# hw9 - m = R R z ρπ r 2 dz m since dm = ρπ r 2 dz = R R...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 9 (2.2, 2.9) 2.2 By symmetry, x = z = 0 y C = m i y i m i = 21( - 1) + 1(1 . 5) + 11(3 . 8) 21 + 1 + 11 = 22 . 3 33 = 0 . 676 f t 2.9 Using the given frame of reference, we need the center of mass to be located at z = 0 z C = m i z i m i Let the density of the material be ρ . The center of mass for a cone and the hemisphere are respectively at 3 H 4 and 3 R 8 , where H and R are the height of the cone and radius of the hemisphere respectively, as shown below. For cone: z C , cone = R H 0 z dm m = R H 0 z ρπ r 2 dz m since dm = ρπ r 2 dz = R H 0 z ρπ K 2 z 2 dz m since r = kz , where k = R / H = ρπ K 2 R H 0 z 3 dz m = ρπ K 2 H 4 dz 4 m = 3 ρπ K 2 H 4 dz 4 ρπ R 2 H since m = 1 / 3 ρπ R 2 H = 3 H 4 i.e., the center of mass of the cone is 3 / 4 H from top of the cone 1

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For sphere: z C , sphere = R R 0 z dm
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Unformatted text preview: m = R R z ρπ r 2 dz m since dm = ρπ r 2 dz = R R z ρπ ( R 2-z 2 ) dz m since r 2 = R 2-z 2 = ρπ R 4 4 m = ρπ R 4 4(2 ρπ R 3 / 3) since m = 1 2 ρ 4 3 π R 3 = 3 R 8 i.e., the center of mass of a hemisphere is 3 R / 8 from the base Using these for ﬁnding the center of mass of the composite body, z C = ∑ m i z i ∑ m i = ± ρ π R 2 H 3 ² ( H 4 ) + ± ρ 2 π R 3 3 ² (-3 R 8 ) ± ρ π R 2 H 3 ² + ± ρ 2 π R 3 3 ² = R 2 H 2 12 = R 4 4 H 2 = 3 R 2 H = R √ 3 2...
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hw9 - m = R R z ρπ r 2 dz m since dm = ρπ r 2 dz = R R...

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