CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    A sphere of mass m and radius r rolls on a horizontal plane without slipping with the speed u. Now, if it rolls up vertically, the maximum height it would attain will be

    A)  \[\frac{3{{u}^{2}}}{4g}\]                                              

    B)  \[\frac{5{{u}^{2}}}{2g}\]              

    C)  \[\frac{7{{u}^{2}}}{10g}\]                           

    D)  \[\frac{{{u}^{2}}}{2g}\]

    E)  \[\frac{11{{u}^{2}}}{9g}\]

    Correct Answer: C

    Solution :

    The rolling sphere has rotational as well as translational kinetic energy. \[\therefore \]Kinetic energy \[=\frac{1}{2}m{{u}^{2}}+\frac{1}{2}I{{\omega }^{2}}\]                                 \[=\frac{1}{2}m{{u}^{2}}+\frac{1}{2}\left( \frac{2}{5}m{{r}^{2}} \right){{\omega }^{2}}\]                                 \[=\frac{1}{2}m{{u}^{2}}+\frac{1}{5}m{{u}^{2}}=\frac{7}{10}m{{u}^{2}}\] \[(\because \,\,u=r\omega )\] From conservation of energy, Potential energy = kinetic energy ie,           \[mgh=\frac{7}{10}m{{u}^{2}}\] or            \[h=\frac{7{{u}^{2}}}{10g}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner