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Unformatted text preview: kg, m2 = 2.5 kg, and m3 = 3.2 kg, that form an equilateral triangle with edge length a = 140 cm. xcm 1 = M ∑m x
i =1 3 ii = m1 x1 + m2 x2 + m2 x3 = 83.0cm m1 + m2 + m2 ycm ˆ ⇒ rcm = (83cm )i + (58cm ) ˆ j 1 = M ∑m y
i =1 i 3 i = 58.0cm What if the the distribu6on of par6cles is con6nuous? What is the center of mass of this object? Ans: We replace the summa6on by integra6on. 1 rcm = M ∑mr
i =1 n ii 1 rcm = r dm M∫ In components: xcm = 1 x dm, M∫ ycm = 1 M ∫ y dm,
1 V zcm = 1 z dm, M∫ If the object has uniform density (made from same materials throughout its body): xcm = 1 x dV , V∫ ycm = ∫ y dV , zcm = 1 z dV V∫ Newton’s second law revisited Ques0on: How could we write down Newton’s second law for this object ? Answer: We relate everything to the center of mass and treat the object as if all the mass of it were concentrated there. Fnet = Macm Here: M: total mass of system or object. Fnet: net, external force ac6ng on system. Internal forces betwee...
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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.
 Spring '08
 BLANK
 Momentum

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