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MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    If the disc shown in figure has mass M and it is free to rotate about its symmetrical axis passing through 0, its angular acceleration is

    A)  \[\frac{6F}{MR}\]                           

    B)  \[\frac{3F}{MR}\]

    C)  \[\frac{6F}{5MR}\]                        

    D)  \[\frac{4F}{MR}\]

    Correct Answer: A

    Solution :

                    If\[\alpha \]be the angular acceleration of the disc, then, using\[{{F}_{net}}=I\alpha ,\] we have, \[\alpha =\frac{{{F}_{net}}}{I}\] \[=\frac{3FR}{(M{{R}^{2}}/2)}=\frac{6F}{MR}\]


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