LECTURES 1 - 10 INTRODUCTION TO CLASSICAL MECHANICS

3.1.2 Kepler’s third lawHow does the period of an orbiting mass depend on its radius?IVariables: period P, central mass M0 , orbitradius r , Gravitational constant GIGuess the form: let P k M0a r b G c(k is a dimensionless constant)IDimensions of G (MLT 2 ).L2 M 2IT 1 M a Lb (M 1 L3 T 2 )c M (a c) Lb 3c T 2cIIGmM0r2a c 0, b 3c 0, 2c 1Iv a 1/2, c 1/2, b 3/2IP2 Compare terms: 1/2 3/2P k M034rG 1/2 P2 k2 3rGM0 mv 2r2πrP4π 2 3GM0 r k 2 4π 2

3.1.3 The range of a cannon ballA cannon ball is fired with Vy upwards and Vx horizontally, assume no air resistance.I Variables: Vx , Vy , distance travelled alongx (range) R, acceleration due to gravity gIFirst with no use of directed length dimensionsILet R kVxa Vyb g c .(k is a dimensionless constant)IDimensionally L (L/T )a b (L/T 2 )cICompare terms:a b c 1 and a b 2c 0, which leaves oneexponent undetermined.IINow use directed length dimensions , then Vx will bedimensioned as Lx /T , Vy as Ly /T , R as Lx and g asLy /T 2Ix vx tIy vy t 21 gt 2 0The dimensional equation becomes:Lx (Lx /T )a (Ly /T )b (Ly /T 2 )c t a 1, b 1 and c 1.IR k35vx vygx 2vyg2vx vyg

3.1.4 Example of limitations of the methodIILet y f (x1 , x2 , . . . xn ) where x1 , x2 , . . . xn haveindependent dimensionsHowever in general y (x1a x2b . . . x n ) φ(u1 , . . . uk ) where uiare dimensionless variablesExtend to how the period of a rigidpendulum depends on length pivot to CM.IIIn actual fact P P(g, , m, I) where Iis the moment of inertia[I] ML2 can define u T q gIm 2φ(u)i.e. Equation is not reproduced36T 2πqImg

3.2 Newton’s Laws of motion37INI: Every body continues in a state of rest or in uniformmotion (constant velocity in straight line) unless acted uponby an external force.INII: The rate of change of momentum is equal to theapplied force; where the momentum is defined as theproduct of mass and velocity (p mv). [i.e. the appliedforce F on a body is equal to its mass m multiplied by itsacceleration a.]INIII: When one body exerts a force on a second body, thesecond body simultaneously exerts a force equal inmagnitude and opposite in direction on the first body [i.e.action and reaction forces are equal in magnitude andopposite in direction.]

3.3 Frames of reference38IA frame of reference is an environment which is used toobserve an event or the motion of a particle.IA coordinate system is associated with the frame toobserve the event (eg the body’s location over time).IThe observer is equipped with measuring tools (eg rulersand clocks) to measure the positions and times of events.IIn classical mechanics, time intervals between events isthe same in all reference frames (time is absolute).IIn relativity, we will need to use space-time frames.IA reference frame in which NI is satisfied is called aninertial reference frame.

Inertial reference framesA frame in which Newton’s first law is satisfied:IIDeep spaceThe Earth? [Only in circumstances where we can ignore gravity& the spin of the Earth.]Principle of Relativity : The laws of Physics are the same in allinertial frames of reference.At t 0, x 0, x 0 0 and S and S 0 are coincident.Galilean Transformation of coordinates:I x 0 x v0 t, y 0 y , z 0 z, t 0 t39IVelocity of a body v in S; velocity measured in S 0v 0 v v0IAcceleration measured in S 0IHence F 0 F (consistent with the principle of relativity)a0 a

3.4 The Principle of Equivalence40IThe Principle of Equivalence dictates that m m .IInertial mass Gravitational massIThis may seem obvious, but it was not an original postulateof Newton

4.1 Newton’s Second LawThe rate of change of momentum of a body is equal to theapplied force on the body.dpdt ma where p mvIF IIn components: (Fx , Fy , Fz ) m(ax , ay , az )IAssuming constant mass, we can define the equation ofmotion in 1D :IF d(mv )dt2 m ddt 2xWe require two initial conditions for a unique solution: e.g.v v0 at t 0 and x x0 at t 0We shall later solve the EOM for three examples:(i) F constant, (ii) F v , (iii) F x41

