Phys06 - Unit 6 Center of Mass 6.1 6.2 6.3 System of...

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1 Unit 6 Center of Mass 6.1 System of particles 6.2 Collision and the Newton’s second law for a system of particles 6.3 Appendix – Centers of mass of uniform bodies 6.1 System of particles (1) Two particles in 1-D Total mass: Mmm =+ 12 We set a reference zero O on the line along the particles. Center of mass: x mx mm M CM + + = + 11 2 2 If mmm == then x x m xx CM = + = + () 22 1 x 2 x d 1 m 2 m x 0
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2 (2) N particles in 1-D x mx m x mmm m M CM NN N ii i N + +⋅⋅⋅⋅⋅⋅+ + + = = 11 22 123 1 (3) N particles in 3-D r 1 1 = (,,) xyz r 2 2 = ⋅⋅⋅⋅⋅⋅ = = = = = = N i i i CM N i i i CM N i i i CM z m M z y m M y x m M x 1 1 1 1 1 1 rr CM i i i N M m = = 1 1 (4) For continuous system x M xm CM i i i = 1 When m i 0 x M xdm CM = 1 y M ym CM i i i = 1 y M ydm CM = 1 In general, = dm M CM r r 1 Remarks: The centre of mass could be outside the object. 1 m 2 m 3 m 1 r G 2 r G 3 r G O M i m i r G x Block
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3 Example Suppose the masses in the following figure are separated by 0.500 m, and that m 1 = 0.260 kg and m 2 = 0.170 kg. What is the distance from m 1 to the center of mass of the system? Answer: Method 1: We know that the center of mass lies on the line joining m 1 and m 2 . Now, let O be the center of mass and x 1 be the distance between m 1 and O . We can write 2 1 1 1 ) 500 . 0 ( m x m x = That is, ) 170 . 0 )( 500 . 0 ( ) 260 . 0 ( 1 1 x x = , and we obtain 198 . 0 1 = x m. This is the location of the center of mass. Method 2: We take an arbitrary point O as the origin which has a distance a from m 1 . Hence, we can write 2 1 2 1 1 ) 500 . 0 ( m m a m a m x a + + + = + , which gives 198 . 0
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This note was uploaded on 09/06/2010 for the course BSC PHY1417 taught by Professor Prof during the Spring '08 term at HKU.

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Phys06 - Unit 6 Center of Mass 6.1 6.2 6.3 System of...

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