2. Summary Lecture 9 Systems of Particles 9.8-11Collisions 9.12Rocket propulsion Rotational Motion 10.1Rotation of Rigid Body 10.2Rotational variables 10.4Rotation with constant acceleration Systems of Particles 9.8-11Collisions 9.12Rocket propulsion Rotational Motion 10.1Rotation of Rigid Body 10.2Rotational variables 10.4Rotation with constant acceleration Problems:Chap. 9 : 27, 40, 71, 73, 78 Chap. 10: 6, 11, 16, 20, 21, 28, Problems:Chap. 9 : 27, 40, 71, 73, 78 Chap. 10: 6, 11, 16, 20, 21, 28, Friday March 24 20-minute test on material in lectures 1-7 during lecture Friday March 24 20-minute test on material in lectures 1-7 during lecture

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

5. In 1 dimension

6. m 1 v 1i m 2 v 2i =0 Before m 1 v 1f m 2 v 2f After Elastic Collision 1D Mom. Cons. m 1 v 1i = m 1 v 1f + m 2 v 2f ………………(1)  m 2 v 2f = m 1 (v 1i - v 1f )…………………(2) Energy Cons ½ m 1 v 1f 2 + ½ m 2 v 2f 2 = ½ m 1 v 1i 2  ½ m 2 v 2f 2 = ½ m 1 (v 1i 2 - v 1f 2 )

7.  m 2 v 2f 2 = m 1 (v 1i - v 1f ) (v 1i + v 1f ) ……(3) Mom. Cons. m 1 v 1i = m 1 v 1f + m 2 v 2f ………………(1)  m 2 v 2f = m 1 (v 1i - v 1f )…………………(2) Energy Cons ½ m 1 v 1f 2 + ½ m 2 v 2f 2 = ½ m 1 v 1i 2  ½ m 2 v 2f 2 = ½ m 1 (v 1i 2 - v 1f 2 ) Mult. by 2 and factorise Divide equ. (3) by (2) V 1i is usually given, so to find v 2f we need to find an expression for v 1f. Get this from equ. (1). m 1 v 1f = m 1 v 1i - m 2 v 2f  Substitute this form of v 1f into equ 4  v 2f = v 1i + v 1i – m 2 /m 1 v 2f  v 2f (1 + m 2 /m 1 ) = 2v 1i  v 2f = v 1i + v 1f …………….…(4)

8. If m 1 >> m 2 v 2f  2v 1i If m 1 = m 2 If m 2 >>m 1 v 1f  0 v 2f  0 v 2f  v 1i v 1f  v 1i v 1f  -v 1i Analyze the equations

9. m 1 v 1i m 2 v 2i =0 v cm CM What is V cm ? Mom of CM = mom of m 1 + mom of m 2 (m 1 + m 2 ) V cm = m 1 v 1i + m 2 v 2i Collision viewed from Lab. Ref. frame

10. m 1 v 1i m 2 v 2i =0 v cm CM Collision viewed from Lab. Ref. frame

11. Note that the CM moves at constant vel Because there is no EXTERNAL force acting on the system Collision viewed from Lab. Ref. frame

12. Note that the CM moves at constant vel Because there is no EXTERNAL force acting on the system Collision viewed from Lab. Ref. frame

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

14. Note that the CM is at rest Collision viewed from CM Ref. frame

15. Note that the CM is at rest Collision viewed from CM Ref. frame

16. In 1 dimension

17. m 1 v 1i m 2 v 2i =0 v cm CM What is V cm ? Mom of CM = mom of m 1 + mom of m 2 (m 1 + m 2 ) V cm = m 1 v 1i + m 2 v 2i

18. m 1 v 1i m 2 v 2i v cm CM Completely inelastic collision Observing from the Lab. reference frame

19. Note that the CM moves at constant vel Because there is no EXTERNAL force acting on the system Observing from the Lab. reference frame

20. Note that the CM moves at constant vel Because there is no EXTERNAL force acting on the system Observing from the Lab. reference frame

21. Let’s observe the inelastic collision from the view point of the centre of mass

22. Observing from the CM reference frame

23. Elastic billiard balls comets  -particle scattering Collisions in 2 dimensions

24. Momentum is conserved Consider x-components m 1 v 1i = m 1 v 1f cos  1 + m 2 v 2f cos  2 Consider y-components 0= -m 1 v 1f sin  1 + m 2 v 2f sin  2 Since elastic collision energy is conserved 7 variables!3 equations Elastic collisions in 2-D m 1 v 1i before 22 11 m 2 v 2f m 1 v 1f after Impact parameter

25. Inelastic Almost any real collision! An example: Automobile collision Collisions in 2 dimensions

26. m B = 550 kg v B = 78 kph  V f = pApA pBpB pfpf P fy = p f sin  P fx = p f cos   m A = 830 kg v a = 62 kph

27. Cons. Momentum ==> p A + p B = p f X component P A = P f cos  m A v A = (m A + m B ) v f cos  ………….(1) Y component P B = P f sin  m B v B = (m A + m B ) v f sin  ………….(2) pApA pBpB pfpf P fx = p f cos  P fy = p f sin   = mAvAmAvA Divide equ (2) by (1) ____________________ m A v A = (m A + m B ) v f cos  Gives  = 39.8 0

