Summary Lecture 9

1. This Friday 20-minute test on material in lectures 1-7 during lecture Systems of Particles 9.12 Rocket propulsion Rotational Motion 10.1 Rotation of Rigid Body 10.2 Rotational variables 10.4 Rotation with constant acceleration Summary Lecture 9 Thursday 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Problems:Chap. 9: 27, 40, 71, 73, 78 Chap. 10: 6, 11,16, 20,21, 28, 28

2. momentum conservation and

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

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

5. Now the force pushing the rocket is F = Note: since m is not constant i.e. F dt = v dm + m dv F dt = v dm - U dm 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. = logex = 1/x dx Thus 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. If I want to find out the TOTAL effect of throwing out gas, from when the mass was mi and velocity was vi, to the time when the mass is mf and the velocity vf, I must integrate.

7. 2 Speed in units of gas velocity Reducing mass (mf = 0) 1 Constant mass (v = at) .8 1 .6 .2 .4 Fraction of mass burnt as fuel

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

9. Rotation

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

12. Y reference line fixed in body q2 q1 X Rotation axis (Z) The orientation of the rigid body is defined by q. (For linear motion position is defined by displacementr.)

13. The unit of  is radian (rad) There are 2 radian in a circle 2 radian = 3600 1 radian = 57.30

14. Y  q2 q1 X Rotation axis (Z) Angular Velocity At time t2 At time t1 w is a vector

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

16. Angular Acceleration Dw w1 w2 angular acceleration a a is a vector direction: same as Dw. Units of a-- rad s-2 ais the analogue of a

17.  Consider an object rotating according to:  = -1 – 0.6t + 0.25 t2 e.g at t = 0  = -1 rad  = d/dt  = - .6 + .5t e.g. at t=0 = -0.6 rad s-1

18. Angular motion with constant acceleration

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

20. That's all folks

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

22. s v  r Relating Linear and Angular variables w and v s = qr w Not quite true. V, r, and w are all vectors. Although magnitude of v = wr. The true relation isv = wx r

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

24. A  B Ay = Asin Ax = Acos Vector Product C = A x B A = iAx + jAyB = iBx + jBy So C = (iAx + jAy) x (iBx + jBy) = iAx x(iBx + jBy)+ jAy x(iBx + jBy) = ixi AxBx + ixj AxBy + jxi AyBx + jxj AyBy now ixi = 0 jxj = 0 ixj = kjxi = -k So C = 0 + kAxBy - kAyBx +0 C= ABsin = 0 - kABsin

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

26. v a r w Since w = v/r this term = v2/r(or w2r) This term is the tangential accel atan. (or the rate of increase of v) Relating Linear and Angular variables a and a The centripetal acceleration of circular motion. Direction to centre

27. a and a a = ar&v2/r Central accelerationPresent even whenais zero! Tangential acceleration (how fast v is changing) Relating Linear and Angular variables The acceleration “a” of a point distance “r” from axis consists of 2 terms: Total linear acceleration a a r

29. The Falling Chimney CM L gcosq g The whole rigid body has an angular accelerationa The tangential accelerationatan distance r from the base isatan = ar q At the CM:atan = aL/2, and at the end:atan = aL But at the CM,atan= g cosq(determined by gravity) The tangential acceleration at the end is twice this, but the acceleration due to gravity of any mass point is onlyg cosq. The rod only falls as a body because it is rigid ………..the chimney is NOT.