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Summary Lecture 8 Systems of Particles 9.2 Centre of mass 9.3 Tomorrow 12 – 2 pm Newton 2 for system of particles Conservation of momentum Week PPP “ExtEnsion” March 20 – 24 lecture. 20-minute test 9.8-11 Collisions on material in 9.12 Rocket propulsion Room 211 podium lectures 1-7 level Turn up any time Problems:Chap. 9: 1, 6, 10 , 15, 20, 9.4-7 27, 40, 71, 73, 78 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 The Centre of Mass Lizzie Borden took an axe The motion of the Centre of And gave her mother forty whacks. Mass isshe a simple And when saw whatparabola. she had done, She gave her father forty-one (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 The Centre of Mass m1 m2 m1 m2 M = m1 + m 2 m1 m1g x d1 = m2g x d2 m1g d1 CM M d2 m2 m2g Centre of Mass (1D) M = m 1 + m2 M m1 0 x1 m2 x2 xcm M xcm = m1 x1 + m2 x2 xcm moment of M = moment of individual masses m1x1 m 2x 2 M In general xcm 1 m i xi M Centre of Mass (3D) For a collection of masses in 3D cm r1 r2 r 3 r cm 1 mi r i M Centre of Mass (3D) For a collection of masses in 3D rcm = ixcm +jycm cm r cm 1 mi r i M 1 xcm mi x i M 1 ycm mi yi M So in a solid body we can find the CM by finding xcm and ycm y 2 1 xcm mi xi M 1 1 xcm ( 3x0 8x1 4x2) 15 15 kg 4 kg 3 kg Xcm = 16/15 = 1.07 m y cm 1 mi y i M y cm rCM 1.33 r cm 1 mi r i M 8 kg 0 1 ( 3x0 8x2 4x1) 15 ycm = 20/15 = 1.33 m 1 1.07 2 x Finding the Centre of Mass For solid bodies the sum: rcm 1 r m M becomes an integral: 1 rcm r dm M For odd shaped objects this probably needs to be determined experimentally For symmetric objects this can often be calculated 1. Look for a symmetry axis 2. Then carry out the integral to find the position of xcm along the axis. Solid cone h R Symmetry line r xcm x 1 x dm M R 2 h 3 xcm 2 0 x dx hM R 2 h4 xcm x dx dm = r2 dx but r = (R/h)x dm = (R/h)2x2 dx Mass of cone M = 1/3 R2 h xcm h2 M 4 R 2 h h xcm M 4 xcm 3M h M 4 xcm = ¾ h cm F Ma cm Fext z Macm z Fext y Macm y Fext x Macm x For a system of particles, the dynamics of the Centre of Mass obeys Newton 2. Sum of all EXTERNAL forces acting on system The total mass of the system The acceleration of the CM of the system cm CM F Ma cm Fext z Macm z Fext y Macm y Fext x Macm x This also applies to a solid For awhere systemthe of individual particles, body, the dynamics of the Centre particles are rigidly of Mass obeys 2. connected. TheNewton dynamics of the Centre of Mass obeys Newton 2 Linear Momentum of system of particles Fext ma You will recall that F dp ma dt Where p=mv is the momentum of each particle cm For a system of particles P == Mv p cm Fext dP dt This also applies to extended objects for system of particles Conservation of Linear Momentum Fext If Fext = 0 dP 0 dt dP dt NO EXTERNAL forces act on the system P is a constant 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. Exploding rocket C of M Why? No external horizontal forces so horiz momentum unchanged m = 3.8 g, n =12 v = 1100 m s-1 M=12 kg Initial momentum Pi = n mv + M Vi Define system = n mv + 0 M= 12 kg + 12 m Final momentum pf = (M + nm) V = Pi = n mv V nmv V M nm V 4.2 m s 1 m = 3.8 g, n =12 v = 1100 m s-1 KE initial M=12 kg ½ n mv2 ½ x 12 x 0.0038 x (1100)2 Initial momentum Pi = n mv + M Vi 27588 J = n mv + 0 KE final M= 12 kg ½ (M + 12m)V2 + 12 m V 2 ½ x (12.0456) x 4.2 Final momentum pf = (M + nm) V = Pi = n mv 163 J nmv V M nm V 4.2 m s 1 Collisions What is a collision? No external forces