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

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Aicharya · Accepted Answer

$T \cos 	heta$ component will cancel $\mathrm{mg}$. $T \sin 	heta$ component will provide necessary centripetal force to the ball towards centre $C$. $ \begin{array}{rlrl} & 	herefore & T \sin 	heta & =m r \omega^2=m(l \sin 	heta) \omega^2 \ & 	ext { or } & T & =m l \omega^2 \ & 	herefore & \omega & =\sqrt{\frac{T}{m l}} \ & 	ext { or } & \omega_{\max } & =\sqrt{\frac{T_{\max }}{m l}}=\sqrt{\frac{324}{0.5 	imes 0.5}} \ & = & 36 \mathrm{rad} / \mathrm{s} \end{array} $ $	herefore$ Correct option is (d). Analysis of Question (i) Question is simple. (ii) This is called the conical pendulum. (iii) The interesting fact in this problem is that $\omega$ or $T$ is independent of $	heta$. $\omega \propto \sqrt{T}$ If $\omega$ is increased, $T$ will also increase

A ball of mass $(m) 0.5 \mathrm{~kg}$ is attached to the end of a string having length $(L) 0.5 \mathrm{~m}$…

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