I Whenever we rotate about an axis of symmetry, for every point A there is a point B which cancels it, and J → Jz zˆ = Izz ω zˆ and where J is parallel to ω along the z axis 24 22.2 Moment of inertia & energy of rotation Particles rotating in circular motion about a common axis of rotation with angular velocity ω (where vi = ω × ri ).

I Kinetic energy of mass mi : 1 2 Ti = 2 mi vi 1 P 2 I Total KE = 2 i mi vi

I vi = ω × ri

vi = ω ri sin φi sin φ = di i ri 1 PN 2 2 I Trot = 2 i mi di ω

1 2 Trot = 2 Iz ω

where Iz is calculated about the axis of rotation

25 22.3 Calculation of moments of inertia 22.3.1 MoI of a thin rectangular plate (a) About the x axis R 2 I Ix = y dm M dm = ρ dx dy ; ρ = ab R 2 I Ix = y ρ dx dy b h 3 i+ a y 2 + 2 = ρ [x] a 3 b − 2 − 2 h 3 3 i h 2 i b b b (c) About the z axis = ρ a 24 + 24 = ρa b 12 R 2 I Iz = ρ d dx dy M b2 I Hence I = x 12 = R ρ (x2 + y 2) dx dy b a h 3 i+ h 3 i+ = ρ y 2 + x 2 (b) About the y axis 3 b 3 a − 2 − 2 M a2 I Iy = a2+b2 12 I Hence Iz = M ( 12 ) 26 22.3.2 MoI of a thin disk perpendicular to plane of disk

R 2 I Iz = r dm = ρ( π ); ρ = M dm 2 r dr πR2 R 3 I Iz = 2πρ r dr R h r 4 i R4 M = 2πρ = (π ) 2 4 0 2 πR 1 2 I Hence Iz = 2 MR

I For a cylinder thickness t :

I Use cylindrical coordinates = ρ( π ); ρ = M dm 2 r dr dt πR2 t RR 3 I Iz = 2πρ r dr dt 1 2 I Iz = 2 MR , the same.

27 22.3.3 MoI of a solid sphere

1 2 I dI = 2 x dm (for a disk) 2 M dm = ρ(πx dz); ρ = 4 3 3 πR πρ 4 → dI = 2 x dz R 1 R +R 4 I I = dI = 2 πρ −R x dz 2 2 2 I But x = R − z 1 R +R 4 2 2 4 I = 2× 2 πρ 0 (R −2R z +z )dz

h 2 3 iR = πρ R4z − 2R z + 1 z5 3 5 0 8 5 8 5 4 3 = πρ 15 R = M(π 15 R )/( 3 πR ) 2 2 I Hence I = 5 MR

28 23.1 Parallel axis theorem

ICM is the moment of inertia of body mass M about an axis passing through its centre of mass. I is the moment of inertia of the body about a parallel axis a distance d from the ﬁrst.

I About the axis through the CM R 2 ICM = r dm 0 I r = d + r 0 2 2 2 I r = d + 2 d · r + r

I About the parallel axis : I = R r 0 2dm Z = R d 2dm + 2 d · r dm + R r 2dm | {z } = 0 (deﬁnition of CM) 2 I Hence I = ICM + Md 29 23.1.1 Example : compound pendulum

Rectangular rod length a width b mass m swinging about axis O, distance ` from the CM, in plane of paper

a2+b2 I ICM = m ( 12 )

I Parallel axis theorem : 2 I = ICM + m`

I Torque about O = r × F : τ = −m g ` sin θ kˆ

I J = I ω = I θ˙ kˆ dJ ¨ ˆ I Differentiate : τ = dt = I θ k

I Equate I θ¨ = −m g ` sin θ

I Small angle approximation ¨ m g ` θ + I θ = 0

30 Compound pendulum continued ¨ m g ` I θ + I θ = 0 a2+b2 2 where I = m ( 12 ) + m ` q I I SHM with period T = 2π mg` qa2+b2+12`2 → T = 2π 12g`

I Measurement of g : 2 I Plot T vs `

I Find values of `1 and `2 that give the same value of T → T1 = T2 T 2 π2 Can show : 1 = 4 (`1+`2) g

31 23.2 Perpendicular axis theorem

Consider a rigid object that lies entirely within a plane. The perpendicular axis theorem links Iz (MoI about an axis perpendicular to the plane) with Ix , Iy (MoI about two perpendicular axes lying within the plane).

I Consider perpendicular axes x , y , z (which meet at origin O) ; the body lies in the xy plane R 2 I Iz = d dm = R x2 + y 2 dm = R x2 dm + R y 2 dm

→ Iz = Ix + Iy This is the perpendicular axis theorem.

