A solid sphere of mass m rolls down an inclin | vedclass

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Question

(D) The total kinetic energy $(KE)$ of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.$K

Vedclass · Accepted Answer

$\frac{7}{10} m v^{2}$

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(D) The total kinetic energy $(KE)$ of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.$KE = KE_{	ext{trans}} + KE_{	ext{rot}}$$KE = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2}$For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^{2}$ and the condition for rolling without slipping is $v = R \omega$,which implies $\omega = \frac{v}{R}$.Substituting these into the equation:$KE = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m R^{2}) (\frac{v}{R})^{2}$$KE = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m v^{2})$$KE = \frac{1}{2} m v^{2} (1 + \frac{2}{5})$$KE = \frac{1}{2} m v^{2} (\frac{7}{5})$$KE = \frac{7}{10} m v^{2}$

$A$ solid sphere of mass $m$ rolls down an inclined plane without slipping,starting from rest at the top of an inclined plane. The linear speed of the sphere at the bottom of the inclined plane is $v$. The kinetic energy of the sphere at the bottom is

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