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Summary Lecture 9 Systems of Particles 9.12 Rocket propulsion Rotational Motion 10.1 Rotation of Rigid Body 10.2 Rotational variables 10.4 This Friday Thursday 12 – 2 pm 20-minute test Rotation with constant acceleration material in PPPon “ExtEnsion” lectures 1-7 lecture. Problems:Chap. 9: 27, 40, 71, 73, 78 Chap. 10: 6, 11, 16, 20, 21, 28, during Room 211lecture podium level Turn up any time Principle of Rocket propulsion: In an ISOLATED System (no external forces) Momentum is conserved Momentum = zero Dm v An example of an isolated system where momentum is conserved! 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 Burns fuel at a rate dm dt 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 Force on Rocket F dt = v dm - U dm Now the force pushing the rocket is F = dprocket dt i.e. d dv (m v) Note: m dt dt since m is not constant dm dv F v m dt dt F F dt = v dm + m dv so that v dm + m dv = v dm - U dm dm dv = -U m 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. 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. dm dv = -U m 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. vf mf 1 dv U mi dm m vi Thus [ v]vfvi U[ln m]mf mi v f v i U(ln mf ln mi ) U(ln mi ln mf ) if v i 0 mi v f Uln mf = logex = 1/x dx e = 2.718281828… Speed in units of gas velocity 2 Reducing mass (mf = 0) 1 Constant mass (v = at) .2 .8 1 .6 .4 Fraction of mass burnt as fuel An example Mi = 850 kg mf = 180 kg Thrust = dp/dt of gas = U dm/dt U = 2800 m s-1 =2.3 x 2800 dm/dt = 2.3 kg s-1 = 6400 N F = ma Initial acceleration F = ma ==> a = F/m Thrust –mg = ma Thrust = 6400 N -2 = 6400/850 = 7.6 m s 6400 – 8500 = ma mg = 8500 N mi Final vel. v f U ln mf 2800 ln a = -2100/850 850 4300 m s 1 180 = -2.5 m s-2 n FIXED axis Rotation of a RIGID body about an 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 Y reference line fixed in body q2 q1 X The orientation of the rigid body is defined by Rotation axis (Z) q. (For linear motion position is r defined by displacement .) The unit of q is radian (rad) There are 2 radian in a circle 2 radian = 3600 1 radian = 57.30 Angular Velocity Y q2 Dq q1 X q q Dq av 2 1 t 2 t1 Dt limit Dq dq inst Rotation axis (Z) Dt 0 Dt is a vector dt Angular Velocity How do we specify its angular velocity? is a vector right hand is the rotational analogue of v is rate of change of q units of …rad s-1 Angular Acceleration D 2 1 2 1 D av t 2 t1 Dt limit D d inst Dt 0 Dt dt angular acceleration is a vector direction: same as D. Units of -- rad s-2 is the analogue of a q Consider an object rotating according to: q = -1 – 0.6t + 0.25 t2 e.g at t = 0 q = -1 rad = dq/dt = - .6 + .5t e.g. at t=0 = -0.6 rad s-1 Angular motion with constant acceleration An example where is constant ω0 100 2π x rad / se c = 3.49 rad s-1 3 60 - t v u0 at 0= 33¹/³ RPM 0 0 t = 8.7 s =+-0.4 rad s-2 Q1 How long to come to rest? 22 1 at q sut t t 2 0 Q2 How many revolutions does it take? = 15.3 rad = 15.3/22.43 rev. 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 q r s s = qr Relating Linear and Angular variables and v s q r Not quite true. V, r, and are all vectors. Although magnitude of v = r. The true relation is v = x r s = qr ds v dt d v (qr) dt dθ v r dt v ωr Direction of vectors Grab first vector () with right hand. Turn to second vector (r) . v=x r Direction of screw is direction of third vector (v). r Vector Product A C =Ax B A = iAx + jAy B = iBx + jBy So C = (iAx + jAy) x (iBx + jBy) Ay = Asinq q Ax = Acosq B = iAx x (iBx + jBy) + jAy x (iBx + jBy) = ixi AxBx + ixj AxBy + jxi AyBx + jxj AyBy now So C= 0 + k AxBy - kAyBx + 0 = 0 - k ABsinq ixi = 0 jxj = 0 ixj = k jxi = -k C= ABsinq Is q a vector? Rule for adding vectors: The sum of the vectors must not depend on the order in which they were added. However Dq is a vector! Relating Linear and Angular variables a and v r vxr a dv d (ω x r ) dt dt dr dω a ωx x r dt dt Since = v/r this term = v2/r (or 2r) The centripetal acceleration of circular motion. Direction to centre a ωxv αxr This term is the tangential accel atan. (or the rate of increase of v) Relating Linear and Angular variables a and The acceleration “a” of a point distance “r” from axis consists of 2 terms: Total linear acceleration a a = r & v2/r r Central Tangential acceleration even when acceleration Present is zero! (how fast v is changing) The Falling Chimney The whole rigid body has an angular acceleration CM gcosq g The tangential acceleration atan distance r from the base is L atan r q At the CM: atan L/2, and at the end: atan = L 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 only g cosq. The rod only falls as a body because it is rigid ………..the chimney is NOT.