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

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(D) The ball moves in a horizontal circle of radius $r = L \sin \theta$.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: $T \cos \theta = mg$Horizontal component: $T \sin \theta = m \omega^2 r = m \omega^2 (L \sin \theta)$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 \times \omega^2 \times 0.5$$324 = 0.25 \times \omega^2$$\omega^2 = \frac{324}{0.25} = 1296$$\omega = \sqrt{1296} = 36 \ rad/s$.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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$A$ 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 a vertical axis. The maximum tension that the string can bear is $324 \ N$. The maximum possible value of angular velocity of the ball (in $rad/s$) is

IIT 2011 All IIT

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$9$
B
$18$
C
$27$
D
$36$

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3-2.Motion in Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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