A) \[\frac{3}{5}\]
B) \[\frac{2}{7}\]
C) \[\frac{2}{5}\]
D) \[\frac{3}{7}\]
Correct Answer: B
Solution :
Key Idea: Total energy is sum of translational and rotational kinetic energy. When spherical ball rolls without slipping, its rotational kinetic energy is \[{{K}_{R}}=\frac{1}{2}I{{\omega }^{2}}\] where J is moment of inertia\[\left( =\frac{2}{5}m\,{{R}^{2}} \right)\]and\[\omega \]is angular velocity. Translational kinetic energy is\[{{K}_{T}}=\frac{1}{2}m\,{{v}^{2}}\] where v is velocity of ball. Total energy = Rotational kinetic energy + Translational energy \[=\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}\left( \frac{2}{5}m{{R}^{2}} \right){{\left( \frac{v}{R} \right)}^{2}}+\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}\] \[=\frac{7}{10}m{{v}^{2}}\] \[\therefore \] \[Required\text{ }fraction=\frac{Rotational\text{ }energy}{Total\text{ }energy}\] \[=\frac{\frac{1}{5}m{{v}^{2}}}{\frac{7}{10}m{{v}^{2}}}=\frac{2}{7}\] Note: Rotational kinetic energy is less than translational kinetic energy.
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