A disc of mass 1,kg and radius R is free of r | vedclass

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Question

$mg 2 R =\frac{1}{2} I _{	ext {disc }} \omega^{2}+\frac{1}{2} I _{	ext {particle }} \omega^{2}$

$mg 2 R =\frac{\omega^{2}}{2}\left[\frac{ mR ^{2}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$mg 2 R =\frac{1}{2} I _{	ext {disc }} \omega^{2}+\frac{1}{2} I _{	ext {particle }} \omega^{2}$

$mg 2 R =\frac{\omega^{2}}{2}\left[\frac{ mR ^{2}}{2}+ mR ^{2}ight]$

$mg 2 R =\frac{\omega^{2}}{2} \frac{3}{2} mR ^{2}$

$\frac{3}{4} \omega^{2}=\frac{2 g }{ R }$

$\omega^{2}=\frac{8 g }{3 R }$

$\omega=\sqrt{\frac{80}{3 R}}$

$16 \frac{ x }{3 R }=\frac{80}{3 R }$

$x =5$

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