A solid sphere rolls without slipping down an inclined plane of angle $\theta$. What fraction of its total kinetic energy is rotational?

## CORE INFORMATION\n✓ **Answer**: [Option C]\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Kinetic Energy of Rolling Motion\n\t**Related Topics**: Rotational Kinetic Energy, Translational Kinetic Energy, Moment of Inertia\n\t**Prerequisites**: Understanding of rotational and translational kinetic energy, moment of inertia for a solid sphere.\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Solid sphere rolling without slipping\n\t* Inclined plane with angle $\theta$\n**Method & Steps**:\n\t1. Define total kinetic energy:\n\t\tReasoning: Total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energies.\n\t\tFormula/Rule: $KE_{total} = KE_{trans} + KE_{rot}$\n\t2. Express translational kinetic energy:\n\t\tReasoning: Translational kinetic energy is given by $\frac{1}{2}mv^2$, where m is the mass and v is the linear velocity.\n\t\tFormula/Rule: $KE_{trans} = \frac{1}{2}mv^2$\n\t3. Express rotational kinetic energy:\n\t\tReasoning: Rotational kinetic energy is given by $\frac{1}{2}I\omega^2$, where I is the moment of inertia and $\omega$ is the angular velocity.\n\t\tFormula/Rule: $KE_{rot} = \frac{1}{2}I\omega^2$\n\t4. Relate linear and angular velocity:\n\t\tReasoning: For rolling without slipping, the linear velocity is related to the angular velocity by v = r$\omega$, where r is the radius of the sphere.\n\t\tFormula/Rule: $v = r\omega$\n\t5. Find the moment of inertia of a solid sphere:\n\t\tReasoning: The moment of inertia of a solid sphere about its center is $\frac{2}{5}mr^2$.\n\t\tFormula/Rule: $I = \frac{2}{5}mr^2$\n\t6. Substitute and calculate rotational kinetic energy:\n\t\tWorking: $KE_{rot} = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{5}mv^2$\n\t\tKey Point: Substituting $\omega = v/r$ and $I = \frac{2}{5}mr^2$\n\t7. Calculate total kinetic energy:\n\t\tWorking: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$\n\t\tKey Point: Sum of the translational and rotational kinetic energy\n\t8. Find the fraction of rotational kinetic energy:\n\t\tWorking: $\frac{KE_{rot}}{KE_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$\n\t\tResult: Fraction of rotational kinetic energy is 2/7\n\t\tVerification: The result is unitless as it is a fraction.\n ## OPTION ANALYSIS\n[A] :x: Incorrect. This would be the fraction if the rotational and translational kinetic energies were equal\n[B] :x: Incorrect. This is the fraction of rotational KE divided by the translational KE.\n[C] ✓: Correct. The fraction of the total kinetic energy that is rotational is 2/7.\n[D] :x: Incorrect. This is not the correct fraction obtained from the calculation.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting to use the correct moment of inertia for a solid sphere. [How to avoid]: Remember the correct formula for the moment of inertia, $I = \frac{2}{5}mr^2$.\n\t\t2.[Error 2]: Not correctly relating linear and angular velocity. [How to avoid]: Always remember the relation $v = r\omega$ when dealing with rolling without slipping.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: The fraction of rotational kinetic energy for a rolling object depends only on the moment of inertia and not on the mass or velocity.\n\t\t2.[Memory aid]: The fraction can be generalized as $\frac{I}{I+mr^2}$ which will give $\frac{2/5}{2/5+1} = \frac{2}{7}$ for a solid sphere.

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