188_Problem CHAPTER 9

188_Problem CHAPTER 9 - PROBLEM 9.141 CONTINUED ( I x )4 =...

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PROBLEM 9.141 CONTINUED () ( ) 2 4 1 0.24662 kg 0.100 m 4 x I = 32 0.61655 10 kg m = ×⋅ Then 23.151 9.7667 5.0868 0.61655 10 kg m x I  = ++− ×  or 37.4 10 kg m x I W And ( ) ( ) ( ) ( ) 1234 yy y y y I IIII =++− where ( ) 22 2 1 1 1.20576 kg 0.320 0.240 m 3 y I =+  64.307 10 kg m = ( ) 2 2 1 0.90432 kg 0.320 m 3 y I = 30.867 10 kg m = ( ) 2 3 1 0.33912 kg 0.240 m 6 y I = 3.2556 10 kg m = 2 2 4 11 6 0.24662 kg 0.100 m 2 9 y I π  =−   2 2 2 40 . 1 0 0 0.160 m 3 × ++ 7.5466 10 kg m = Then 64.307 30.067 3.2556 7.5466 10 kg m y I = or 90.9 10 kg m y I W And ( ) ( ) ( ) ( ) zz z z z I where ( ) 2 1 1 1.20576 kg 0.320 m 41.157 10 kg m 3 z I =
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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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