Two beads of mass 2m and m are connected by a rod of length $l$ and of negligible mass are free to move in a circular wire frame of radius l as shown. Both are given speed ‘v’ tangentially in anti clock wise direction (when the rod is in horizontal position). The speed ‘v’ so that the rod becomes vertical is $\sqrt{(\frac{n \sqrt{n} - 1}{n}) g l}$ . Find the value of ‘ $n$ ’

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Detailed Solution

The speeds given to 2m will also be possessed by ‘m’
K.E in horizontal position gets converted in P.E in vertical position
$\frac{1}{2} 2 m v^{2} + \frac{1}{2} m v^{2} =$ change in P.E in vertical position
$Δ P . E = 2 m g [l \cos 30^{0} - l \cos 60^{0}] + m g [l \cos 30^{0} + \frac{l}{2}]$
$2 m g [\frac{l \sqrt{3}}{2} - \frac{l}{2}] + m g [\frac{l \sqrt{3}}{2} + \frac{l}{2}]$

$∴ m g l [\sqrt{3} - 1] + m g l [\frac{\sqrt{3} + 1}{2}] ∴ m g l [\sqrt{3} - 1 + \frac{\sqrt{3}}{2} + \frac{1}{2}] = m g l [\frac{3 \sqrt{3}}{2} - \frac{1}{2}] K . E = \frac{1}{2} 3 m v^{2} = m g l [\frac{3 \sqrt{3} - 1}{2}] ∴ v = \sqrt{(\frac{3 \sqrt{3} - 1}{3}) g l}$