A uniform sphere of mass 500; g rolls without | vedclass

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Question

$\mathrm{m}=0.5 \mathrm{kg}, \mathrm{v}=5 \mathrm{cm} / \mathrm{s}$

$\mathrm{KE}$ in rolling $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \

Vedclass · Accepted Answer

$8.75 	imes 10^{-4} \;\mathrm{J}$

Vedclass · Answer

$\mathrm{m}=0.5 \mathrm{kg}, \mathrm{v}=5 \mathrm{cm} / \mathrm{s}$

$\mathrm{KE}$ in rolling $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$

$=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}ight)$

$=8.75 	imes 10^{-4} \mathrm{J}$

A uniform sphere of mass $500\; g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00\; \mathrm{cm} / \mathrm{s}$. Its kinetic energy is

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