AP Physics C Mechanics · Question of the Day

AP Physics C Mechanics Question of the Day

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Friday, September 18, 2026

A thin hoop of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. Its rotational inertia is $I = MR^2$ . What fraction of its total kinetic energy is rotational kinetic energy?

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Question of the Day

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A thin hoop of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. Its rotational inertia is $I = MR^2$ . What fraction of its total kinetic energy is rotational kinetic energy?

$1/4$
$1/3$
$1/2$ (correct answer)
$2/3$

Explanation: The total kinetic energy is $K_{total} = K_{trans} + K_{rot}$ . For a hoop rolling without slipping ( $v=R\omega$ ), $K_{trans} = \frac{1}{2}Mv^2$ and $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2$ . So, $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2$ . The fraction that is rotational is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}Mv^2}{Mv^2} = \frac{1}{2}$ .