4.2 Newton’s Third LawAction and reaction forces are equal in magnitude and oppositein direction.Compressed springElectrostatic interactionF12 F2142

Conservation of momentumCompressed springdP1dtIF12 m1 a1 IF12 F21 ITherefore (P1 P2 ) constantddt (P1andF21 m2 a2 dP2dt P2 ) 0 (Newton III)In an isolated system, the total momentum is conserved.43

Newton II : Example 1. E.O.M. under constant forceHow fast should we accelerate the triangular wedge to keep theblock m stationary on the wedge?Forces on block:Forces on wedge:IHorizontal: Fex Fi sin θ MAxIHorizontal: Fi sin θ maxIVertical: R Fi cos θ Mg 0IVertical: Fi cos θ mg mayFor block to remain at the same place Ax ax and ay 044mgcos θandax g tan θ AxIFi IHence Fex Mg tan θ mg tan θ (m M)g tan θ

Example constant force, continuedWhat is the internal force that the blocks apply on each otherand the reaction force by the ground on M?From before:45mgcos θIFi IR Fi cos θ Mg 0IHence: R Fi cos θ Mg (m M)g

Example 2. Force proportional to velocitySolve the equation of motion for the case F βv (β 0)with x x0 and v v0 at t 0IIIIIIIm dvdt βvβdvdt αv where α mR v dvRt αtv0 v α 0 dt v v0 eRxRtRt αt dtv dxdt x0 dx 0 vdt 0 v0 ex x0 vα0 e αt vα0x x0 vα0 (1 e αt )When t , x x0 vα046

Example 3. Force proportional to position: simpleharmonic oscillator2Solving the equation of motion for the case F m ddt 2x kx2Im ddt 2x kmx 0 ; trial solution x A cos ωt B sin ωt ẋ Aω sin ωt Bω cos ωt ; ẍ Aω 2 cos ωt Bω 2 sin ωtkmIẍ ω 2 x ω 2 IAlternatively x x0 cos(ωt φ) (or x x0 Re[ei(αt φ) ] )IExpand : x x0 (cos(ωt) cos φ sin(ωt) sin φ)A x0 cos φ ; B x0 sin φ x02 A2 B 2 ; tan φ B/AI47x0 amplitude, φ phase, ω angular frequency (T 2πω )

III48x x0 cos(ωt φ)qkω mφ ω t

4.3 Energy conservation in one dimensionWork done on a body by a force FR x2R x2 dvI W F(x)dx mx1x1 dt dxI49dxWe can write: dvdt dx dt dv vdvR x2R v2hence x1 F (x)dx m v1 vdv 21 m(v22 v12 ) T2 T1INow introduce an arbitrary reference point x0R x2R x2R x1x1 Fdx x0 Fdx x0 Fdx defines a conservative forceRxRxhence T2 [ x02 Fdx] T1 [ x01 Fdx]IWe define the potential energy U(x) at a point x :RxU(x) U(x0 ) x0 Fdx and henceT2 U2 T1 U1 (total energy PE KE conserved)INote the minus sign. The potential energy (relative to areference point) is always the negative of the work done bythe force F (x) dUdx

5.1 Conservative forcesZbF . dr U(a) U(b)Wab aFor a conservative field of force, the work done depends onlyon the initial and final positions of the particle independent ofthe path.The conditions for a conservative force (all equivalent) are:IThe force is derived from a (scalar) potential function:F(r) U F (x) dUdx etc.IThere is zero net work by the forcewhen moving a particleHaround any closed path: W c F . dr 0IIn equivalent vector notation F 0For any force: Wab 12 mvb2 21 mva2Only for a conservative force: Wab U(a) U(b)50

Conservative force: example 1. Constant accelerationConsider a particle moving under constant force (in 1-D).IF ma. Say at t 0 x x0 and v v0IT2 U2 T1 U1 (the total energy is conserved)II11222 mv (ma)x 2 mv0v 2 v02 2a(x x0 )III51 (ma)x0 constantGravitational potential energyRyU( y) y12 F (y )dyRyU( y) y12 ( mg)dy mg(y2 y1 )

Example 2. Simple harmonic oscillator2Equation of motion: F m ddt 2x kxITotal energy: E T (x) U(x) 12 mẋ 2 0( kx)dx kx 22kx 22Check conservation of energy:EOM : mẍ kx 0 [multiply by ẋ] mẍ ẋ kx ẋ 0 I0Fdx RxPotential energy: U(x) I52RxI1d22 m dt (ẋ )d 12 k dt(x 2 ) 0Integrate wrt t: 12 mẋ 2 12 kx 2 constant i.e. energyconserved.

