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A smooth semicircular wire of radius $R$ is fixed in a vertical plane. One end of a massless spring of natural length $\frac{3 R}{4}$ is attached to the lower point $O$ of the wire track. A small ring of mass $m$ which can slide on the track, is attached to the lower end of the spring. The ring help stationary at point $P$ such that the spring makes an angle $60 °$ with the vertical.
The spring constant $k = \frac{m g}{R}$ . Consider the instant when the spring is released, determine the tangential acceleration of the ring and the normal reaction.

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$\frac{(4 \sqrt{3} + 1) g}{8} m / s^{2}$

$\frac{(4 \sqrt{3} + 2) g}{8} m / s^{2}$

$\frac{(5 \sqrt{3} + 1) g}{8} m / s^{2}$

$\frac{(4 \sqrt{3} + 1) g}{10} m / s^{2}$

answer is C.

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Detailed Solution

Shown that, $∠ C P O = 60 ° = ∠ O P C, ∴ O P = R$

The extension of spring $= R - \frac{3 R}{4} = \frac{R}{4}$ .

$∴ F = k x = \frac{m g}{R} \times \frac{R}{4} = \frac{m g}{4}$

The tangential force

$F_{1} = m g \cos 30 ° + F \sin 30 ° = m g \cos 30 ° + \frac{m g}{4} \sin 30 ° = \frac{(4 \sqrt{3} + 1) m g}{8}$

Thus tangential acceleration $a_{1} = \frac{F_{1}}{m} = \frac{(4 \sqrt{3} + 1) g}{8} m / s^{2}$ .

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