problem14_48

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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14.48: The cylindrical rod has mass , M radius , R and length L with a density that is proportional to the square of the distance from one end, 2 Cx = ρ . a) . 2 dV Cx dV M = = ρ The volume element . 2 dx πR dV = Then the integral becomes . 2 2 0 dx R Cx M L π = Integrating gives . 3 3 2 2 0 2 L R C dx x R C M L π π = = Solving for . 3 , 3 2 L R M C C π = b) The density at the L
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Unformatted text preview: ( 29 ( 29 ( 29 . 2 3 2 3 2 3 2 L πR M L πR M L Cx ρ = = = The denominator is just the total volume , V so , 3 V M = or three times the average density, . V M So the average density is one-third the density at the L x = end of the rod....
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• average density, volume element dV, Cx R dx, 3M 3M

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