32 23.2.1 Perpendicular axis theorem : example

I Consider a thin circular disk lying in a plane

I Perpendicular axis theorem : Iz = Ix + Iy

1 2 I Iz = 2 MR (R = radius of disk)

1 2 I Hence Ix = Iy = 4 MR (due to symmetry)

33 23.3 Example 1 : solid ball rolling down slope

[Energy of ball] = [Rotational KE in CM] + [KE of CM] + [PE]

1 2 1 2 E = 2 Iω + 2 Mv + Mgy

I Ball falls a distance h from rest → at y = 0 : 1 2 1 2 M g h = 2 Iω + 2 Mv 1 v 2 1 2 = 2 I R + 2 Mv 2 2 I Solid sphere: I = 5 MR 1 2 2 I Mgh = 2 Mv 5 + 1 q 10 → v = 7 gh q 1 2 4 Compare with a solid cylinder I = 2 MR → v = 3 gh The ball gets to the bottom faster !

34 23.4 Example 2 : where to hit a ball with a cricket bat b We want the bat handle (Point A, 2 from the CM) to remain stationary after the ball has hit. When the ball hits, there is rotation of the bat about the CM, plus motion of the CM of the bat. The velocity to the ω b right (vCM ) must equal the velocity from the left 2 due to rotation.

35 Where to hit a ball with a cricket bat continued

I Ball hits at point x from centre F I Force to the CM : F = ma → a = m

I Moment of inertia wrt CM : 1 2 ICM = 12 mb ¨ I Torque (couple) about O = ICM θ ¨ mb2 ¨ F I Hence ICM θ = 12 θ = x × 2 × 2 → θ¨ = 12Fx mb2

I Require acceleration at A to be zero. b ¨ I Acceleration at A due to rotation = 2 θ

I Equate accelerations : F 6Fx b m = mb → x = 6

2 Need to hit the bat 3 from the top 36 23.5 Example 3 : an aircraft landing The landing wheel of an aircraft may be approximated as a uniform circular disk of diameter 1 m and mass 200 kg. The total mass of the aircraft including that of the 10 wheels is 100, 000 kg. When landing the touch-down speed is 50 ms−1. Assume that the wheels support 50% of the total weight of the aircraft. Determine the time duration of wheel-slip if the coefﬁcient of friction between the wheels and the ground is 0.5, assuming that the speed of the plane is not changed signiﬁcantly.

37 An aircraft landing continued 0 I Torque |τ| = |r × F| = aµM g 0 M about O, where M = 20 (ie. 10 wheels, supporting 50% of mass)

I Angular momentum J = Iω 1 2 where I = 2 ma (MoI of solid disk where m is mass of a wheel) dJ dω I τ = dt = I dt where u = a ω u is the speed of the wheel rim I du 0 → τ = a dt = aµM g 2 0 R tf a µM g R v0 I Integrate : 0 I dt = 0 du v0 is the speed of the aeroplane

v0 I v0 m → tf = a2µM0g = 2µM0g

50×200 I Putting in numbers: tf = 1×105 ∼ 0.2 s 2×0.5× 20 ×9.8 38 An aircraft landing continued

Conﬁrm the assumption that the speed of the plane is not changed by calculating the speed at the end of wheel-slip in the absence of other braking processes.

I Energy expended in getting the wheels 1 2 up to speed: Ewheels = 2 Iω × 10 (ie. 5 2 10 wheels) = 2 mv0 1 2 [ Remember v0 = aω , I = 2 ma ]

I Total energy of the aeroplane 1 2 E = 2 Mv 5 2 → Energy loss: δE = Mv δv = 2 mv0 5 mv 2 5 m → δv = 2 0 → δv = 2 Mv0 v0 M 5 δv 2 ×200 I Putting in numbers : = 5 v0 1×10 → a 0.5% effect

39 An aircraft landing continued Calculate the work done during wheel slip. R I Work done : W = τ dθ = τ θf (τ is constant, θf is the total turning angle) dθ 2µM0g I From before : u = a dt = m t 0 R θf R tf 2µM g I Integrate : 0 a dθ = 0 m t dt 0 µM g 2 v0 m → a θf = m tf where tf = 2µM0g

I Putting it all together: 0 µ 1 0 M g v0 m 2 1 v0 2 I W = τ θf = a (aµM g)( )( 0 ) = 2 I( a ) | {z } m 2µM g τ | {z } θf 1 2 I Hence W = Ewheel = 2 I ω as expected !

40 Lagrange and Hamilton

I Joseph-Louis Lagrange (1736-1810)

I Sir William Rowan Hamilton (1805-1865)

41 24.1 Lagrangian mechanics : Introduction

I Lagrangian Mechanics: a very effective way to ﬁnd the equations of motion for complicated dynamical systems using a scalar treatment → Newton’s laws are vector relations. The Lagrangian is a single scalar function of the system variables

I Avoid the concept of force → For complicated situations, it may be hard to identify all the forces, especially if there are constraints

I The Lagrangian treatment provides a framework for relating conservation laws to symmetry

I The ideas may be extended to most areas of fundamental physics (special and general relativity, electromagnetism, quantum mechanics, quantum ﬁeld theory .... )

42 24.2 Introductory example : the energy method for the E of M

I For conservative forces in 1D motion : 1 ˙ 2 dE Energy of system : E = 2 mx + U(x) [Note dt = 0] ˙ ¨ ∂U ˙ I Differentiate wrt time: mxx + ∂x x = 0 ¨ ∂U → m x = − ∂x = F → This is the E of M for a conservative force

I Take a simple 1d spring undergoing SHM : 1 ˙ 2 1 2 E = 2 mx + 2 kx = constant dE ˙ ¨ ˙ dt = 0 → mxx + kxx = 0 → mx¨ + kx = 0

I Hence we derived the E of M without using NII directly.

