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Summary Lecture 8

Tomorrow 12 – 2 pm PPP “ Extension” lecture. Room 211 podium level Turn up any time. Summary Lecture 8. Systems of Particles 9.2 Centre of mass 9.3 Newton 2 for system of particles 9.4-7 Conservation of momentum 9.8-11 Collisions 9.12 Rocket propulsion. Week March 20 – 24

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Summary Lecture 8

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  1. Tomorrow 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Summary Lecture 8 Systems of Particles 9.2 Centre of mass 9.3 Newton 2 for system of particles 9.4-7 Conservation of momentum 9.8-11 Collisions 9.12 Rocket propulsion Week March 20 – 24 20-minute test on material in lectures 1-7 10 Problems:Chap. 9: 1, 6, 10 , 15, 20, 27, 40, 71, 73, 78

  2. System of Particles

  3. So far we have considered the motion of POINT PARTICLES FINITE OBJECTS can move as a whole (translational motion) and also rotate about the “Centre of Mass” The “Centre of Mass” is that point where if we apply a force, the object will not rotate. What happens depends on where we apply the force

  4. Lizzie Borden took an axeAnd gave her mother forty whacks.And when she saw what she had done,She gave her father forty-one The Centre of Mass The motion of the Centre of Mass is a simple parabola. (just like a point particle) The motion of the entire object is complicated. • This motion resolves to • motion of the CM • motion of points around the CM

  5. The Centre of Mass m1 m1 m1 m1 m1 m2 m2 m2 m2 m2 d1 d2 CM m1g m2g M = m1 + m2 M m1g x d1 = m2g x d2

  6. M m1 m2 0 x1 x2 xcm Centre of Mass (1D) M = m1 + m2 M xcm = m1 x1 + m2 x2 moment of M = moment of individual masses In general

  7. cm r1 r2 r3 Centre of Mass (3D) For a collection of masses in 3D

  8. cm Centre of Mass (3D) For a collection of masses in 3D rcm = ixcm +jycm So in a solid body we can find the CM by finding xcm and ycm

  9. y 8 kg 2 15 kg 4 kg 1 rCM 1.33 3 kg 0 1 2 x 1.07 Xcm = 16/15 = 1.07 m ycm = 20/15 = 1.33 m

  10. For solid bodies the sum: becomes an integral: Finding the Centre of Mass For odd shaped objects this probably needs to be determined experimentally For symmetric objects this can often be calculated • Look for a symmetry axis • Then carry out the integral to find the position of xcm along the axis.

  11. h xcm R r Symmetry line x x dx Solid cone dm = r2 dx but r = (R/h)x dm =  (R/h)2x2 dx Mass of cone M = 1/3 R2 h xcm = ¾ h

  12. cm Sum of all EXTERNAL forces acting on system The total mass of the system The acceleration of the CM of the system For a system of particles, the dynamics of the Centre of Mass obeys Newton 2.

  13. cm CM This also applies to a solid body, where the individual particles are rigidly connected. The dynamics of the Centre of Mass obeys Newton 2 For a system of particles, the dynamics of the Centre of Mass obeys Newton 2.

  14. You will recall that cm Where p=mv is the momentum of each particle for system of particles Linear Momentum of system of particles For a system of particles P = p = Mvcm This also applies to extended objects

  15. Pis a constant Conservation of Linear Momentum NO EXTERNALforces act on the system If Fext = 0 That is: Px, Py and Pz remain constant if Fext-x, Fext-y and Fext-z are zero In an isolated system, momentum is conserved.

  16. C of M Exploding rocket Why? No external horizontal forces so horiz momentum unchanged

  17. m = 3.8 g, n =12 v = 1100 m s-1 M=12 kg Define system M= 12 kg + 12 m V Initial momentum Pi = n mv + M Vi = n mv + 0 = Pi Final momentum pf = (M + nm) V = n mv

  18. m = 3.8 g, n =12 v = 1100 m s-1 M=12 kg M= 12 kg + 12 m V KE initial ½ n mv2 ½ x 12 x 0.0038 x (1100)2 27588 J Initial momentum Pi = n mv + M Vi = n mv + 0 KE final ½ (M + 12m)V2 ½ x (12.0456) x 4.22 163 J Final momentum pf = (M + nm) V = Pi = n mv

