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The simple $2 kg$ pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at B and continues in the smaller arc in the vertical plane. Calculate the magnitude of the force R supported by the pin at B when the pendulum passes the position $θ = 30 °$ . (Take, $g = 9.8 m / s^{2}$ )

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$Tcos 30 °$

$Tsin 60 °$

$Tcos 45 °$

$Tsin 30 °$

answer is A.

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Detailed Solution

Speed of bob in the given position,

Here, $v = \sqrt{2 gh}$

$h = (400 + 400 \cos 30 °) mm$

$= 746 mm$

$= 0.746 m$ v

$= \sqrt{2 \times 9.8 \times 0.746}$

$= 3.82 m / s$

Now $T - mgcosθ = \frac{{mv}^{2}}{r}$

$T = 2 \times 9.8 \times \cos 30 ° + \frac{2 \times (3.82)^{2}}{(0.4)}$

or $T = 90 N$

$∴ R = Tsin 30 ° = 45 N$

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