Unformatted text preview: the triangle? CoM of each side: Side a: CoM = ½a = ( ½(x A +x B ) , ½(y A +y B ) ) Side b: CoM = ½b = ( ½(x B +x C ) , ½(y B +y C ) ) Side c: CoM = ½c = ( ½(x C +x A ) , ½(y C +y A ) ) CoM of triangle (average of 3 centers of masses above): CoM = ( 1 / 3 [½(x A +x B )+½(x B +x C )+½(x C +x A )] , 1 / 3 [½(y A +y B )+ ½(y B +y C )+½(y C +y A )] ) 3. If mass is uniformly distributed over the interior of ABC, where is the center of mass? For a triangle of uniform interior mass, the center of mass is also located at the centroid of the triangle CoM = 1 / 3 (A+B+C) CoM = ( 1 / 3 (x A +x B +x C ) , 1 / 3 (y A +y B +y C ) )...
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 Spring '00
 GOODE
 Math, Center Of Mass, Mass, Fundamental physics concepts

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