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 rad/s) is [image:4_image43.emf]

Question

Dr. Abubacker · Accepted Answer

(d) $T\cos	heta$ component will cancel $mg.$
[image:4_image131.emf]
$T\sin	heta$ Component will provide necessary centripetal force the ball towards center C.
$	herefore T\sin{	heta = mr\omega^{2} = m(l\sin{	heta)\omega^{2}}\ }$
$or\ T = ml\omega^{2} \Longrightarrow \omega = \sqrt{\frac{T}{ml}}rad/s\ $
${or\ \omega}_{\max} = \sqrt{\frac{T_{\max}}{ml} = \sqrt{\frac{324}{0.5 	imes 0.5} = 36\ rad/s}}$

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 rad/s) is

Explanation

Reviewed by Dr. Abubacker

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