A body of mass m hangs at one end of a string | vedclass

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(A) Let the mass be $m$ and the length of the string be $l$. The body is given a horizontal velocity $v$ at the mean position (lowest point).At the hi

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$2\,mg$

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(A) Let the mass be $m$ and the length of the string be $l$. The body is given a horizontal velocity $v$ at the mean position (lowest point).At the highest point of its trajectory,the angle with the vertical is $	heta = 60^{\circ}$. At this point,the velocity of the body becomes zero.Using the principle of conservation of mechanical energy between the lowest point and the point at $60^{\circ}$:$\frac{1}{2}mv^2 = mg(l - l\cos 60^{\circ})$$\frac{1}{2}mv^2 = mgl(1 - 0.5) = 0.5mgl$$v^2 = gl$At the mean position (lowest point),the tension $T$ is given by $T - mg = \frac{mv^2}{l}$.Substituting $v^2 = gl$:$T = mg + \frac{m(gl)}{l} = mg + mg = 2\,mg$.

$A$ body of mass $m$ hangs at one end of a string of length $l$,the other end of which is fixed. It is given a horizontal velocity such that the string just reaches a position where it makes an angle of $60^{\circ}$ with the vertical. The tension in the string at the mean position is:

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