CP1 REVISION LECTURE 1 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford TT 2017

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1 CP1 REVISION LECTURE 1 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford TT

2 OUTLINE : CP1 REVISION LECTURE 1 : INTRODUCTION TO CLASSICAL MECHANICS 1. Force and work 1.1 Newton s Laws of motion 1.2 Work done and conservative forces 2. Projectile motion 2.1 Constant acceleration 2.2 Resistive force F R v 2.3 Resistive force F R v 2 3. Rocket motion 3.1 The rocket : vertical launch 4. Two-body collisions 4.1 The Centre of Mass frame 4.2 Two-body elastic collision in 1D : Lab to CM system 4.3 Solving collision problems in the CM frame 4.4 Inelastic collisions 2

3 Outline of revision lectures Three revision lectures: Today: Force and work Projectile motion Rocket motion Two-body collisions Tomorrow: Central forces Effective potential Circular motion and orbits Tuesday Week 2: Rotational motion Lagrangian mechanics 3

4 1. Force and work 1.1 Newton s Laws of motion NI: Every body continues in a state of rest or in uniform motion (constant velocity in straight line) unless acted upon by an external force. NII: The rate of change of momentum is equal to the applied force: F = ma NIII: Action and reaction forces are equal in magnitude and opposite in direction. Problems of particle motion involve solving the equation of motion in 3D: F = m dv 4

5 Work done from A B 1.2 Work and conservative forces For any conservative force: W ab = b a F dx = 1 2 mv 2 b 1 2 mv 2 a W ab = U(a) U(b) For a conservative field of force, the work done depends only on the initial and final positions of the particle independent of the path. Equivalent definitions: The force is derived from a (scalar) potential function: F(r) = U F(x) = du dx etc. There is zero net work by the force when moving a particle around any closed path: W = c Fdx = 0 In equivalent vector notation F = 0 5 For any force: W ab = 1 2 mv b mv a 2 For a conservative force: W ab = U(a) U(b) If these are different, energy is dissipated to the environment

6 2. Projectile motion 2.1 Constant acceleration in 2D, no resistive force a = dv = constant v v 0 dv = t 0 a v = v 0 + at r 0 dr = t 0 (v 0 + at) r = v 0 t at2 Under gravity: a = gŷ a x = 0; a y = g vx = v 0 cos θ x = (v0 cos θ)t vy = v 0 sin θ gt y = (v0 sin θ)t 1 2 gt2 Trajectory: y = (tan θ)x g (sec 2 θ)x 2 2v0 2 6

7 2.2 Resistive force F R v Example of body falling vertically downwards under gravity with air resistance velocity. v = 0 at x = 0 and t = 0 Equation of motion: m dv = mg βv v 0 dv g αv = t [ 1 α log e(g αv) ] v g αv g 0 where α = β m 0 = t = exp ( αt) v = g α (1 exp( αt)) Calculate distance travelled x = As t, v t g 0 α (1 exp( αt)) Terminal velocity g α 7

8 2.3 Resistive force F R v 2 Body falls vertically downwards under gravity with air resistance [velocity] 2, v = 0, x = 0 at t = 0 Equation of motion: m dv = mg βv 2 Terminal velocity when dv = 0 : v T = Equation of motion becomes dv v Integrate dv 0 g(1 v 2 /vt) = t 2 0 Standard integral : mg β 1 1 z 2 dz = 1 2 log e = g ( ) 1 v 2 /vt 2 ( 1+z 1 z ) [ ( )] v vt 2g log 1+v/vT e 1 v/v T = t 1+v/v T 1 v/v 0 T = exp(t/τ), where τ = v T 2g (1 v v T ) = (1 + v v T ) exp( t τ ) Velocity as a function of time: [ ] v = v 1 exp( t/τ) T 1+exp( t/τ) 8

9 Velocity as a function of distance F R v 2 Equation of motion: dv = g ( 1 v 2 /vt 2 ) Write dv = dv dx dx = v dv dx v 0 v dv g(1 v 2 /vt) = x 2 0 dx [ v T 2 2g log ( e 1 v 2 /vt 2 ) ] v = x 0 ( 1 v 2 /v 2 T ) = exp ( x/xt ), where x T = v 2 T 2g v 2 = v 2 T [1 exp ( x/x T )] 9