43 24.3 Becoming familiar with the jargon

24.3.1 Generalised coordinates

A set of parameters qk (t) that speciﬁes the system conﬁguration. qk may be a geometrical parameter, x, y, z, a set of angles ··· etc

24.3.2 Degrees of Freedom

The number of degrees of freedom is the number of independent coordinates that is sufﬁcient to describe the conﬁguration of the system uniquely.

44 Examples of degrees of freedom

I Pendulum whose pivot can I Ball rolling down an incline move freely in x direction E = 1 mx˙ 2 + 1 Iφ˙2 − mgx sin α 2 2 I Pivot coordinates : (x, 0) ˙ ˙ I But x = Rφ → x = Rφ I Pendulum coordinates :

I The problem is reduced to (x + ` sin θ, −` cos θ) 2 a 1-coordinate variable 1 ˙ 2 1 d E = 2 m1x + 2 m2 dt (x + ` sin θ) + 2 q ≡ x and q˙ ≡ x˙ 1 d 1 1 + 2 m2 dt (−` cos θ) − mg` cos θ

I System has only 1 degree I This system has 2 degrees of freedom : x of freedom : x and θ 45 24.3.3 Constraints

I A system has constraints if its components are not permitted to move freely in 3-D. For example : → A particle on a table is restricted to move in 2-D → A mass on a simple pendulum is restricted to oscillate at an angle θ at a ﬁxed length ` from a pivot

I The constraints are Holonomic if : → The constraints are time independent → The system can be described by relations between general coordinate variables and time → The number of general coordinates is reduced to the number of degrees of freedom

46 24.3.4 Conﬁguration Space

I The conﬁguration space of a mechanical system is an n-dimensional space whose points determine the spatial position of the system in time. This space is parametrized by generalized coordinates, q = (q1 ··· , qn)

I Example 1. A point in space determines where the system is; the coordinates are simply standard Euclidean coordinates: (x, y, z) = (q1, q2, q3)

I Example 2. A rod location x, angle θ - as it moves in 2D space is passes through points (x, θ) in the conﬁguration space

47 25.1 The Lagrangian : simplest illustration The Lagrangian : L = T − U

1 ˙ 2 I In 1D : Kinetic energy T = 2 mx No explicit dependence on x Potential energy U = U(x) No explicit dependence on x˙ 1 ˙ 2 I Deﬁne the Lagrangian in 1D : L = 2 mx − U(x) ∂L ˙ ∂L ∂U I ∂x˙ = mx and ∂x = − ∂x gives force F d ∂L ¨ I Differentiate wrt time : dt ∂x˙ = mx = F

I Hence we get the Euler - Lagrange equation for x : d ∂L ∂L dt ∂x˙ = ∂x

I Now generalize : the Lagrangian becomes a function of 2n variables (n is the dimension of the conﬁguration space). Variables are the positions and velocities

L(q1, ··· , qn, q˙ 1, ··· , q˙ n) d ∂L = ∂L Next we expand on this concept. dt ∂q˙ k ∂qk 48 25.2 The calculus of variations

I Take 2 points A(x0, y0) and B(x1, y1)

I Curve joining them is represented by equation y = y(x) such that y(x) satisﬁes the boundary conditions :

→ y(x0) = y0 , y(x1) = y1

I We want to ﬁnd the function y = y(x) subject to the above conditions which makes the closest path between the points a minimum. (note that this differs from what we are used to. We are not minimizing a set of variables here but a function).

I This is the calculus of variations. A branch of mathematics that deals with functionals as opposed to functions.

49 The calculus of variations, continued (1)

I We assume the unknown function f is a continuously differentiable scalar function, and the functional to be minimized depends on y(x) and at most upon its ﬁrst derivative y 0(x).

I We then wish to ﬁnd the stationary values of the path between points: an integral of the form I = R x1 f (y, y 0, x)dx x0 → f (y, y 0, x) is a function of x, y and y 0 (the ﬁrst derivative of y)

I Consider a small change δy(x) in the function y(x) subject to the conditions that the endpoints are unchanged :

→ δy(x0) = 0 and δy(x1) = 0 0 I To ﬁrst order, the variation in f (y, y , x) is ∂f ∂f 0 2 02 δf = ∂y δy + ∂y 0 δy + O(δy , δy ) 0 d where δy = dx δy

I Thus the variation in the integral I is

R x1 h ∂f ∂f d i δI = δy + 0 δy dx x0 ∂y ∂y dx 50 The calculus of variations, continued (2)

R x1 h ∂f ∂f d i I δI = δy + 0 δy dx x0 ∂y ∂y dx

I Integrate the second term by parts

x1 R 2nd h ∂f i R x1 d ∂f = 0 δy − 0 δy dx term ∂y x0 dx ∂y x0 x h ∂f i 1 The ∂y0 δy term = 0 due to the x0 conditions on the end points

