Answer we relate everything to the center of mass and

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Unformatted text preview: by a distance d, the center of mass can be found from: xcm = m2 d m1 + m2 m1 d m2 x where x is measured from the loca6on of m1. Exampes: 1. Just one par0cle: 0 d=0 m+0 i.e., the CM is in the center of the par6cle m1 = m, m2 = 0 ⇒ xcm = m1 = 5m, m2 = m ⇒ xcm = m 1 d= d m + 5m 6 2. One heavy par0cle (5m), one light par0cle (m): i.e., the center of mass is closer to the heavy par6cle m d d= 3. Two equally heavy par0cle: m1 = m2 = m ⇒ xcm = m+ m 2 i.e., the center of mass is at the midpoint between the par6cles (as one would expect). General expression for the center of mass What if the system consists of more than two parts, distributed in 3D space ? where M=m1+m2+… In terms of components: Example xcm = 1 M m2 r2 m3 m4 r4 ∑m z i =1 n 1 rcm = M ∑ mi ri i =1 n r1 r3 ∑ mi xi , i =1 n ycm = 1 M ∑ mi yi , i =1 n zcm = 1 M ii Find the center of mass for three par;cles with masses m1 = 1.2...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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