A solid sphere of mass M rolls without slippi | vedclass

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(D) The total kinetic energy $(KE)$ of a solid sphere rolling without slipping is the sum of its translational and rotational kinetic energies.$KE_{to

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

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(D) The total kinetic energy $(KE)$ of a solid sphere rolling without slipping is the sum of its translational and rotational kinetic energies.$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$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:$KE_{total} = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$When the sphere compresses the spring by a maximum distance $x_m$,the entire kinetic energy is converted into the potential energy of the spring $(U = \frac{1}{2} kx_m^2)$.Equating the two:$\frac{1}{2} kx_m^2 = \frac{7}{10} Mv^2$$x_m^2 = \frac{14}{10} \frac{Mv^2}{k} = \frac{7Mv^2}{5k}$$x_m = v \sqrt{\frac{7M}{5k}}$

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

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