A ball of mass ( m )=0.5 kg is attached to t | vedclass

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Question

REF.Image.

$m \omega^2 L \sin 	heta \cos 	heta= mg \sin 	heta$

$\Rightarrow \cos 	heta=\frac{ g }{\omega^2 L}$

$	herefore \sin 	heta=\f

Vedclass · Accepted Answer

$36$

Vedclass · Answer

REF.Image.

$m \omega^2 L \sin 	heta \cos 	heta= mg \sin 	heta$

$\Rightarrow \cos 	heta=\frac{ g }{\omega^2 L}$

$	herefore \sin 	heta=\frac{1}{\omega^2 L} \sqrt{\left(\omega^2 Light)^2-g^2}$

$T= mg \cos 	heta+ m \omega^2 L \sin ^2 	heta$

$= mg \frac{ g }{\omega^2 L}+\frac{ m \omega^2 L}{\left(\omega^2 Light)^2}\left(\left(\omega^2 Light)^2- g ^2ight)$

$=\frac{ m }{\omega^2 L}\left[g^2+\left(\omega^2 Light)^2- g ^2ight]$

$= m \omega^2 L=324 	ext { (given) }$

$\Rightarrow \omega=\sqrt{\frac{324}{0.5 	imes 0.5}}$

$\omega=36 \ rad / s$

A ball of mass $( m )=0.5 \ kg$ is attached to the end of a string having length $(L)$ $=0.5 m$. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is $324 \ N$. The maximum possible value of angular velocity of ball (in radian/s) is

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