Solid spherical ball is rolling on a friction | vedclass

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Question

$K _{	ext {total }}= K _{	ext {rotational }}+ K _{	ext {Translational }}$

$K _{ total }=\frac{1}{2} I _{ cm } \omega^{2}+\frac{1}{2} mV _{ cm }^

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

$K _{	ext {total }}= K _{	ext {rotational }}+ K _{	ext {Translational }}$

$K _{ total }=\frac{1}{2} I _{ cm } \omega^{2}+\frac{1}{2} mV _{ cm }^{2}$

$v _{ cm }= R \omega$ for pure rolling

$I _{ cm }=\frac{2}{5} mR ^{2}$

$K _{ Rot }=\frac{1}{2} I _{ cm } \omega^{2}=\frac{1}{2} 	imes \frac{2}{5} mR ^{2} 	imes \frac{ v _{ cm }^{2}}{ R ^{2}}=\frac{1}{5} mv _{ cm }^{2}$

$K _{	ext {Total }}=\frac{1}{5} mv _{ cm }^{2}+\frac{1}{2} mv _{ cm }^{2}=\frac{7}{10} mv _{ cm }^{2}$

$\frac{ K _{ Rot }}{ K _{	ext {Total }}} \frac{\frac{1}{5} mv _{ cm }^{2}}{\frac{7}{10} mv _{ cm }^{2}}=\frac{2}{7}$

Solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is ................

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