  19. coll isions

  20. No external forces Momentum is conserved p = F dt Collisions What is a collision? An isolated event involving 2 or more objects Usually interact (often strongly) for short time Equal and opposite impulses are exerted on each other

  21. But Energy is always conserved??? Collisions Elastic collisions Energy and momentum are conserved Inelastic collisions Only momentum is conserved

  22. Elastic collision In 1 dimension

  23. Elastic Collision 1D Before m1 v1i m2 v2i = 0 After m1 v1f m2 v2f We want to find V1f and V2f

  24. Mom. Cons. m1v1i = m1v1f + m2v2f………………(1)  m2v2f = m1(v1i- v1f)…………………(2) Energy Cons ½ m1v1f2 + ½ m2v2f2 = ½ m1v1i2 ½ m2v2f2 = ½ m1(v1i2 - v1f2) Mult. by 2 and factorise  m2v2f2 = m1(v1i- v1f) (v1i+ v1f) ……(3) Divide equ. (3) by (2)  v2f = v1i + v1f…………….…(4) V1i is usually given, so to find v2f we need to find an expression for v1f. Get this from equ. (1). m1v1f = m1v1i - m2v2f  Substitute this form of v1f into equ 4  v2f = v1i + v1i – m2/m1 v2f  v2f(1 + m2/m1) = 2v1i

  25. If m1>> m2 v2f 2v1i v1f v1i If m2>>m1 v2f 0 v1f -v1i If m1= m2 v2f v1i v1f 0

  26. That's all folks

  27. CM Motion of the C of M m1 v1i m2 v2i =0 vcm What is Vcm? Mom of CM = mom of m1 + mom of m2 (m1 + m2 ) Vcm = m1v1i + m2v2i

  28. vcm CM Motion of the C of M m1 v1i m2 v2i =0

  29. Let’s observe the elastic collision from the view point of the centre of mass

  30. inelastic collision In 1 dimension

  31. vcm CM m1 v1i m2 v2i =0 What is Vcm? Mom of CM = mom of m1 + mom of m2 (m1 + m2 ) Vcm = m1v1i + m2v2i

  32. vcm CM m1 v1i m2 v2i

  33. Let’s observe the elastic collision from the view point of the centre of mass

  34. Collisions in 2 dimensions Elastic billiard balls comets a-particle scattering

  35. Impact parameter m2v2f m1v1i 2 1 m1v1f before after Elastic collisions in 2-D Momentum is conserved Consider x-components m1v1i= m1v1f cos 1 + m2v2f cos 2 Consider y-components 0= -m1v1f sin 1 + m2v2f sin 2 Since elastic collision energy is conserved 7 variables! 3 equations

  36. Collisions in 2 dimensions Inelastic Almost any real collision! an example Automobile collision

  37. mA= 830 kg Vf = va = 62 kph q mB = 550 kg vB = 78 kph pf Pfy= pf sinq pB q pA Pfx= pf cosq

  38. = pB q mAvA pf Pfy= pf sinq pA Pfx= pf cosq Cons. Momentum ==> pA + pB = pf X component PA = Pf cosq mAvA = (mA+ mB) vf cosq………….(1) Y component PB = Pf sinq mBvB = (mA+ mB) vf sinq………….(2) ____________________ mAvA = (mA+ mB) vf cosq Divide equ (2) by (1) Gives q = 39.80

  39. mAvA Cons. Momentum ==> pA + pB = pf X component PA = Pf cosq mAvA = (mA+ mB) vf cosq………….(1) Y component PB = Pf sinq mBvB = (mA+ mB) vf sinq………….(2) pf = Pfy= pf sinq pB q = 39.80 pA q Pfx= pf cosq Gives Vf = 48.6 kph Use equ 2 to find Vf

  40. Can the investigators determine who was speeding? http://www.physics.ubc.ca/~outreach/phys420/p420_96/danny/danweb.htm

  41. momentum conservation and

  42. Burns fuel at a rate v v+D v U = Vel. of gas rel. to rocket Dm IN THE EARTH REF. FRAME Vel of gas rel me = vel of gas rel. rocket- vel of rocket rel me = U - v Mom. of gas = m(U - v) = -change in mom. of rocket (impulse) i.e. F dt = m(v - U) = v dm - U dm

  43. Now the force pushing the rocket is F = i.e. Note since m is not constant Fdt = v dm + m dv F dt = m(v - U) = v dm - U dm

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