Center of Mass

# Center of Mass - the triangle CoM of each side Side a CoM =...

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Centers of Mass for a Triangle Let ABC be a triangle with vertices {A,B,C} and sides {a,b,c}. 1. If mass is equally distributed on the three vertices, where is the center of mass? CoM is located at the median of the vertices of the triangle, the centroid CoM = 1 / 3 (A+B+C) CoM = ( 1 / 3 (x A +x B +x C ) , 1 / 3 (y A +y B +y C ) ) 2. If mass is uniformly distributed over the sides, where is the center of mass? For a triangle with uniform linear mass, I want to find the centers of mass of each individual side and then find the intersection of those points, but I’m not sure if this will actually get me the correct center of mass or if it is just the same as the above problem, at the centroid? It seems like the center of mass of a triangle with uniform linear mass could be at the centroid sometimes, but not always, depending on
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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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