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

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(A) Let the mass $m$ be at the lowest point (mean position) $A$ and it is given a horizontal velocity $v$. The string reaches a point $B$ where it mak

Vedclass · Accepted Answer

$2\,mg$

Vedclass · Answer

(A) Let the mass $m$ be at the lowest point (mean position) $A$ and it is given a horizontal velocity $v$. The string reaches a point $B$ where it makes an angle $	heta = 60^o$ with the vertical.By the law of conservation of energy between point $A$ and point $B$:$\frac{1}{2}mv^2 = mg(l - l\cos	heta) = mgl(1 - \cos 60^o)$$\frac{1}{2}mv^2 = mgl(1 - 0.5) = 0.5mgl$$v^2 = gl$At the mean position (lowest point),the tension $T$ is given by $T - mg = \frac{mv^2}{l}$.Substituting $v^2 = gl$ into the equation:$T = mg + \frac{m(gl)}{l} = mg + mg = 2mg$.

$A$ 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 so that the string just reaches a position where it makes an angle of $60^o$ with the vertical. The tension in the string at the mean position is:

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