A solid sphere of mass 500 gm and radius | vedclass

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Question

since it is pure rolling, $V=\omega R$

$\Rightarrow \omega=\frac{V}{R}=\frac{0.2}{0.1}=2 \mathrm{rad} / \mathrm{s}$

$K E_{	ext {Translational}}

Vedclass · Accepted Answer

$0.014$

Vedclass · Answer

since it is pure rolling, $V=\omega R$

$\Rightarrow \omega=\frac{V}{R}=\frac{0.2}{0.1}=2 \mathrm{rad} / \mathrm{s}$

$K E_{	ext {Translational}}=\frac{1}{2} m v^{2}=\frac{1}{2} 	imes \frac{1}{2} 	imes(0.2)^{2}$

$=10^{-2} J$

$K E_{	ext {Rotational}}=\frac{1}{2} J \omega^{2}=\frac{1}{2}\left(\frac{2}{5} M R^{2}ight) \omega^{2}$

$=\frac{\left(\frac{1}{2}ight)(0.1)^{2}}{5}(2)^{2}=4 	imes 10^{-3} J$

$K E_{T o t a l}=K E_{T r a n s l a t i o n a l}+K E_{r o a t a t i o n a l}$

$=10^{-2} J+4 	imes 10^{-3} J$

$=0.014 J$

A solid sphere of mass $500\ gm$ and radius $10\ cm$ rolls without slipping with the velocity $20\ cm/s$. The total kinetic energy of the sphere will be ........ $J$

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