Smooth wire shaped as circular arc AC and CB are of same radius but centre of curvature are on opposite sides as shown. A small bead is projected horizontally from A with speed such that normal force become zero at point C . Speed of the particle when it reach to B will be [Take $θ = cos^{- 1} (5/6)$ ]:

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arc figure

TestHub · Accepted Answer

At C, normal is zero: m g cos θ = R m V 1 2 ​ ​ . Energy A to C: 2 1 ​ m V 1 2 ​ = 2 1 ​ m U 2 − m g ( R − R cos θ ) , with cos θ = 5/6 giving U = g R ​ . Energy A to B (drop of R): m g R = 2 1 ​ m V 2 − 2 1 ​ m U 2 . So V = 2 5 ​ g R ​ . Hence option (C).

Physics - Circular Motion Question with Solution | TestHub

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