SHM potential energy curve53IE U(x) 12 mv 2IThe particle can only reach locations x that satisfy U E

Example 3. Minimum approach of a chargeA particle of mass m andcharge Q1 starts fromx with velocity v0 . Itapproaches a fixed charge Q.Calculate its minimum distanceof approach xmin .IQQ1( ve direction)Force on charge Q1 : F (x) 4π x20IPotential energy at point x: U(x) Rx F (x)dxQQ1 4π 0x(where PE 0 at x )54IConservation of energy :IMin. dist. when v 0:122 mv0122 mv0 0 12 mv 2 U(x)QQ14π 0 xmin xmin QQ12πm 0 v02

5.2 Potential with turning points55IU is a maximum: unstable equilibriumIU is a minimum: stable equilibrium

5.2.1 Oscillation about stable equilibriumIIFor SHM : U(x) 12 k (x x0 )2Taylor expansion about x0 : 2 dU1 d UU(x) U(x0 ) (x x0 ) 2!(x x0 )2 . . .dx x x0dx 2 x x0 {z } {z} 056 k

Example: The Lennard-Jones potentialThe Lennard-Jones potentialdescribes the potential energybetween two atoms in a molecule:U(x) [(x0 /x)12 2(x0 /x)6 ]( and x0 are constants and x isthe distance between the atoms).Show that the motion for small displacements about the minimum issimple harmonic and find its frequency.hi 1 d 2UI U(x) U(x ) dU(x x) (x x0 )2 . . .002dx x x2! dx0IU(x0 ) [(x0 /x)IdU(x)dx x x0212x x0 2(x0 /x)6 ]x x0 12 [ x10 (x0 /x)13 17x0 (x0 /x) ]x x0 0 as expected.Id U(x) dx 2 x x0IHence U(x) I172 F (x) dUdx 2 2( x 2 )(x x0 ) k (x x0 ) SHM about x0IAngular frequency of small oscillations : ω 2 12 [13(x0 /x)14x0272 (x2!x02 7(x0 /x)8 ]x x0 72 x02 x0 )2057km 72 mx02

5.2.2 Bounded and unbounded potentials58IBounded motion : E E1 : x constrained x1 x x2IUnbounded motion : E E2 : x unconstrained at high xx0 x

6.1 Lab & CM frames of referenceFrom hereon we will deal with 2 inertial frames:II59The Laboratory frame: this is the framewhere measurements are actually madeThe centre of mass frame: this is the framewhere the centre of mass of the system is atrest and where the total momentum of thesystem is zero

6.2 Internal forces and reduced massIInternal forces only:F12 m1 r 1 ; F21 m2 r 2Then r 2 r 1 F21m2 F12m1NIII : F21 F12 FintIIIDefine r r2 r1 ṙ ṙ2 ṙ1 11Fint m1 m2 r̈11µ m1m1 m2m1 m2 is1m2 Fint µr̈Define µ the reduced mass of the systemThis defines the equation of motion of the relative motion of theparticles under internal forces, with position vector r & mass µ60

6.3 The Centre of MassThe centre of mass (CM) is the point where the mass-weightedposition vectors (moments) relative to the point sum to zero ;the CM is the mean location of a distribution of mass in space.ITake a system of n particles,each with mass mi locatedat positions ri , the positionvector of the CM is definedby:Pni 1 mi (ri rcm ) 0ISolve for Prcm :rcm M1 ni 1 mi riPwhere M ni 1 mi61

