Phys 0174 Fall 2008 - Chapter 09, 4 slides

Phys 0174 Fall 2008 - Chapter 09, 4 slides - Chapter 9 The...

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1 Chapter 9 Center of Mass and Linear Momentum ¾ Center of mass (com) for a system of particles ¾ The velocity and acceleration of the center of mass ¾ Linear momentum for a single particle and a system of particles ¾ Conservation of linear momentum ¾ Collisions in one and two dimensions ¾ Rockets 2 The center of mass is a point that represents the average location for the total mass of a system. 2 1 2 2 1 1 m m x m x m x cm + + = 11 2 2 33 1 123 1 ... 1 ... ... n nn com i i i n n ni i mx x mmm m M Mmmm m m = = +++ + == ++++ =++++= 1 1 n com i i i m M = = rr GG 3 1 1 The position vector for the center of mass is given by the equation: ˆˆ ˆ The position vector can be written as: The components of are given by the equ n com i i i com com com com com m M xyz = = =++ ri j k r G G G G 1 ations: 1 n com i i com i i com i i ii i xm x ym y zm z MM M = = ∑∑ 4 Solid bodies can be considered as systems with continuous distribution of matter and so the center of mass equation becomes an integ The Center of Mass for Solid Bodies 1 r a l: com com xx d m y M com ydm z zdm M M = ∫∫ . C C A simpler special case is that of uniform objects in which the mass density i 1 s co nstant and equal to com com com d Vy dm M dV ydV z zdV VV V V ρ = =
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5 6 O m 1 m 3 m 2 F 1 F 2 F 2 x y z 123 , Consider a system of particles of masses , , ..., and position vectors , , ,. .. Newton's Second Law for , , respectively. T a System of Particle he position vector of the center of mass s n n nm m m m rrr r GGG G 1 is given by: 1 n com i i i m M = = rr GG () 11 2 2 33 1 1 1 Center of mass velo Take the time derivative of bo cit th sides: 1 y 1 n com i i i n com i i n com i ii nn i dd m dt dt mm m MM M m dt M dt = = = == + +  =  + = + r vv v v v v r G G G G G G 1 1 n com i i i m M = = 7 O m 1 m 3 m 2 F 1 F 2 F 2 x y z 1 2 1 1 2 1 Take the time derivative of 1 Center o both sides: 1 1 f mass acceleration n com i i i n com i n com i i i n com i i n n i i i m M m m dt dt M m dt M dt = = = = =  = = + + + + aa a a a v a v G G G G G G 1 1 n com i i i m M = = 8 1 1 1 Multiply both side 1 s by : n com i n com i i i i n co n i i n n m i M Mm m m M m M = = = + + + + = =++++= a a aF F F F F G G G G G G G G G G int The force can be decomposed into two components: applied and internal i app i =+ F FF F G GG G ( ) int int int int 22 int int int int 12 3 The above equation takes the form: ... ... ... app app app app com app app app app com n n M a FF FF FF FF M a FFF F FFF F =++++++ ++ = +++ + ++++ + G G G G G G G G G
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9 ( ) ( ) int int int int 12 3 3 ... ... app app app a n pp com n M a FFF FFF F F = +++ + +++ + + GGG G G G The sum in the second parenthesis vanishes by virtue of Newto The sum in the first parenthesis of the equation above is just . The equation of motion for the cent n's t er of hird mass becom a. es lw net F G ,, : In terms of compone nts we have: com net net x com x net y com y net z com z M FM a a a = == = aF G G com net M = G G net x com x net y com y net z com z a a a = = = 10 0 2 1 2 2 1 1 = + + = m m v m v m v cm BEFORE AFTER () ( ) 88 kg 1.5m s 54 kg 2.5m s 88 kg 54 kg 0.002 0 cm v −+ + + 11 12 A comet is moving through interstellar space
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Phys 0174 Fall 2008 - Chapter 09, 4 slides - Chapter 9 The...

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