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 [image:14_image47.emf]

Question

Dr. Abubacker · Accepted Answer

(d) [image:14_image88.emf]
$T\sin	heta = mL\sin{	heta\omega^{2}}$
$324 = 0.5 	imes 0.5 	imes \omega^{2}$
$\Rightarrow \omega^{2} = \frac{324}{0.5 	imes 0.5}$
$\Rightarrow \omega = \sqrt{\frac{324}{0.5 	imes 0.5}}$
$\Rightarrow \omega = \frac{18}{0.5} = 36\ rad/sec$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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