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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 ! 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 ! 1; 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 ! 1. I h is known as the impact parameter I Angular momentum J = m r × v ! jJj = mvr sin γ = mv0h (as r ! 1) 1 2 ! Total energy E = 2 mv0 (again as r ! 1) 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 θ1 (1) I By symmetry, integrated change in vy = 0 : ∆vy = 0 2 _ R +1 R +1 mr θ I Change in ∆p : m∆v = F dt = F ( ) dt x x −∞ x −∞ x J ! m∆v = ( m ) R +θ1 F r 2dθ | {z } x J −θ1 x = 1 α α I F = −( )^r ! = − θ But r 2 Fx r 2 cos mα R θ1 ! m∆vx = −2( J ) 0 cos θdθ 2α ! ∆vx = −( J ) sin θ1 (2) 2α I From (1) & (2) ! −( J ) sin θ1 = 2v0 cos θ1 Jv0 I tan θ1 = − α ! θ1 + β = π ; φ + 2β = π φ π φ π φ I θ1 = 2 + 2 ; tan θ1 = 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 ! 1 ; cos θ1 = − e φ π φ 1 I θ1 + β = π ; φ + 2β = π ! 2 = θ1 − 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 ! 1. 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 ! = ! z^ I Definition of ! for circular motion : r_ = ! × r I Total angular momentum of the system of particles about O J = r1 × (m1v1) + r2 × (m2v2) I v1 = ! × r1 v2 = ! × r2 I Since ri ? vi 2 2 J = (m1 r1 + m2 r2 ) ! I J = I! (J is parallel to !) 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 ! Now consider the same system but with rotation tilted wrt rotation axis by an angle φ.
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