Example : SHM of two connected masses in 1DSHM between two masses m1 and m2 connected by a springIx x2 x1 ; Natural length LIFint k (x L) µ ẍ(µ Im1 m2m1 m2 reduced mass)ẍ µk (x L) 0Solution: x x0 cos(ωt φ) Lqwhere ω µkWith respect to the CM:m1 x1 m2 x2MIxCM Ix10 x1 xcm Ix20 x2 xcm where M m1 m2Mx1 m1 x1 m2 x2MMx2 m1 x1 m2 x2MEg. take m1 m2 m ω 62 mM2 x q2kmm1 xM;x10 21 x, x20 12 x

6.3.1 CM of a continuous volumeIf the mass distribution is continuous with density ρ(r) inside avolume V , then:PnIi 1 mi (ri rcm ) 0becomesRV ρ(r)(r rcm )dV 0where dm ρ(r)dVISolve for rcmRrcm M1 V ρ(r) r dVwhere M is the total mass inthe volume63

Example: the CM of Mount RanierMount Ranier has approximately the shape of a cone (assumeuniform density) and its height is 4400 m. At what height is the centreof mass?We have cylindrical symmetry - just need to consider the y direction.Integrate from top (y 0) to bottom (y h)RhIycm 0ydmMRh R0 h0ydmdmdm ρ(πr 2 )dy ρ(πy 2 tan2 θ)dyRhIycm yρ(πy 2 tan2 θ)dyR0 h220 ρ(πy tan θ)dyRhycm R0h0y 3 dyy 2 dy 3h44h3 3h4(measured from the top)I64h 4400m ycm 1100mabove the base

6.3.2 Velocity in the Centre of Mass frameIThe position of the centre ofmass is givenP by:rcm M1 ni 1 mi riPwhere M ni 1 miIThe velocity of the CM:Pvcm ṙcm M1 ni 1 mi ṙiIIn the Lab frame, total momentum ptot :Pptot ni 1 mi ṙi MvcmHence the total momentum of a system in the Lab frameis equivalent to that of a single particle having a mass Mand moving at a velocity vcm65

6.3.3 Momentum in the CM framevcm ṙcm Pnm ṙi 1P i ii mi Pnm vi 1P i ii miIVelocity of the CM:IVelocity of a body in the CM relative to LabIThe total momentum of the system of particles in the CM:IPv0i vi vcmPPp0i i mi v0i i mi (vi vcm )PPPPPmv i mi vi i mi Pj mj j j i mi vi j mj vj 0ijHence the total momentum of a system of particles in theCM frame is equal to zeroI66In addition, the total energy of the system is a minimumcompared to all other inertial reference frames.

6.3.4 Motion of CM under external forcesIIForce on particle i: mi r̈i Fi ext Fi intnnnXXXPextmi r̈i Fi Fi int ni Fi ext i {z }all massesP i {z }external forcesinternal forces zeromi riMIrCM IPwhere M i miP m r̈r̈CM i Mi iP M r̈CM i Fi exti i {z }The motion of the system is equivalent to that of a singleparticle having a mass M acted on by the sum of external forces(The CM moves at constant velocity if no external forces)67

6.4 Kinetic energy and the CMILab kinetic energy :TLab 12Pmi v2i ; v0i vi vCMwhere v0i is velocity of particle i in the CMIITLab But12PPmi v02i mi v0iPmi v0i · vCM 02 TLab TCM 12 MvcmThe kinetic energy in the Labframe is equal to the kineticenergy in CM the kineticenergy of the CM68Pmi v2CMmi v0i· M vCM 0{z } MP· vCM 12

7.1 Two-body collisions - general conceptsNo external forces. Collision via massless springs or other force type.69Iti : collision starts. Allenergy is kinetic.Ti 21 m1 u12 12 m2 u22It : collision peaks. Somekinetic is converted intopotential (of the spring).E 21 m1 x 1 2 21 m2 x 2 2 EintItf : collision ends. Allenergy is kinetic again.Tf 12 m1 v12 12 m2 v22 ;Ti Tf E ( inelastic)

7.1.1 Momentum exchange and impulsedpDuring collision: internal force causes change of momentum F dtIAt ti : total momentump p1 p2 m1 u1 m2 u2IAt t : m1 dp1 F12 dtm2 dp2 F21 dtII70At tf : m1 v1 and m2 v2RtImpulse p1 I1 ti f F12 dtRt p2 I2 ti f F21 dtISince F12 F21 0 p1 p2 0IMomentum conserved.

