Two point mass m are connected the light rod of length t and it is free to rotate in vertical plane as shown.Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in verticalfane

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Solution

$v = 0$
$v_{2} - u_{2} = - 2 g h$
$0 = u_{2} - 2 g h (h = l)$
$u = \sqrt{2 g l}$

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Two point mass m are connected the light rod of length t and it is free to rotate in vertical plane as shown.Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in verticalfane

v=0v2−u2=−2gh0=u2−2gh(h=l)u=√2gl

$v = 0$
$v_{2} - u_{2} = - 2 g h$
$0 = u_{2} - 2 g h (h = l)$
$u = \sqrt{2 g l}$