I Hence R x1 h ∂f d ∂f i δI = − 0 δy dx x0 ∂y dx ∂y

I For I to be stationary, δI = 0 for any small arbitrary variation δy(x)

I This is only possible if the integrand vanishes identically

I Hence we get out the Euler-Lagrange Equation

∂f d ∂f ∂y − dx ∂y 0 = 0

51 The calculus of variations, continued (3)

I So far we have used x as the independent variable with a functional f which is a function of (y(x), y 0, x)

I Throughout we could have used other variables, in particular time t and generalized coordinates q1, ··· , qn and derivatives q˙ 1, ··· , q˙ n. The principles would have been the same.

I The integral R t1 I = f [q1(t), ··· , qn(t), q˙ 1(t), ··· , q˙ n(t)] dt t0 must be stationary wrt variations in any one & all of the variables q1(t), ··· , qn(t) subject to the conditions δqi (t0) = δqi (t1) = 0

I We get the n Euler-Lagrange equations for i = 1, ··· , n ∂f − d ∂f = 0 ∂qi dt ∂q˙ i The E-L equations give the conditions for the closest path between points 52 25.3 A sanity check

The shortest distance between 2 points.

I Consider 2 neighboring points on the curve y(x) subject to boundary conditions y(x0) = y0 , y(x1) = y1 p 2 2 I Distance between the points d` = dx + dy p 02 p 02 I d` = 1 + y dx f ≡ 1 + y ∂f d ∂f I The Euler-Lagrange Equation ∂y − dx ∂y 0 = 0

∂f ∂f y 0 d ∂f I = 0 , 0 = √ , 0 = 0 ∂y ∂y 1+y 02 dx ∂y y 0 0 I Hence √ = constant, hence y is constant 1+y 02 → y = mx + c

I We have proved that the shortest distance between 2 points is a straight line !

53 25.4 Fermat’s Principle & Snell’s Law

Fermat : The actual path that a light ray propagating between one point to another will take is the one that makes the time travelled between the two points stationary. Question: at which point (x, 0) will the ray hit the interface between the two media to propagate from A to B?

I Time taken from A to B : 1 1 1 2 2 2 1 2 2 2 t(x) = (x − x1) + y + (x2 − x) + y v1 1 v2 2

I The Euler-Lagrange Equation

∂t d ∂t ∂x − dy ∂x0 = 0 (where the second term = 0)

∂t 1 x−x1 1 x2−x I ∂x = 0 = v 1 − v 1 1 ( − )2+ 2 2 2 ( − )2+ 2 2 [ x x1 y1 ] [ x2 x y2 ]

sin θ1 sin θ2 I − = 0 v1 v2

n1 sin θ1 = n2 sin θ2 Snell’s Law 54 25.5 Hamilton’s principle (Principle of Stationary Action)

I Consider for example a particle of mass m at point (xA, yA) moving under the inﬂuence of a force in the x − y plane. We want to ﬁnd the path that the particle will follow to reach a point (xB, yB).

I Hamilton’s principle: the path that the particle will take from A to B is the one that makes the following functional stationary : R B ˙ ˙ I = A L (q1(t), ··· , qn(t), q1(t), ··· , qn(t)) dt where L is the Lagrangian, I is called the action integral

I Hence the action integral I is stationary under arbitrary variations q1(t), q2(t) ··· which vanish at the limits of integration ie. A and B.

55 26.1 Conjugate momentum and cyclic coordinates

d ∂L ∂L I The E-L equation is = with L = T − U dt ∂q˙ k ∂qk ∂L I Deﬁne conjugate (generalized) momentum : pk = ∂q˙ k Note this is not necessarily linear momentum ! 1 2 ˙2 → eg. simple pendulum L = 2 m` θ + mg` cos θ → ∂L = m`2θ˙ : which is angular momentum ∂θ˙ ∂L I Following on, E-L equation reads p˙ = k ∂qk I If the Lagrangian L does not explicitly depend on qk , the coordinate qk is called cyclic or ignorable

I With no qk dependence : ∂L ∂L = 0 and pk = = constant ∂qk ∂q˙ k The momentum conjugate to a cyclic coordinate is a constant of motion 56 26.2 Example : rotating bead A bead slides on a wire rotating at constant angular speed ω in a horizontal plane

I Polar coordinates v = r˙ ˆr + rθ˙ˆθ

I L = T − U with U = 0 1 ˙2 1 2 2 I L = 2 mr + 2 mr ω

I Single variable qk → r d ∂L ∂L I E-L dt ∂r˙ = ∂r˙ ∂L ˙ d ∂L ¨ ∂r˙ = mr → dt ∂r˙ = mr ∂L 2 ∂r = mrω 2 I E-L → m¨r − mrω = 0 2 Central force Fcentral = mω r ωt −ωt I r = Ae + Be 57 Example : rotating bead continued