7.1.2 An off-axis collision in 2DIII71Impulse is along line ofcentresRt p1 ti f F12 dtRt p2 ti f F21 dtRtv1 m11 ti f F12 dt u1Rtv2 m12 ti f F12 dt

7.2 Elastic collisions in the Lab frameBeforeAfterConservation of momentum: m1 u1 m2 u2 m1 v1 m2 v2Conservation of energy:122 m1 u1 12 m2 u22 12 m1 v21 12 m2 v22[Note that the motion is in a plane, and the 2D representationcan be trivially extended into 3D by rotation of the plane].72

7.2.1 Elastic collisions in 1D in the Lab frameIMomentum : II122 m1 u1 122 m2 u2 (1)(2)122 m1 v1 12 m2 v22m1 (v1 u1 )(v1 u1 ) m2 (u2 v2 )(u2 v2 )Divide (2) & (3) : (v1 u1 ) (u2 v2 ) (u1 u2 ) (v2 v1 ) 73m1 (v1 u1 ) m2 (u2 v2 )Energy : m1 u1 m2 u2 m1 v1 m2 v2(3)(4)Relative speed before collision Relative speed after

7.2.2 Special case in 1D where target particle is at restIu2 0 ; From (1) & (4) :m1 u1 m1 v1 m2 (u1 v1 )Iv1 ISimilarly :m1 u1 m1 (v2 u1 ) m2 v2Iv2 (m1 m2 )u1m1 m22m1 u1m1 m2v1 m1 m2m1 m2 u1and v2 2m1m1 m2 u1Special cases:74Im1 m2 : v1 0, v2 u1(complete transfer of momentum)Im1 m2 : Gives the limits v1 u1 , v2 2u1(m2 has double u1 velocity)Im1 m2 : Gives the limits v1 u1 , v2 0(“brick wall” collision)

Example: Newton’s cradleConsider here just 3 ballsIIf the balls are touching, the most general case is:Momentum after collision : mu mv1 mv2 mv3Energy after collision :122 mu 21 mv12 12 mv22 12 mv322 equations, 3 unknownsIThe obvious solution: v1 v2 0, v3 uIBut other solution(s) possible:Momentum : mu 31 mu 32 mu 23 muEnergy :I75122 mu 1218 mu 4218 mu 4218 muSo why does the simple solution always prevail?

7.2.3 Collision in 2D : equal masses, target at restm1 m2 m , u2 0IMomentum: mu1 mv1 mv2 u1 v1 v2Squaring u21 v21 v22 2v1 .v2122 mu1 21 mv21 21 mv22 u21 v21 v22IEnergy:IHence 2v1 .v2 0π290 EITHER v1 0 & v2 u1 OR (θ1 θ2 ) I76Either a head-on collision or opening angle is

Relationship between speeds and anglesm1 m2 m , u2 0IIIFrom cons. mom. (v2 )2 (u1 v1 )2 v22 v12 u12 2u1 v1 cos θ1Energy: u12 v12 v22 v22 u12 v12Equate : 2v12 2u1 v1 cos θ1cos θ1 Iv1u1and by symmetrycos θ2 v2u1Note we can also do this via components of momentum : u1 v1 cos θ1 v2 cos θ2 and v1 sin θ1 v2 sin θ2 (u1 v1 cos θ1 )2 v22 cos2 θ2 and v12 sin2 θ1 v22 sin2 θ2 Add : v22 v12 sin2 θ1 u12 2u1 v1 cos θ1 v12 cos2 θ1 Gives : v22 v12 u12 2u1 v1 cos θ177

8.1 Elastic collisions in the CM frameLab frame:Centre of mass frame (zero momentum frame)78IConservation of momentum in CM:m1 u0 1 m2 u0 2 0 ; m1 v0

1.2 Book list II Introduction to Classical Mechanics A P French & M G Ebison (Chapman & Hall) I Introduction to Classical Mechanics D. Morin (CUP) (good for Lagrangian Dynamics and many examples). I Classical Mechanics : a Modern Introduction, M W McCall (Wiley 2001) I Mechanics Berkeley Physics Course Vol I C Kittel e