A vertical frictionless semicircular track of radius $1 m$ is fixed on the edge of a movable trolley (figure). Initially, the system is at rest and a mass m is kept at the top of the track. The trolley starts moving to the right with a uniform horizontal acceleration a=2 g / 9. The mass slides down the track, eventually losing contact with it and dropping to the floor $1.3 m$ below the trolley. This $1.3 m$ is from the point, where mass loses contact. (Take, $g = 10 m / s^{2}$ )
Calculate the angle $θ$ at which it loses contact with the trolley and

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$37 °$

$27 °$

$30 °$

$53 °$

answer is C.

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Detailed Solution

Let y, be the velocity of mass relative to track at angular position $θ$ .

From work-energy theorem, KE of particle relative to track

= Work done by force of gravity + work done by pseudo force

$∴ \frac{1}{2} {mv}_{y}^{2} = mg (1 - cosθ) + m (\frac{2 g}{9}) sinθ$

or $v_{r}^{2} = 2 g (1 - cosθ) + \frac{4 g}{9} sinθ$

...(i)

Particle leaves contact with the track, where N=0

or $mgcosθ - m (\frac{2 g}{9}) sinθ = {mv}_{r}^{2}$

or $gcosθ - \frac{2 g}{9} sinθ = 2 g (I - cosθ) + \frac{4 g}{9} sinθ$

or $3 cosθ - \frac{6}{9} sinθ = 2$

Solving this, we get

$θ = 37 °$

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