A particle of mass m is placed on a fixed, smooth sphere of radius R at a position where the radius passing through the particle makes an angle of 30^ with the vertical. The particle is released from this position. Determine the distance travelled by the particle before it leaves contact with the sphere.

Step-by-step solution

Let be the angle made by the radius vector of the particle with the vertical when it leaves the sphere. At the point of leaving the sphere, the normal reaction N = 0 . The centripetal force is provided by the component of gravity: mg = mv^2 R v^2 = Rg By conservation of mechanical energy from the initial position ₀ = 30^ to : mgR( 30^ - ) = 1 2 mv^2 Substituting v^2 = Rg : mgR( 30^ - ) = 1 2 mgR 30^ = 3 2 = 2 3 30^ = 2 3 3 2 = 1 3 = ⁻¹ ( 1 3 ) 54.74^ The angle described by the particle is = - 30^ = 54.74^ - 30^ = 2