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JEE Main & Advanced Physics Two Dimensional Motion Question Bank Uniform Circular Motion

  • question_answer
    If a particle of mass \[m\] is moving in a horizontal circle of radius \[r\]  with a centripetal force \[(-k/{{r}^{2}})\], the total energy is                     [EAMCET (Med.) 1995; AMU (Engg.) 2001]

    A)             \[-\frac{k}{2r}\]         

    B)             \[-\frac{k}{r}\]

    C)             \[-\frac{2k}{r}\]         

    D)             \[-\frac{4k}{r}\]

    Correct Answer: A

    Solution :

                    \[\frac{m{{v}^{2}}}{r}=\frac{k}{{{r}^{2}}}\] Þ \[m{{v}^{2}}=\frac{k}{r}\]\ K.E.= \[\frac{1}{2}m{{v}^{2}}=\frac{k}{2r}\]             P.E.\[=\int{F\,dr}\]\[=\int{{}}\frac{k}{{{r}^{2}}}dr=-\frac{k}{r}\]                   \ Total energy = K.E. + P.E. \[=\frac{k}{2r}-\frac{k}{r}=-\frac{k}{2r}\]


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