10 A body of mass m + δm has velocity v. In time δt it ejects mass δm, which is moving with velocity u along the line of v The change in mass is m + δm m, the change in velocity is v v + δv 10 Case 1: No external force 3. Rocket motion Change of momentum: δp = m(v + δv) + uδm (m + δm)v = 0 }{{}}{{} After Before δp = mv + mδv + uδm mv vδm = mδv (v u) δm = 0 δp = m δv (v u) }{{} δm = 0 Relative velocity=w

11 Divide by δt : δp δt Total mass conserved d No external force Now apply an external force F = m δv w δm = 0 : Let δt 0 δt δt (m + δm) = 0 : δm δt m dv + w dm = 0 dm Change of momentum = δp = Fδt = mδv w δm Divide by δt, let δt 0 and δm δt dm m dv + w dm = F [Rocket equation] 11

12 The rocket : vertical launch Rocket equation: m dv + w dm = F Rocket rises against gravity F = mg Eject mass with constant relative velocity to the rocket w Rocket ejects mass uniformly: m = m 0 αt dm = α Now consider upward motion: vf mdv = ( mg + wα) v i dv = t f [ ] (m vf v i = g(t f t i ) w log 0 αt f ) e (m 0 αt i ) = [ g(t f t i ) w log e (m f /m i )] t i ( g + ) wα m 0 αt

13 Rocket vertical launch, continued The rocket starts from rest at t = 0; half the mass is fuel. What is the velocity and height reached by the rocket at burn-out at time t = T? ] ( )] v = (m [ gt w log 0 αt) e (m 0 ) = [ gt w log e 1 α m 0 t = dx What is the condition for the rocket to rise? dv > 0 dm At t = 0, m = m 0, = α : αw m 0 g > 0 w > m 0g α m = m0 αt ; at burnout t = T, m = m 0 2 α = m 0 2T 13 Maximum velocity is at the burn-out of the fuel: At t = T : v max = gt + w log e 2 Height : x 0 dx = T 0 x = gt w mo α [ gt w log e ( 1 α m 0 t )] Standard integral : loge z dz = z log e z z [( ) ( 1 α m 0 t (log e 1 α m 0 t After simplification : x = gt 2 )) 2 + wt (1 log e 2) ( )] T 1 α m 0 t 0

14 4. Two-body collisions Conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Conservation of energy: 1 2 m 1u m 2u 2 2 = 1 2 m 1v m 2v E (= 0 if elastic) We deal with 2 inertial frames: 14 The Laboratory frame: this is the frame where measurements are actually made The centre of mass frame: this is the frame where the centre of mass of the system is at rest and where the total momentum of the system is zero

15 Elastic collisions in 1D in the Lab frame, m 2 at rest Solve the conservation of energy & momentum equations in 1D Special cases: 15 v 1 = m 1 m 2 m 1 +m 2 u 1 and v 2 = 2m 1 m 1 +m 2 u 1 m1 = m 2 : v 1 = 0, v 2 = u 1 (complete transfer of momentum) m1 >> m 2 : Gives the limits v 1 u 1, v 2 2u 1 (m 2 has double u 1 velocity) m1 << m 2 : Gives the limits v 1 u 1, v 2 0 ( brick wall collision)

16 Elastic collisions in 2D in the Lab frame: equal masses, target at rest m 1 = m 2 = m, u 2 = 0 Momentum: mu1 = mv 1 + mv 2 u 1 = v 1 + v 2 Squaring u 2 1 = v2 1 + v v 1.v 2 Energy: 1 2 mu2 1 = 1 2 mv mv2 2 u2 1 = v2 1 + v2 2 Hence 2v1.v 2 = 0 16 EITHER v 1 = 0, v 2 = u 1 OR θ 1 + θ 2 = π 2 Either a head-on collision or opening angle is 90

