PROBLEM 9.144 CONTINUED ()222210.3600.20045.216 kg0.3600.200m1222yI=+++22.5562 kg m=⋅()()222311640.10014.7969 kg0.1000.1000.600m239yIπ×=−+++26.3024 kg m22417.3985 kg0.0500.1000.600m2yI+22.7467 kg mThen ( )215.07202.55626.30242.7467 kg myI+−⋅221.1839 kg m2or 21.2 kg myIW
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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.