Question:

A small disc is on the top of a smooth hemisphere of radius 'R'. The smallest horizontal velocity 'V' that should be imparted to the disc so that the disc leaves the hemisphere surface without sliding down (there is no friction) is

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When solving problems involving motion on curved surfaces, it's critical to analyze the forces acting at the point of separation, where the normal force becomes zero.
Updated On: Mar 22, 2025
  • \(V = \sqrt{g2R}\)
  • \(V = \sqrt{2gR}\)
  • \(V = \sqrt{gR}\)
  • \(V = \sqrt{\frac{g}{R}}\)
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The Correct Option is C

Solution and Explanation

For the disc to leave the surface of the hemisphere, the normal force must become zero at the point of leaving. Using the conservation of mechanical energy and Newton's laws, we can set the centripetal force required to keep the disc on the hemisphere equal to the gravitational component acting perpendicular to the surface at the point of leaving, which is at the very top of the hemisphere. 
The required velocity \( V \) for this to occur can be calculated by setting the gravitational force \( mg \) equal to the required centripetal force \( \frac{mV^2}{R} \) at the top, where \( m \) is the mass of the disc: \[ mg = \frac{mV^2}{R} \Rightarrow V^2 = gR \Rightarrow V = \sqrt{gR} \]

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