What happens if the angular speed is now a free coordinate ? 1 ˙2 1 2 ˙2 I L = 2 mr + 2 mr θ

I Two variables qk → r, θ

I r variable: as before → m¨r − mrθ˙2 = 0 d ∂L ∂L I θ variable: = dt ∂θ˙ ∂θ ∂L 2 I = mr θ˙ ∂θ˙ ∂L I ∂θ = 0 2 ¨ d 2 ˙ I E-L : mr θ = dt mr θ = 0 → Conservation of angular momentum

58 26.3 Example : simple pendulum Evaluate simple pendulum using Euler-Lagrange equation

I Single variable qk → θ

I v = ` θ˙ 1 2 ˙2 I T = 2 m` θ

I U = −mg` cos θ 1 2 ˙2 I L = T − U = 2 m` θ + mg` cos θ ∂L 2 d ∂L 2 I = m` θ˙ → = m` θ¨ ∂θ˙ dt ∂θ˙ ∂L I ∂θ = −mg` sin θ 2 I E-L → m` θ¨ + mg` sin θ = 0 ¨ g → θ + ` sin θ = 0 This is great, but note that the method does not get the tension in the string since59 ` is a constraint (see next slide). 26.3.1 Dealing with forces of constraint For the simple pendulum using Euler-Lagrange equation. The method did not get the tension in the string since ` was constrained. If we need to ﬁnd the string tension, we need to include the radial term into the Lagrangian and to include a potential function to represent the tension:

1 ˙2 I ` → r , add 2 mr , add V (r) 1 ˙2 1 2 ˙2 I L = 2 mr + 2 mr θ + mgr cos θ − V (r) ∂L ˙ d ∂L ¨ I ∂r˙ = mr → dt ∂r˙ = mr ∂L ˙2 ∂V (r) I ∂r = mrθ + mg cos θ − ∂r ∂V (r) n I − ∂r = (−T ) with T in the −ˆr dir . 2 I E-L → m¨r = mrθ˙ + mg cos θ − T

I Reintroduce ¨r = 0 and r = `; v = rθ˙ mv 2 = T − mg cos θ as expected from NII r |{z} | {z } | {z } Tension Weight Centripetal force 60 26.3.2 The Lagrange multiplier method An alternative method of dealing with constraints. Back to the simple pendulum using Euler-Lagrange equation ···

Before : single variable qk → θ. This time take TWO variables x, y but introduce a constraint into the equation. L = T − U 0 1 ˙ 2 ˙ 2 1 2 2 2 I L = 2 m(x +y )+mgy + 2 λ(x +y −` ) λ is the Lagrange multiplier d ∂L0 ∂L0 I = (including λ) dt ∂q˙ i ∂qi x coord. → mx¨ = λx (1) y coord. → my¨ = mg + λy (2) λ coord. → x2 + y 2 − `2 = 0 (3) (which reproduces the constraint) ¨ Tx ¨ Ty Comparing with Newton II : mx = − ` ; my = mg − ` . We see from the NII approach the Lagrange multiplier λ is T proportional to the string tension λ = − ` and introduces force 61 27.1 The Lagrangian in various coordinate systems

I Cartesian coordinates

1 ˙ 2 ˙ 2 ˙ 2 L = 2 m(x + y + z ) − U(x, y, z) Already shown that E-L gives ¨ ∂U ¨ ∂U ¨ ∂U mx = − ∂x ; my = − ∂y ; mz = − ∂z → m ¨r = −OU

I Cylindrical coordinates x = r cos φ ; x˙ = r˙ cos φ − r sin φ φ˙ y = r sin φ ; y˙ = r˙ sin φ + r cos φ φ˙ z = z ; z˙ = z˙ T = 1 m (x˙ 2 + y˙ 2 + z˙ 2) 2 φ is cyclic if U = U(r) only = 1 m[(r 2 cos2 φ + r 2 sin2 φφ˙2 − 2rr˙ cos φ sin φφ˙)] 2 ∂L 2 2 → p = = mr φ˙ +(r 2 sin φ + r 2 cos2 φφ˙2 + 2rr˙ cos φ sin φφ˙) + z˙ 2] φ ∂φ˙ L = 1 m(r˙2 + r 2 φ˙2 + z˙ 2) − U(r, φ, z) → constant angular 2 momentum 62 The Lagrangian in various coordinate systems continued

I Spherical coordinates x = r sin θ cos φ ; x˙ = r˙ sin θ cos φ + r cos θ cos φ θ˙ − r sin θ sin φ φ˙ y = r sin θ sin φ ; y˙ = r˙ sin θ sin φ + r cos θ sin φ θ˙ + r sin θ cos φ φ˙ z = r cos θ ; z˙ = r˙ cos θ − r sin θ θ˙

1 ˙ 2 ˙ 2 ˙ 2 T = 2 m (x + y + z )

1 ˙2 2 ˙2 2 ˙2 T = 2 m (r + r θ + (r sin θ) φ ) + cross terms which all sum to zero

1 ˙2 2 ˙2 2 ˙2 L = 2 m(r + r θ + (r sin θ) φ ) − U(r, θ, φ)

63 27.2 Example 1: the rotating bead

A bead of mass m slides on a frictionless wire which rotates about a vertical axis at an angular velocity ω. The wire is tilted away from the vertical by an angle α. Describe the motion of the bead.

