A mass is performing vertical circular motion | vedclass

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(B) The tension $T$ at any point in a vertical circular motion is given by:$T = \frac{mv^2}{l} + mg \cos 	heta$where $	heta$ is the angular displace

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$B$

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(B) The tension $T$ at any point in a vertical circular motion is given by:$T = \frac{mv^2}{l} + mg \cos 	heta$where $	heta$ is the angular displacement from the lowest point,$l$ is the length of the string,and $m$ is the mass of the particle.At the lowest point $(B)$,$	heta = 0^\circ$,so $\cos 	heta = 1$,and the tension is $T_B = \frac{mv_B^2}{l} + mg$.At the highest point $(A)$,$	heta = 180^\circ$,so $\cos 	heta = -1$,and the tension is $T_A = \frac{mv_A^2}{l} - mg$.It is clear that the tension is maximum at the lowest point $(B)$ because the centripetal force and the component of gravity both contribute to the tension. If the average velocity of the particle is increased,the tension at the lowest point will increase the most. Therefore,the string has the maximum probability of breaking at point $B$.

$A$ mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased,then at which point will the string break?

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