(A) In a horizontal circular motion,the tension $T$ in the string provides the necessary centripetal force required for the circular motion of the bal

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(A) In a horizontal circular motion,the tension $T$ in the string provides the necessary centripetal force required for the circular motion of the ball.For a ball of mass $m$ moving in a horizontal circle of radius $r$ with angular velocity $\omega$,the centripetal force is given by $F_c = m r \omega^2$.Here,the length of the string $\ell$ acts as the radius of the circular path $(r = \ell)$.Therefore,the tension $T$ is equal to the centripetal force:$T = m \ell \omega^2$Rearranging the formula to solve for $\omega$:$\omega^2 = \frac{T}{m \ell}$$\omega = \sqrt{\frac{T}{m \ell}}$

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 $m$ is attached to the free end of an inextensible string of length $\ell$. Let $T$ be the tension in the string. The ball is moving in a horizontal circular path about the vertical axis. The angular velocity of the ball at any particular instant will be

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A
$\sqrt{\frac{T}{m \ell}}$
B
$\sqrt{\frac{T \ell}{m}}$
C
$\sqrt{\frac{m \ell}{T}}$
D
$\sqrt{\frac{Tm}{\ell}}$

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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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$A$ ball of mass $m$ is attached to the free end of an inextensible string of length $\ell$. Let $T$ be the tension in the string. The ball is moving in a horizontal circular path about the vertical axis. The angular velocity of the ball at any particular instant will be

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$A$ ball of mass $0.5 \ kg$ is attached to a string of length $50 \ cm$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400 \ N$. The maximum possible value of angular velocity of the ball in $rad/s$ is:

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