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(A) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_

Vedclass · Accepted Answer

$\frac{7}{10}mv^2$

Vedclass · Answer

(A) The total kinetic energy $(K_{total})$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.$K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia about its center of mass is $I = \frac{2}{5}mR^2$ and the rolling condition is $v = R\omega$ (or $\omega = v/R$).Substituting these values:$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mv^2)$$K_{total} = \frac{1}{2}mv^2(1 + \frac{2}{5})$$K_{total} = \frac{1}{2}mv^2(\frac{7}{5}) = \frac{7}{10}mv^2$.

$A$ solid sphere is rolling with a linear velocity $v$. Its total kinetic energy is:

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