A solid sphere of mass m rolls without slippi | vedclass

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(D) The total kinetic energy of a rolling solid sphere is the sum of its translational kinetic energy and rotational kinetic energy.Total kinetic ener

Vedclass · Accepted Answer

$v\sqrt {\frac{{7m}}{{5k}}} $

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(D) The total kinetic energy of a rolling solid sphere is the sum of its translational kinetic energy and rotational kinetic energy.Total kinetic energy $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and for rolling without slipping,$\omega = \frac{v}{r}$.Substituting these values,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.By the law of conservation of energy,this kinetic energy is converted into the potential energy of the spring at maximum compression $x$.$\frac{7}{10}mv^2 = \frac{1}{2}kx^2$.Solving for $x$,$x^2 = \frac{14mv^2}{10k} = \frac{7mv^2}{5k}$.Therefore,$x = v\sqrt{\frac{7m}{5k}}$.

$A$ solid sphere of mass $m$ rolls without slipping with velocity $v$ and compresses a spring of spring constant $k$ as shown in the figure. The maximum compression in the spring will be:

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