I Use spherical coordinates

I From before : 1 ˙ 2 2 2 2 ˙2 T = 2 m R + R α˙ + (R sin α) φ

I But α˙ = 0 , φ˙ = ω = constant 1 ˙ 2 2 2 2 I T = 2 m R + R ω sin α U = mgR cos α (Take U = 0 at R = 0)

I L = T − U 1 ˙ 2 2 2 2 L = 2 m R + R ω sin α − mgR cos α

64 The rotating bead, continued

1 ˙ 2 2 2 2 I L = 2 m R + R ω sin α − mgR cos α

I Single generalized coordinate R

I E-L equation: d ∂L = ∂L dt ∂R˙ ∂R ∂L 2 2 ∂R = mR ω sin α − mg cos α d ∂L = d mR˙ = mR¨ dt ∂R˙ dt

2 2 I R¨ − R ω sin α = −g cos α −λt +λt I Solution : R = Ae + Be + R0 where λ = ω sin α [ P.I. → R¨ = 0 → R = g cos α ] 0 ω2 sin2 α ˙ 1 If R = 0 at t = 0 , A = B ; then if R = R1 at t = 0, A = B = 2 (R1 − R0) g cos α I If R¨ = 0 → R = R = → circular motion 0 ω2 sin2 α 65 27.3 Example 2: bead on rotating hoop

A vertical circular hoop of radius R rotates about a vertical axis at a constant angular velocity ω. A bead of mass m can slide on the hoop without friction. Describe the motion of the bead.

I Use spherical coordinates again

I From before : 1 ˙ 2 2 ˙2 2 ˙2 T = 2 m R + R θ + (R sin θ) φ

I But R˙ = 0 , φ˙ = ω = constant 1 2 ˙2 2 2 ˙ T = 2 m(R θ + (R sin θ) ω ) (NB. R = 0) ◦ I U = −mgR cos θ (U = 0 at θ = 90 )

I L = T − U 1 2 ˙2 2 2 L = 2 m(R θ + (R sin θ) ω ) + mgR cos θ One single generalized coordinate : θ

66 Bead on rotating hoop, continued 1 2 ˙2 2 2 2 L = 2 m(R θ + R sin θ ω ) + mgR cos θ d ∂L ∂L I E-L equation: = dt ∂θ˙ ∂θ d ∂L d 2 2 I = m R θ˙ = m R θ¨ dt ∂θ˙ dt ∂L 2 2 ∂θ = m R sin θ cos θ ω − mgR sin θ ¨ 2 g → θ = sin θ cos θ ω − R sin θ ¨ 2 2 → θ + ω0 − ω cos θ sin θ = 0 2 g where ω0 = R ¨ 2 I If ω = 0, θ + ω0 sin θ = 0 → SHM, back to pendulum formula

I If ω 6= 0, for equilibrium, a necessary condition : θ¨ = 0 → θ = 0 (stable equilibrium provided ω2R ≤ g ), → θ = π (unstable equilibrium) 2 ω0 g → cos θ = ω2 = ω2R (stable equilibrium about a circle provided ω2R ≥ g ) 67 28.1 Hamilton mechanics

I Lagrangian mechanics : Allows us to ﬁnd the equations of motion for a system in terms of an arbitrary set of generalized coordinates

I Now extend the method due to Hamilton → use of the conjugate (generalized) momenta p1, p2, ··· , pn replace the generalized velocities q˙ 1, q˙ 2, ··· , q˙ n

I This has advantages when some of conjugate momenta are constants of the motion and it is well suited to ﬁnding conserved quantities ∂L I From before, conjugate momentum : pk = ∂q˙ k and E-L equation reads for coordinate k : p˙ = ∂L k ∂qk d ∂L ∂L (since E-L is p˙ k = ( ) = ) dt ∂q˙ k ∂qk

68 The Hamiltonian, continued

I An aside: use rules of I Lagrangian L = L(qk , q˙ k , t) =⇒ partial differentiation: dL ∂L P ∂L dqk ∂L dq˙ k I dt = ∂t + k ∂q dt + ∂q˙ dt k k I If f = f (x, y, z) ∂L P ∂L ∂L ∂ ∂ dy ∂ = + k ( q˙ k + ˙ q¨k ) df f f f dz ∂t ∂qk ∂qk I dx = ∂x + ∂y dx + ∂z dx ∂L ∂L I Conjugate momentum deﬁnition : pk = , p˙ k = ∂q˙ k ∂qk dL ∂L P ˙ ˙ ¨ I Therefore dt = ∂t + k (pk qk + pk qk ) | {z } d ˙ dt (pk qk ) d X ˙ ∂L dH ∂L I dt (L − pk qk ) = ∂t dt = − ∂t k | {z } -H P I Deﬁne Hamiltonian H = k pk q˙ k − L

