15_5_23 - x = y = 0. The centre of mass is on the axis of...

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23. The cone z = h 1 - p x 2 + y 2 a ! has normal n = - z x i - z y j + k = - h a x i + y j p x 2 + y 2 ! + k , so its surface area element is dS = s h 2 a 2 + 1 dx dy = a 2 + h 2 a dx dy . The mass of the conical shell is m = σ ZZ x 2 + y 2 a 2 dS = σ a 2 + h 2 a a 2 ) = πσ a p a 2 + h 2 . The moment about z = 0 is M z = 0 = σ ZZ x 2 + y 2 a 2 h 1 - p x 2 + y 2 a ! a 2 + h 2 a dx dy = 2 πσ h a 2 + h 2 a Z a 0 ± 1 - r a ² r dr = πσ ha a 2 + h 2 3 . Thus z = h 3 . By symmetry,
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Unformatted text preview: x = y = 0. The centre of mass is on the axis of the cone, one-third of the way from the base towards the vertex. x y z h a a z = h-h a √ x 2 + y 2 Fig. 5-23...
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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