A ball of mass m is attached to the free end | vedclass

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Question

(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

Vedclass · Accepted Answer

$\sqrt{\frac{T}{m \ell}}$

Vedclass · Answer

(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}}$

$A$ 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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