17 4.1 The Centre of Mass frame The position of the centre of mass is given by: r cm = 1 M n i=1 m i r i where M = n i=1 m i Velocity of the CM: vcm = ṙ cm = n i=1 m i ṙ i i m i = n i=1 m i v i i m i Velocity of a body in the CM w.r.t. the Lab v i = v i v cm The total momentum in the CM: i p i = i m iv i = i m i(v i v cm ) = 0 17 The total momentum of a system of particles in the CM frame is equal to zero

18 4.2 Two-body elastic collision in 1D : Lab to CM system vcm = (m 1u 1 +m 2 u 2 ) (m 1 +m 2 ) Before in CM : m 1 u 1 + m 2u 2 = 0 After in CM : m 1 v 1 + m 2v 2 = 0 If elastic: u 1 u 2 = v 2 v 1 Solving: In CM, total momentum = 0, incoming and outgoing velocities are equal magnitudes and opposite direction. v 1 = u 1 v 2 = u 2 18

19 Collisions in the CM frame in 2D Conservation of momentum in CM: m 1 u 1 + m 2 u 2 = 0 ; m 1 v 1 + m 2 v 2 = 0 Conservation of energy in CM: 1 2 m 1u m 2u 2 2 = 1 2 m 1v m 2v 2 2 Solve the above equations : v 1 = u 1 ; v 2 = u 2 In CM, speeds before = speeds after 19

20 4.3 Solving collision problems in the CM frame 1) Find centre of mass velocity v CM (u1 v CM )m 1 + (u 2 v CM )m 2 = 0 vcm = m 1u 1 +m 2 u 2 m 1 +m 2 2) Transform initial Lab velocities to CM u 1 = u 1 v CM, u 2 = u 2 v CM 3) Get final CM velocities 20 v 1 = u 1 ; v 2 = u 2

21 4) Transform vectors back to the Lab frame v1 = v 1 + v CM ; v 2 = v 2 + v CM 5) Can then use trigonometry to solve Also note: T Lab = T Mv 2 cm The kinetic energy in the Lab frame is equal the kinetic energy in CM + the kinetic energy of CM 21

22 Example: Elastic collision, m 2 = 2m 1, θ 1 = 30 Find the velocities v 1 and v 2 and the angle θ 2 Magnitude of velocities: vcm = m 1u 1 +m 2 u 2 m 1 +m 2 = u 0 3 u 1 = u 0 v CM = 2u 0 3 u 2 = v CM = u 0 3 v 1 = u 1 = 2u 0 3 v 2 = u 2 = u

23 Relationships between angles and speeds Sine rule: (sin 30/ 2u 0 3 ) = (sin α/ u 0 3 ) sin α = 1 4 α = 14.5 β = 30 + α = 44.5 sin 30/ 2u 0 3 = sin( )/v 1 v 1 = 0.93u 0 23 Cosine rule: v 2 2 = ( u 0 3 ) 2 + ( u 0 3 ) 2 2( u 0 3 ) 2 cos β v 2 = 0.25u 0 Sine rule: (sin 44.5/v 2 ) = (sin θ 2 / u 0 3 ) θ 2 = 68.0

24 An inelastic collision is where energy is lost (or there is internal excitation). Coefficient of restitution 4.4 Inelastic collisions Defined as e = v 2 v 1 u 1 u 2 = Speed of relative separation Speed of relative approach We can show e = 1 E T (was derived in lectures) 24 where T = 1 2 µu2 1 with µ = m 1m 2 m 1 +m 2 (the reduced mass) T is the initial energy in the centre of mass frame, hence e is related to the fractional energy loss in this frame e = 1 completely elastic; e = 0 completely inelastic, in general 0 < e < 1

25 Completely inelastic collision in the CM vs. Lab Before collision: After collision: 25 KE in CM: T = T LAB 1 2 (m 1 + m 2 )vcm 2 Differentiate: Loss in KE T = T LAB (obvious) Max. energy that can be lost in the CM : T = T Max. energy can be lost in Lab = 1 2 m 1u (m 1 + m 2 )v 2 CM