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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    A thin uniform rod AB of mass m and length l is hinged at one end A to the level floor. Initially if stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity of the rod, when its end B strikes the floor is {g is acceleration due to gravity):

    A)  \[\left( \frac{mg}{l} \right)\]                                     

    B)  \[{{\left( \frac{mg}{3l} \right)}^{1/2}}\]

    C)  \[\left( \frac{g}{l} \right)\]                                         

    D)  \[{{\left( \frac{3g}{l} \right)}^{1/2}}\]

    Correct Answer: D

    Solution :

                     According to the equation. \[mg\frac{l}{2}=\frac{1}{2}I{{\omega }^{2}}\] \[mg\frac{l}{2}=\frac{1}{2}\frac{m{{l}^{2}}}{3}{{\omega }^{2}}\] \[{{\omega }^{2}}=\frac{2\times 3mg\times l}{m{{l}^{2}}\times 2}\] \[\omega =\sqrt{\frac{3g}{l}}\]


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