I If L does not depend explicitly on time, H is a constant of motion

69 28.2 The physical signiﬁcance of the Hamiltonian P I From before : H = k pk q˙ k − L ∂L ∂L I Where conjugate momentum : pk = , p˙ k = ∂q˙ k ∂qk 1 ˙ 2 ˙ 2 ˙ 2 I Take kinetic energy T = 2 m x + y + z 1 ˙ 2 ˙ 2 ˙ 2 I L = 2 m x + y + z − U(x, y, z) P ˙ 1 ˙ ˙ ˙ ˙ ˙ ˙ I H = k pk qk − L = 2 m (2x.x + 2y.y + 2z.z) − (T − U) = 2T − (T − U) = T + U = E → total energy dH ∂L I From before dt = − ∂t dH → If L does not depend explicitly on time dt = 0 → energy is a constant of the motion

I Can show by differentiation : Hamilton Equations → ∂H = q˙ ; ∂H = −p˙ ∂pk k ∂qk k If a coordinate does not appear in the Hamiltonian it is cyclic or ignorable 70 28.3 Example: re-visit bead on rotating hoop First take the case of a free (undriven) system

1 2 ˙2 2 2 ˙2 I L = 2 m (R θ + R sin θ φ ) + mgR cos θ P ∂L I H = pk q˙ k − L ; pk = k ∂q˙ k ∂L 2 2 2 I p = = m R θ˙ ; p = mR sin θ φ˙ θ ∂θ˙ φ 2 2 2 2 2 I H = m R θ˙ + mR sin θ φ˙ − L 1 2 ˙2 2 2 ˙2 = 2 m R θ + R sin θ φ −mgR cos θ → H = T + U = E L does not depend explicitly on t , H, E conserved → Hamiltonian gives the total energy

Hamilton Equations : q˙ = ∂H ; p˙ = − ∂H k ∂pk k ∂qk ˙ ∂H → pφ = − ∂φ = 0 (ignorable) 2 2 → pφ = mR sin θ φ˙ = Jz = constant of the motion 71 Example continued

Now consider a DRIVEN system - hoop rotating at constant angular speed ω 1 2 ˙2 2 2 2 I L = 2 m (R θ + R ω sin θ) + mgR cos θ P ∂L I H = pk q˙ k − L ; pk = k ∂q˙ k ∂L 2 I p = = m R θ˙ ; a single coordinate θ θ ∂θ˙ 2 2 I H = m R θ˙ − L 1 2 ˙2 2 2 2 = 2 m R θ − R ω sin θ −mgR cos θ dH ∂L 1 2 ˙2 2 2 2 I dt = − ∂t I E = 2 m (R θ + R ω sin θ) − mgR cos θ IH is a constant Hence E = H + mR2ω2 sin2 θ of the motion, → E (= T + U) 6= H E is not const. So what’s different ? In this case the hoop has been forced to rotate at an angular velocity ω. External energy is being supplied to the system. 72 28.4 Noether’s theorem The theorem states : Whenever there is a continuous symmetry of the Lagrangian, there is an associated conservation law.

I Symmetry means a transformation of the generalized coordinates qk and q˙ k that leaves the value of the Lagrangian unchanged.

I If a Lagrangian does not depend on a coordinate qk (ie. is cyclic) it is invariant (symmetric) under changes qk → qk + δqk ; the ∂L corresponding generalized momentum pk = is conserved ∂q˙k 1. For a Lagrangian that is symmetric under changes t → t + δt, P ∂L the total energy H is conserved → H = k ∂q˙ q˙ − L 2. For a Lagrangian that is symmetric under changes r → r + δr, the linear momentum p is conserved 3. For a Lagrangian that is symmetric under small rotations of angle θ → θ + δθ about an axis nˆ such a rotation transforms the Cartesian coordinates by r → r + δθ nˆ × r , the conserved quantity is the component of the angular momentum J along the nˆ axis 73 29.1 Re-examine the sliding blocks using E-L

A block of mass m slides on a frictionless inclined plane of mass M, which itself rests on a horizontal frictionless surface. Find the acceleration of the inclined plane.

I Reduce the problem to two generalized coordinates, x and s

I Motion of the inclined plane : 1 ˙ 2 TM = 2 Mx

I Motion of the block : 1 ˙ 02 ˙ 02 Tm = 2 m(x + y ) where 0 0 I x = x + s cos α ; y = −s sin α 0 0 I x˙ = x˙ + s˙ cos α ; y˙ = −s˙ sin α 1 ˙ ˙ 2 1 ˙ 2 I Tm = 2 m (x + s cos α) + 2 m(s sin α) 1 ˙ 2 1 ˙ 2 ˙ ˙ I T = Tm + TM = 2 (m + M)x + 2 m s + 2xs cos α

I U = −m g s sin α

74 Sliding blocks, continued

I Lagrangian L = T − U 1 ˙ 2 1 ˙ 2 ˙ ˙ I L = 2 (m + M)x + 2 m s + 2xs cos α + mgs sin α

