A stone of mass m is tied to a string and is | vedclass

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(D) At the lowest point of a vertical circle,the tension $T$ in the string is given by the equation $T = mg + m\omega^2 r$.Here,the angular velocity $

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$m\{g + (\pi^2 n^2 r)/900\}$

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(D) At the lowest point of a vertical circle,the tension $T$ in the string is given by the equation $T = mg + m\omega^2 r$.Here,the angular velocity $\omega$ in radians per second is calculated from the frequency $n$ revolutions per minute as $\omega = \frac{2\pi n}{60} = \frac{\pi n}{30} 	ext{ rad/s}$.Substituting this into the tension formula:$T = mg + m\left(\frac{\pi n}{30}ight)^2 r$$T = mg + m\left(\frac{\pi^2 n^2}{900}ight)r$$T = m\left\{g + \frac{\pi^2 n^2 r}{900}ight\}$.

$A$ stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n$ revolutions per minute. The total tension in the string when the stone is at its lowest point is

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