1. Summary Lecture 9 Systems of Particles 9.12Rocket propulsion Rotational Motion 10.1Rotation of Rigid Body 10.2Rotational variables 10.4Rotation with constant acceleration Systems of Particles 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, This Friday 20-minute test on material in lectures 1-7 during lecture This Friday 20-minute test on material in lectures 1-7 during lecture Thursday 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Thursday 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time

3. Principle of Rocket propulsion: In an ISOLATED System (no external forces) Momentum is conserved Momentum = zero

4. We found that the impulse (  p = Fdt) given to the rocket by the gas thrown out the back was v mm U = Vel. of gas rel. to rocket Burns fuel at a rate F dt = v dm - U dm v+  v Force on Rocket An example of an isolated system where momentum is conserved! The impulse driving the rocket, due to the momentum, of the gas is given by

5. 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: Every time 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.

6. 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. 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.

7. 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)

8. 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 Thrust = 6400 N mg = 8500 N

10. 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

12. 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.)

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

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

15.   is a vector  is the rotational analogue of v Angular Velocity  is rate of change of  units of  …rad s -1 How do we specify its angular velocity? right hand

16.    is a vector direction: same as . Units of  -- rad s -2  is the analogue of a  Angular Acceleration  angular acceleration

17. Consider an object rotating according to:  = -1 – 0.6t + 0.25 t 2  = d  /dt e.g at t = 0  = -1 rad e.g. at t=0  = -0.6 rad s -1   = -.6 +.5t

18. Angular motion with constant acceleration

19.  0 = 33¹ / ³ RPM  = -0.4 rad s -2 Q1 How long to come to rest? Q2 How many revolutions does it take? = 3.49 rad s -1 = 8.7 s = 15.3 rad = 15.3/2   2.43 rev. An example where  is constant 0 - + +

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

22. 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 Relating Linear and Angular variables

23. v =  x r  r v Grab first vector (  ) with right hand. Direction of screw is direction of third vector (v). Turn to second vector (r). Direction of vectors

24. 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 = 0 + k A x B y - kA y B x + 0 = 0 - k ABsin  now i x i = 0 j x j = 0 i x j = k j x i = -k

25. Is  a vector? However  is a vector! Rule for adding vectors: The sum of the vectors must not depend on the order in which they were added. Rule for adding vectors: The sum of the vectors must not depend on the order in which they were added.

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

27. Total linear acceleration a The acceleration “a” of a point distance “r” from axis consists of 2 terms: Tangential acceleration (how fast v is changing)  a and  a =  r & v 2 /r Central acceleration Present even when  is zero! r Relating Linear and Angular variables

29. 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, But at the CM, a tan = g cos   (determined by gravity) The tangential acceleration at the end is twice this, but the acceleration due to gravity of any mass point is only g cos   The rod only falls as a body because it is rigid The Falling Chimney L and at the end: a tan =  L gcos  ………..the chimney is NOT. ………..the chimney is NOT.