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3-2.Motion in Plane
hard
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
A$9$
B$18$
C$27$
D$36$
(IIT-2011)
Solution
REF.Image.
$m \omega^2 L \sin \theta \cos \theta= mg \sin \theta$
$\Rightarrow \cos \theta=\frac{ g }{\omega^2 L}$
$\therefore \sin \theta=\frac{1}{\omega^2 L} \sqrt{\left(\omega^2 L\right)^2-g^2}$
$T= mg \cos \theta+ m \omega^2 L \sin ^2 \theta$
$= mg \frac{ g }{\omega^2 L}+\frac{ m \omega^2 L}{\left(\omega^2 L\right)^2}\left(\left(\omega^2 L\right)^2- g ^2\right)$
$=\frac{ m }{\omega^2 L}\left[g^2+\left(\omega^2 L\right)^2- g ^2\right]$
$= m \omega^2 L=324 \text { (given) }$
$\Rightarrow \omega=\sqrt{\frac{324}{0.5 \times 0.5}}$
$\omega=36 \ rad / s$
$m \omega^2 L \sin \theta \cos \theta= mg \sin \theta$
$\Rightarrow \cos \theta=\frac{ g }{\omega^2 L}$
$\therefore \sin \theta=\frac{1}{\omega^2 L} \sqrt{\left(\omega^2 L\right)^2-g^2}$
$T= mg \cos \theta+ m \omega^2 L \sin ^2 \theta$
$= mg \frac{ g }{\omega^2 L}+\frac{ m \omega^2 L}{\left(\omega^2 L\right)^2}\left(\left(\omega^2 L\right)^2- g ^2\right)$
$=\frac{ m }{\omega^2 L}\left[g^2+\left(\omega^2 L\right)^2- g ^2\right]$
$= m \omega^2 L=324 \text { (given) }$
$\Rightarrow \omega=\sqrt{\frac{324}{0.5 \times 0.5}}$
$\omega=36 \ rad / s$
Standard 11
Physics
