A solid sphere rolls without slipping at speed v. What fraction of its total kinetic energy is rotational?

2/7 KE{\text{trans}} = \tfrac{1}{2}m v^2 and KE{\text{rot}} = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(\tfrac{2}{5}m R^2)(v/R)^2 = \tfrac{1}{5}m v^2. Total = \tfrac{7}{10}mv^2. Rotational fraction = (1/5)/(7/10) = 2/7.

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Acemy

Rotational Kinetic Energy

AP Physics 1· difficulty 4/5

v

A
$2/7$
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B
$1/2$
C
$5/7$
D
$2/5$

Explanation

KE_{\text{trans}} = \tfrac{1}{2}m v^2

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