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text  A composite rod comprising two rods of mass  end text straight m text  and  end text 2 straight m text  and each of length  end text l equals 1 straight m text  as shown in the  end text
text  figure. Assume  end text omega equals 10 rad divided by straight s text  and  end text straight m equals 1 kg comma text  the  end text KE text  of the rotating composite rod is  end text

  1. 125J    
  2. 150J    
  3. 250J
  4. 175J 

The correct answer is: 250J


    M o m e n t space o f space i n e r t i a space o f space c o m p o s i t e space r o d space a b o u t space P
I subscript P equals fraction numerator m ell squared over denominator 3 end fraction plus fraction numerator left parenthesis 2 straight m right parenthesis ell squared over denominator 12 end fraction plus 2 m open parentheses 3 over 2 ell close parentheses squared equals 5 m ell squared
table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em rowspacing 3 pt end attributes row cell text  Total kinetic energy  end text k text  total  end text == 1 half I subscript P omega squared end cell row cell equals 1 half open parentheses 5 straight m ell squared close parentheses omega squared equals 5 over 2 m ell squared omega squared equals 250 straight J end cell end table

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