A solid sphere is rolling down an inclined pl | vedclass

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Question

(A) The translational kinetic energy is given by $K_{T} = \frac{1}{2} mv^{2}$.The rotational kinetic energy is given by $K_{R} = \frac{1}{2} I \omega^

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$2.5$

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(A) The translational kinetic energy is given by $K_{T} = \frac{1}{2} mv^{2}$.The rotational kinetic energy is given by $K_{R} = \frac{1}{2} I \omega^{2}$.For a solid sphere,the moment of inertia $I = \frac{2}{5} mR^{2}$ and the condition for pure rolling is $v = R\omega$,so $\omega = \frac{v}{R}$.Substituting these into the rotational kinetic energy formula: $K_{R} = \frac{1}{2} (\frac{2}{5} mR^{2}) (\frac{v}{R})^{2} = \frac{1}{5} mv^{2}$.Now,the ratio of translational kinetic energy to rotational kinetic energy is $\frac{K_{T}}{K_{R}} = \frac{\frac{1}{2} mv^{2}}{\frac{1}{5} mv^{2}} = \frac{5}{2} = 2.5$.

$A$ solid sphere is rolling down an inclined plane. What is the ratio of its translational kinetic energy to its rotational kinetic energy?

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