28. Cons. Momentum ==> p A + p B = p f X component P A = P f cos  m A v A = (m A + m B ) v f cos  ………….(1) Y component P B = P f sin  m B v B = (m A + m B ) v f sin  ………….(2) pApA pBpB pfpf P fx = p f cos  P fy = p f sin   = mAvAmAvA  = 39.8 0 Use equ 2 to find V f Gives V f = 48.6 kph

29. Can the investigators determine who was speeding? http://www.physics.ubc.ca/~outreach/phys420/p420_96/danny/danweb.htm Conservation of Energy ½ mv f 2 = f.d d f =  N =  m A + m B ) g

32. IN THE EARTH REF. FRAME Vel of gas rel me = vel of gas rel. rocket - vel of rocket rel me V = U - v v mm U = Vel. of gas rel. to rocket Burns fuel at a rate Mom. of gas =  mV =  m(U - v) F dt = v dm - U dm v+  v i.e. F dt =  m(v - U) Impulse is mom. transfer (  p) So since F = dp/dt,  p = Fdt = - change in mom. of rocket (impulse or  p ) Force on Rocket An example of an isolated system where momentum is conserved!

33. Note: since m is not constant Now the force pushing the rocket is F = i.e. F dt = v dm - U dm F dt = v dm + m dv This means: If I throw out a mass dm of gas with a velocity U, when the rocket has a mass m, the velocity of the rocket will increase by an amount dv.

34. If I want to find out the TOTAL effect of throwing out gas, from when the mass was m i and velocity was v i, to the time when the mass is m f and the velocity v f, I must integrate. v f = U ln Thus = log e x =  1/x dx e = 2.718281828… This means: If I throw out a mass dm of gas with a velocity U, when the rocket has a mass m, the velocity of the rocket will increase by an amount dv.

35. Fraction of mass burnt as fuel Speed in units of gas velocity 1 2.2.4.6.8 1 Constant mass (v = at) Reducing mass (m f = 0)

36. An example M i = 850 kg m f = 180 kg U = 2800 m s -1 dm/dt = 2.3 kg s -1 Thrust = dp/dt of gas =2.3 x 2800 = 6400 N Initial accelerationF = ma ==> a = F/m = 6400/850 = 7.6 m s -2 Final vel.  F = ma Thrust –mg = ma 6400 – 8500 = ma a = -2100/850 = -2.5 m s -2 = U dm/dt

38. Rotation of a body about an axisRIGID n FIXED Every point of body moves in a circle Not fluids,. Every point is constrained and fixed relative to all others The axis is not translating. We are not yet considering rolling motion

40. reference line fixed in body   X Y Rotation axis (Z) The orientation of the rigid body is defined by . (For linear motion position is defined by displacement r.)

41. The unit of  is radian (rad) There are 2  radian in a circle 2  radian = 360 0 1 radian = 57.3 0

42. X Y Rotation axis (Z)  is a vector Angular Velocity At time t 1 At time t 2   

43. Angular velocity   is a vector  is rate of change of  units of  …rad s -1  is the rotational analogue of v

44.    is a vector direction of change in . Units of  -- rad s -2  is the analogue of a  Angular Acceleration 

45.  = -1 – 0.6t +.25 t 2  = d  /dt  = -.6 +.5t e.g at t = 0  = -1 rad e.g. at t=0  = -0.6 rad s-1 

46. Rotation at constant acceleration

47.  0 = 33¹ / ³ RPM  = -0.4 rad s -2 How long to come to rest? How many revolutions does it take? =3.49 rad s -2 = 8.7 s = 45.5 rad = 45.5/2   7.24 rev. An example where  is constant

48. Relating Linear and Angular variables  r s s =  r Need to relate the linear motion of a point in the rotating body with the angular variables  and s

49. Relating Linear and Angular variables s =  r  v r  and v V, r, and  are all vectors. Although magnitude of v =  r. The true relation is v =  x r Not quite true.  s

50. v =  x r  v r

51. So C = (iA x + jA y ) x (iB x + jB y ) = iA x x (iB x + jB y ) + jA y x (iB x + jB y ) = i x i A x B x + i x j A x B y + j x i A y B x + j x j A y B y  A y = Asin  A x = Acos  A B C = A x B Vector Product A = iA x + jA y B = iB x + jB y C= ABsin  So C = k A x B y - kA y B x = 0 - k ABsin  now i x i = 0j x j = 0 i x j = kj x i = -k

52. This term is the tangential acceln a tan. (or the rate of increase of v) Since  = v/r this term = v 2 /r (or  2 r) The centripital acceln of circular motion. Direction to centre r  a and  Relating Linear and Angular variables

53. Total linear acceleration a Thus the magnitude of “a” a =  r - v 2 /r Tangential acceleration (how fast V is changing) Central acceleration  r Relating Linear and Angular variables a and 

54. CM g  The whole rigid body has an angular acceleration  The tangential acceleration a tan distance r from the base is a tan  r at the CM, a tan  L/2, and at end a tan  L Yet at CM, a tan = g cos  (determined by gravity) gcos  At the end, the tangential acceleration is twice this, yet the maximum tangential acceleration of any mass point is g cos  The rod only falls as a body because it is rigid…the chimney is NOT. The Falling Chimney L