Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity $u = \sqrt{3 g l}$ , where $l$ is the length of the pendulum. Find the angle rotated by the string before it becomes slack.

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Accepted Answer

Given:
Length of the string $= l$ ,
Speed of the bob at the lowermost position, $u = \sqrt{3 g l}$
Let the mass of the bob be $m$ , the angle made by the string with horizontal be $θ$ and the velocity be $v$ when the string slacks as shown in the figure.

At the instant, the string slacks, tension in the string will be zero $(T = 0)$ .
Applying Newton's second law in radial direction, we get,
$m g \cos (90 ° - θ) + T = \frac{m v^{2}}{l}$
$\Rightarrow m v^{2} = m g l \sin θ \dots (1)$

Applying conservation of mechanical energy between the lowermost position and the position where the strings slacks, we get,
$Δ U + ΔKE = 0 \Rightarrow m g (l + l \sin θ) + \frac{1}{2} m v^{2} - \frac{1}{2} m {(\sqrt{3 g l})}^{2} = 0 \Rightarrow g l + g l \sin θ + \frac{1}{2} v^{2} - \frac{3 g l}{2} = 0 \dots (2)$
From equations $(1)$ and $(2)$ ,
$g l + g l \sin θ + \frac{1}{2} g l \sin θ - \frac{3 g l}{2} = 0 \Rightarrow \frac{3 \sin θ}{2} - \frac{1}{2} = 0 \Rightarrow \sin θ = \frac{1}{3}$
Total angle rotated by the string,
$θ_{T} = θ + \frac{π}{2} \Rightarrow θ_{T} = \sin^{- 1} (\frac{1}{3}) + \frac{π}{2} = \cos^{- 1} (- \frac{1}{3})$

B M Sharma Solutions for Chapter: Circular Motion, Exercise 48: ILLUSTRATION 9.45

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 48: ILLUSTRATION 9.45

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