161_Problem CHAPTER 9

161_Problem CHAPTER - PROBLEM 9.128 CONTINUED Now and I = x 2dm 2 I = x 2dm = 2 0 x 2 t dx a a h 2 = 2 t 0 x 2 a 2 x dx a h a 1 2 = 2 t x3 x 4 a 3

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PROBLEM 9.128 CONTINUED Now 2 I xdm ζ =∫ and () 2 22 0 2 a I x t dx ρζ = ∫ 2 2 0 a h tx a xd x a ρ 2 34 0 1 2 a ha txx a =− 1 2 32 42 haa a t a    11 48 24 ta h ma == ( b ) Have 2 sin yy I rdm x dm ζθ + 2 sin x dm dm θζ + Now sin xy x Id m I I I θ = + 2 sin 24 6 ma mh =+ 2 or 4 sin 24 y m Ia h W ( c ) Have ( ) 2 zz I x y dm + 2 2 cos x dm + 2 cos x dm dm + 2 cos x II 2 cos 24 6 ma
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This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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