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(D) For a particle performing vertical circular motion,the tension $T$ at the top point is given by $T = \frac{mv^2}{r} - mg$.For the tension to be ze

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$7$

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(D) For a particle performing vertical circular motion,the tension $T$ at the top point is given by $T = \frac{mv^2}{r} - mg$.For the tension to be zero at the top point,we set $T = 0$,which gives $\frac{mv^2}{r} = mg$.This simplifies to $v^2 = rg$,or $v = \sqrt{rg}$.Since the angular velocity $\omega$ is related to linear velocity $v$ by $v = r\omega$,we have $\omega = \frac{v}{r} = \frac{\sqrt{rg}}{r} = \sqrt{\frac{g}{r}}$.Given $r = 20\,cm = 0.2\,m$ and $g = 9.8\,m/s^2$,we calculate:$\omega = \sqrt{\frac{9.8}{0.2}} = \sqrt{49} = 7\,rad/s$.

$A$ particle is tied to a $20\,cm$ long string. It performs circular motion in a vertical plane. What is the angular velocity of the string when the tension in the string at the top is zero? (Take $g = 9.8\,m/s^2$)

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