A solid sphere of radius R is rolled by a for | vedclass

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(B) The force $F$ acts at the top of the sphere,which is at a distance $2R$ from the point of contact. When the centre of mass $(CM)$ moves by a dista

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Speed of the $CM$ when $CM$ moves a distance $S$ is $\sqrt{\frac{20FS}{7M}}$.

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(B) The force $F$ acts at the top of the sphere,which is at a distance $2R$ from the point of contact. When the centre of mass $(CM)$ moves by a distance $S$,the point of application of force moves by $2S$. Thus,the work done by force $F$ is $W_F = F(2S) = 2FS$. Since the sphere rolls without slipping,the work done by static friction is zero. Using the work-energy theorem: $W_F + W_{friction} = \Delta K$. $2FS = \frac{1}{2}Mv^2 + \frac{1}{2}I_{cm}\omega^2$. For a solid sphere,$I_{cm} = \frac{2}{5}MR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$. $2FS = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$. $2FS = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$. $v^2 = \frac{20FS}{7M} \implies v = \sqrt{\frac{20FS}{7M}}$. Therefore,option $B$ is correct.

$A$ solid sphere of radius $R$ is rolled by a force $F$ acting at the top of the sphere as shown in the figure. The sphere rolls without slipping on a rough stationary surface. Initially,the sphere is at rest. Then:

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