A ball of mass m = 0.5 kg is attached to the | vedclass

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(D) The ball moves in a horizontal circle of radius $r = L \sin 	heta$.The forces acting on the ball are tension $T$ along the string and weight $mg$

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) The ball moves in a horizontal circle of radius $r = L \sin 	heta$.The forces acting on the ball are tension $T$ along the string and weight $mg$ acting downwards.Resolving the tension $T$ into components:Vertical component: $T \cos 	heta = mg$Horizontal component: $T \sin 	heta = m \omega^2 r = m \omega^2 (L \sin 	heta)$From the horizontal component,we get $T = m \omega^2 L$.Given that the maximum tension $T_{max} = 324 \ N$,$m = 0.5 \ kg$,and $L = 0.5 \ m$,we have:$324 = 0.5 	imes \omega^2 	imes 0.5$$324 = 0.25 	imes \omega^2$$\omega^2 = \frac{324}{0.25} = 1296$$\omega = \sqrt{1296} = 36 \ rad/s$.

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