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(B) For a ball rolling without slipping,the rotational kinetic energy is given by $K_{rot} = \frac{1}{2} I \omega^2$. Since $I = MK^2$ and $\omega = \

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$\frac{K^2}{K^2 + R^2}$

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(B) For a ball rolling without slipping,the rotational kinetic energy is given by $K_{rot} = \frac{1}{2} I \omega^2$. Since $I = MK^2$ and $\omega = \frac{v}{R}$,we have $K_{rot} = \frac{1}{2} MK^2 \frac{v^2}{R^2}$.The translational kinetic energy is $K_{trans} = \frac{1}{2} Mv^2$.The total kinetic energy $E$ is the sum of rotational and translational kinetic energy:$E = K_{rot} + K_{trans} = \frac{1}{2} MK^2 \frac{v^2}{R^2} + \frac{1}{2} Mv^2 = \frac{1}{2} Mv^2 \left( \frac{K^2}{R^2} + 1 ight) = \frac{1}{2} Mv^2 \left( \frac{K^2 + R^2}{R^2} ight)$.The fraction of total energy associated with rotational energy is $\frac{K_{rot}}{E}$:$	ext{Fraction} = \frac{\frac{1}{2} MK^2 \frac{v^2}{R^2}}{\frac{1}{2} Mv^2 \left( \frac{K^2 + R^2}{R^2} ight)} = \frac{K^2}{K^2 + R^2}$.

$A$ ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If the radius of the ball is $R$,then the fraction of total energy associated with its rotational energy is:

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