A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity $\frac{7 g L}{2}$ . The thread will slack after swinging through an angle $θ$ , such that $θ$ equal.

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Answer 1

As shown in figure, $h = L + sin ϕ$
By energy conservation,
$\frac{1}{2} m v^{2} = m g h + \frac{1}{2} m v^{' 2}$
$\frac{1}{2} \times (\frac{7 g l}{2})^{2} = g (L + L sin ϕ) + \frac{1}{2} v^{' 2}$
$v^{' 2} = - 2 g L - 2 g L sin ϕ + \frac{7}{2} g L$
$= \frac{3}{2} g L - 2 g L sin ϕ$
If string slack then,
$m g sin ϕ = \frac{m v ^{' 2}}{L}$
$g sin ϕ = \frac{1}{L} (v^{'})^{2} = \frac{1}{L} (\frac{3}{2} g L - 2 g L sin ϕ)$
$g sin ϕ = \frac{3}{2} g - 2 g sin ϕ$
$3 g sin ϕ = \frac{3}{2} g$
$sin ϕ = \frac{1}{2}$
$ϕ = \frac{π}{6}$
$θ = \frac{π}{2} + \frac{π}{6} = \frac{4 π}{6} = 12 0^{\circ}$

Answer 2

As shown in figure,

h = L + sin ϕ

By energy conservation,

\frac{1}{2} m v^{2} = m g h + \frac{1}{2} m v^{' 2}

\frac{1}{2} \times (\frac{7 g l}{2})^{2} = g (L + L sin ϕ) + \frac{1}{2} v^{' 2}

v^{' 2} = - 2 g L - 2 g L sin ϕ + \frac{7}{2} g L

= \frac{3}{2} g L - 2 g L sin ϕ

If string slack then,

m g sin ϕ = \frac{m v ^{' 2}}{L}

g sin ϕ = \frac{1}{L} (v^{'})^{2} = \frac{1}{L} (\frac{3}{2} g L - 2 g L sin ϕ)

g sin ϕ = \frac{3}{2} g - 2 g sin ϕ

3 g sin ϕ = \frac{3}{2} g

sin ϕ = \frac{1}{2}

ϕ = \frac{π}{6}

θ = \frac{π}{2} + \frac{π}{6} = \frac{4 π}{6} = 12 0^{\circ}