A bob of heavy mass is suspended by a light string of length . The bob is given a horizontal velocity as shown in the figure. If the string gets slack at some point P making an angle from the horizontal, the ratio of the speed of the bob at point P to

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Solution

Q.
A bob of heavy mass $m$ is suspended by a light string of length $l$ . The bob is given a horizontal velocity $v_{0}$ as shown in the figure. If the string gets slack at some point P making an angle $θ$ from the horizontal, the ratio of the speed $v$ of the bob at point P to its initial speed $v_{0}$ is [2025]

1 ${(\frac{\cos θ}{2 + 3 \sin θ})}^{1 / 2}$

2 ${(\frac{\sin θ}{2 + 3 \sin θ})}^{1 / 2}$

3 ${(\sin θ)}^{1 / 2}$

4 ${(\frac{1}{2 + 3 \sin θ})}^{1 / 2}$

Ans.
(2)

Let speed at P be $v .$

$P M = l \sin θ$

$h = l + l \sin θ$

Using conservation of energy,

$\frac{1}{2} m v_{0}^{2} + 0 = m g h + \frac{1}{2} m v^{2}$

$\frac{1}{2} v_{0}^{2} = g l (1 + \sin θ) + \frac{1}{2} v^{2}$ ...(i)

At $P, T = 0$

So, $T - m g \cos (90 ° + θ) = \frac{m v^{2}}{l}$

$m g \sin θ = \frac{m v^{2}}{l}$

or $\frac{v^{2}}{\sin θ} = g l$

Putting in (i),

$\frac{1}{2} v_{0}^{2} = \frac{2 . v^{2}}{2 . \sin θ} (1 + \sin θ) + \frac{1}{2} v^{2}$

$\frac{1}{2} v_{0}^{2} = \frac{v^{2}}{2} [\frac{2}{\sin θ} (1 + \sin θ) + 1]$

$v_{0}^{2} = v^{2} [\frac{2 + 2 \sin θ + \sin θ}{\sin θ}]$

$v_{0}^{2} = v^{2} \times [\frac{2 + 3 \sin θ}{\sin θ}]$

$\frac{v^{2}}{v_{0}^{2}} = \frac{\sin θ}{2 + 3 \sin θ}$

$\frac{v}{v_{0}} = \sqrt{\frac{\sin θ}{2 + 3 \sin θ}}$

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