A solid sphere of mass 1,kg rolls without sli | vedclass

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(A) The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} Mv^2 + \frac{1}{2} I\

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$10$

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(A) The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5} MR^2$.Since the sphere rolls without slipping,$\omega = \frac{v}{R}$.Substituting these into the energy equation: $K = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$.Given $K = 7 	imes 10^{-3}\,J$ and $M = 1\,kg$,we have $\frac{7}{10} (1) v^2 = 7 	imes 10^{-3}$.$v^2 = 10^{-2} \implies v = 0.1\,m/s$.Converting to $cm/s$: $v = 0.1 	imes 100 = 10\,cm/s$.

$A$ solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........\,cm\,s^{-1}$.

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