System of Particles: Center-of-Mass The center-of-mass of a system of particles is the point that moves as though (1) all of the systems mass were concentrated there and (2) all external forces were applied there. • Location of the Center-of-Mass: M tot rcm Center-of-Mass N 1 M tot M tot N i 1 M tot v cm Ptot v cm i m1 m 2 m 3 m N (total mass of the system) m 1 r1 m 2 r2 m 3 r3 m N rN m i ri M tot • Velocity of the Center-of-Mass: Ptot m i 1 m i ri m 1 r1 m 2 r2 m 3 r3 m N rN i 1 rcm N N (total momentum of the system) p i m 1 v1 m 2 v 2 m 3 v 3 m N v N i 1 M tot N 1 M tot i 1 m 1 v1 m 2 v 2 m 3 v 3 m N v N pi M tot • Acceleration of the Center-of-Mass: (net force acting on the system of particles) F net M tot a cm N m i a i m 1 a1 m 2 a 2 m 3 a 3 m N a N i 1 R. Field 10/10/2013 University of Florida PHY 2053 Fnet d Ptot dt M tot a cm Page 1 Center-of-Mass: Superposition • Superposition: U n iform d en sity r A If object A has mass MA and its center-of-mass is located at rcm B and object B has mass MB and its center-of-mass is located at rcm then the center of mass of the system A+B is located at R A B M A rcm M B rcm A B rcm M A M 2R cm h ole D isk A B • Example: What is the x-coordinate of the center-of-mass of the circular disk of radius 2R with a circular hole of radius R as shown in the figure? Assume the disk has a uniform mass density r. Answer: R/3 x 0 D isk A + B M A x cm M B x cm A A B cm M B A M B Let object A+B be a uniform disk of radius 2R, height h, and center at x=y=0. Let object A be the disk with the hole in it. Let object B be a uniform disk with radius R, height h, and center at y=0, x=-R. x cm A (M A M B) M A B A R. Field 10/10/2013 University of Florida R D isk B 0 x cm 2R MB M A x cm B MB M A R hr 2 x cm B ( ( 2 R ) R ) h r 2 PHY 2053 2 R R 3 Page 2 Center-of-Mass: Example z • Fall 2011 Exam 2 Problem 22: The figure shows a cubical box with each side consisting of a uniform metal plate of negligible thickness. Each of the four sides have mass, M, and the bottom has mass Mbottom. The box is open at the top (at z = L) and has edge length L. If the the z-coordinate of the center-of-mass is at zCM = L/3, what is Mbottom? y x Answer: 2M % Right: 24% 0 z cm cm 1 1 M z M 2z cm 2 M 3z cm 3 M 4z cm 4 M bottom z cm bottom M 1 M 2 M 3 M 4 M bottom 4 M ( L / 2) 4 M M bottom 2 ML 4 M M bottom z cm ( 4 M M bottom ) 2 ML M bottom R. Field 10/10/2013 University of Florida 2 ML z cm 4M 2 ML 4M 2M ( L / 3) PHY 2053 Page 3 Collisions in 1 Dimension: Elastic • Elastic Collision: Lab Frame Before An elastic collision is one in which the kinetic energy is conserved (i.e. the initial total kinetic energy is equal to the final total kinetic energy). If a projectile with mass M1 and speed v1 traveling to the right along the x-axis collides with a target particle at rest with mass M2, what are the velocities of the two particles after they undergo an elastic collision? (momentum conservation) 0 p initial M 1 v1 M 2 v 2 p final M 1 v '1 M 2 v ' 2 v1 v2 =0 M1 M2 x-axis Lab Frame After v'1 v'2 M1 M2 x-axis M 1 ( v1 v '1 ) M 2 v ' 2 (1) Note: a2 - b2 = (a-b)(a+b) (energy conservation) 0 KE initial 1 2 1 2 M 1 v1 2 M 2 v 2 KE 2 1 2 M 1 ( v1 ( v '1 ) ) 2 2 1 2 v '2 R. Field 10/10/2013 University of Florida 1 2 M 1 ( v '1 ) 2 M 1 ( v1 v '1 )( v1 v '1 ) 2M 1 M1 M 1 2 1 2 M 2 ( v '2 ) 2 2 M 2 ( v ' 2 ) (2) v1 v ' 2 v '1 (4) v1 v '1 v ' 2 (3) Divide eq. 2 by eq. 1 and multiply by 2 Multiply eq. 3 by M1 and add it to eq. 1 final v1 2 Insert eq. 5 into eq.4 (5) PHY 2053 v '1 M1 M 2 M1 M 2 v1 Page 4 1d Collisions: Completely Inelastic • Completely Inelastic Collision (Lab Frame): A projectile with mass M1 and speed v1 traveling to the right along the x-axis collides with a target particle at rest with mass M2. If the two particles stick together to form a single particle of mass M1+M2, what is its velocity? What is the velocity of the center-of-mass of the two particles system before the collision? What is the change in the kinetic energy before and after the collision? 0 p initial M 1 v1 M 2 v 2 p final ( M 1 M 2 )V V V cm M1 M1 M 2 v1 KE KE f v1 v2 =0 M1 M2 x-axis Lab Frame After V M1+M2 x-axis (momentum conservation) KE i 1 M 1M 2 2 M1 M 2 2 v1 • Completely Inelastic Collision (CM Frame): CM CM 0 CM CM cm p1 p 2 M tot v cm Ptot 0 p 1 p 2 cm CM CM CM Ptot p1 p 2 p f ( M 1 M 2 )V 0 In the CM frame the initial two particles have equal and opposite momentum. They stick together and are at rest with zero final momentum and zero final kinetic energy. This corresponds to the maximal loss of kinetic energy consistent with momentum conservation. R. Field 10/10/2013 University of Florida Lab Frame Before PHY 2053 CM Frame Before v1 v2 M1 M2 x-axis CM Frame After V=0 M1+M2 x-axis Page 5