I 2 generalized coordinates → x and s ∂L d ∂L I The E-L equation − = 0 ∂qk dt ∂q˙ k ∂L d ∂L d ˙ ˙ I E-L for x : ∂x = 0 ; dt ∂x˙ = dt [(m + M)x + ms cos α] → (m + M)x¨ + ms¨ cos α = 0 (1) ∂L d ∂L d ˙ ˙ I E-L for s : ∂s = mg sin α ; dt ∂s˙ = dt [m (s + x cos α)] → s¨ + x¨ cos α = g sin α (2)

I Rearranging (1) & (2) x¨ = −g sin α cos α ; s¨ = g sin α(1+M/m) sin2 α+M/m sin2 α+M/m

I From (1) Mx˙ + m(x˙ + s˙ cos α) = const. → Mx˙ + mx˙ 0 = const. Conservation of momentum. 75 29.2 Normal modes of coupled identical springs

Coupled identical springs mounted horizontally. x1 and x2 measure displacements from the respective equilibrium positions. Assume the springs are unstretched at equilibrium.

I The problem has two generalized coordinates, x1 and x2 1 ˙ 2 1 ˙ 2 T = 2 Mx1 + 2 mx2 1 2 1 2 1 2 U = 2 kx1 + 2 kx2 + 2 k (x2 − x1)

I L = T − U

∂L d ∂L I E-L equation for x1 : − = 0 ∂x1 dt ∂x˙1 ∂L d ∂L I = −kx1 + k(x2 − x1) = k(x2 − 2x1) ; = Mx¨1 ∂x1 dt ∂x˙1

→ Mx¨1 = k(x2 − 2x1); mx¨2 = k(x1 − 2x2) M x¨1 2k −k x1 I = − m x¨2 −k 2k x2 76 Coupled identical springs, continued

M x¨1 2k −k x1 I = − (1) m x¨2 −k 2k x2 x1 a1 I SHM solutions = exp( i ω t) x2 a2

I Substitute into (1) M 0 a 2k −k a −ω2 1 = − 1 0 m a2 −k 2k a2 | {z } | {z } M K 2 −1 a1 a1 I Putting ω = λ → M K = λ a2 a2

I Eigenvalue equation; homogeneous solutions

I Etc etc ....

77 29.3 Final example: a rotating coordinate system

I Lagrangian of a free particle : 1 2 L = 2 mr˙ , r = (x, y, z) (with U = 0)

I Measure the motion w.r.t. a coordinate system rotating with angular velocity ω = (0, 0, ω) about the z axis. 0 0 0 0 I r = (x , y , z ) are coordinates in the rotating system

 x 0   cos ωt sin ωt 0   x  0 I  y  =  − sin ωt cos ωt 0   y  z0 0 0 1 z

I Take the inverse :  x   cos ωt − sin ωt 0   x 0   y  =  sin ωt cos ωt 0   y 0  z 0 0 1 z0

I Substitute these expressions into the Lagrangian above → ﬁnd L in terms of the rotating coordinates 78 A rotating coordinate system, continued

1 ˙ 0 0 2 ˙ 0 0 2 ˙ 02 I L = 2 m (x − ω y ) + (y + ω x ) + z 1 0 0 2 = 2 m(˙r + ω × r ) in the general case

I In this rotating frame, we can use Lagrange equations to derive the equations of motion. Taking derivatives, we have ∂L 0 0 I ∂r0 = m [˙r × ω − ω × (ω × r )] ∂ ∂ ∂ ∂ where ∂r0 = ∂x0 , ∂y 0 , ∂z0

d ∂L d 0 0 0 0 → dt ∂r˙ 0 = m dt (˙r + ω × r ) = m (¨r + ω × r˙ )

I So the Lagrange equation becomes d ∂L ∂L 0 0 0 dt ∂r˙ 0 − ∂r0 = m [ ¨r + ω × (ω × r ) + 2 ω × r˙ ] = 0 |{z} | {z } | {z } radial force Centrifugal force Coriolis force

79 A rotating coordinate system, continued

I Centrifugal and Coriolis forces are examples of “ﬁctitious forces” : → called “ﬁctitious” since they are a consequence of the reference frame, rather than any interaction. The forces do not exist in an inertial frame.

I The centrifugal force 0 Fcent = m ω × (ω × r ) points outwards in the plane perpendicular to ω with 2 0 magnitude m ω |r⊥| (⊥ is the projection perpendicular to ω )

0 I The Coriolis force Fcor = 2m ω × r˙ acts in a direction perpendicular to the rotation axis ω and to the velocity of the body in the rotating frame 80 A rotating coordinate system, continued

I Coriolis force responsible for the circulation Australia of oceans and the atmosphere.

I A projectile thrown in the northern hemisphere rotates in a clockwise direction

I A projectile thrown in the southern hemisphere rotates in an anti-clockwise direction.

I For a particle moving along the equator, ω ⊥ r˙ 0 , the Coriolis force tends to zero → Iceland no effect on the projectile

I The Coriolis force is responsible for the formation of hurricanes. These rotate in different directions in the northern and southern hemisphere. They never form within 500 miles of the equator where the Coriolis force is too weak.

81 THE END