Momentum is conserved An isolated event involving 2 or more objects Usually interact (often strongly) for short time Equal and opposite impulses are exerted on each other p = F dt Collisions Elastic collisions Energy and momentum are conserved But Energy is always conserved??? Inelastic collisions Only momentum is conserved In 1 dimension Elastic Collision 1D Before After m1 m2 v1i v2i = 0 m1 m2 v1f v2f We want to find V1f and V2f 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 v1 f m1v1i m2 v 2 f v1i m2 v 2 f m1 m1 Substitute this form of v1f into equ 4 v2f = v1i + v1i – m2/m1 v2f 2m1 v2 f ( )v1 i m1 m 2 m1 m2 v1 f ( )v1i m1 m2 v2f(1 + m2/m1) = 2v1i m1 m2 2m1 )v1i v2 f ( )v1 i v1 f ( m1 m2 m1 m 2 If m1>> m2 v2f 2v1i v1f v1i If m2>>m1 v2f 0 v1f -v1i If m1= m2 v2f v1i v1f 0 Motion of the C of M m1 vcm v1i m2 v2i =0 CM What is Vcm? Mom of CM = mom of m1 + mom of m2 (m1 + m2 ) Vcm = m1v1i + m2v2i Vcm m1 v1i m1 m 2 Motion of the C of M m1 vcm v1i m2 v2i =0 CM Let’s observe the elastic collision from the view point of the centre of mass In 1 dimension m1 vcm v1i m2 v2i =0 CM What is Vcm? Mom of CM = mom of m1 + mom of m2 (m1 + m2 ) Vcm = m1v1i + m2v2i Vcm m1 v1i m1 m 2 m1 v1i m2 vcm CM v2i Let’s observe the elastic collision from the view point of the centre of mass Collisions in 2 dimensions Elastic billiard balls comets a-particle scattering Elastic collisions in 2-D Momentum is conserved Impact parameter 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 m1v1i 2 1 before 1 1 1 2 2 m1 v 1i m1 v 1f m 2 v 22f 2 2 2 7 variables! after 3 equations Collisions in 2 dimensions Inelastic Almost any real collision! an example Automobile collision mA= 830 kg va = 62 kph mB = 550 kg pB pA pf Pfx= pf cos Pfy= pf sin vB = 78 kph = Pfy= pf sin pB pf mAvA Pfx= pf cos Cons. Momentum ==> pA + pB = pf pA X component PA = Pf cos mAvA = (mA+ mB) vf cos………….(1) Y component PB = Pf sin Divide equ (2) by (1) mBvB = (mA+ mB) vf sin………….(2) ____________________ mAvA = (mA+ mB) vf cos m v tan B B m Av A Gives = 39.80 = mAvA Pfx= pf cos Cons. Momentum ==> pA + pB = pf pA Pfy= pf sin pB pf X component PA = Pf cos mAvA = (mA+ mB) vf cos………….(1) Y component PB = Pf sin tan = mBvB m Av A 39.80 mBvB = (mA+ mB) vf sin………….(2) Use equ 2 to find Vf vf mB vB (m A m B ) sin Gives Vf = 48.6 kph Can the investigators determine who was speeding? mB vB vf (m A m B ) sin mA= 830 kg mB = 550 kg mBvB tan m Av A http://www.physics.ubc.ca/~outreach/phys420/p420_96/danny/danweb.htm m v v+ v U = Vel. of gas rel. to rocket Burns fuel at a rate dm dt 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 F dt = m(v - U) = v dm - U dm Now the force pushing the rocket is F = dprocket i.e. dt Note m dv since dt m is not constant d F (mv ) dt dm dv v m dt dt Fdt = v dm + m dv so that v dm + m dv = v dm - U dm dm dv = -U m This means that if I throw out dm of gas with vel. U, I will increase rocket velocity by dv. dm dv = -U m This means that if I throw out dm of gas with vel. U, I will increase rocket velocity by 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. vf mi 1 dm Thus dv U vi mi m 1 mf [ v ]vf U [ln m ] note dm ln m vi mi m vf -vi = - U (ln mf - ln mi) = + U (ln mi - ln mf) if vi = 0 vf = mi U ln mf mi vf = U ln mf 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 = dm/dt U U = 2800 m s-1 =2.3 x 2800 dm/dt = 2.3 kg s-1 = 6400 N Initial acceleration F = ma ==> a = F/m = 6400/850 = 7.6 m s-2 Final vel. mi v f U ln mf 850 2800 ln 4300 m s 1 180