A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is __________ m.
(Assume there is no friction between the block and the hemisphere)

Solution

Physics sir · 4 steps

Contact is lost when gravity's radial part alone just supplies the centripetal force, so set N = 0.

1. Speed from energy. With no friction, mechanical energy is conserved. At point A the block has dropped R - h, so v² = 2g(R - h).
2. Radial forces. Along the radius towards O, the net force is mg cos θ - N, and it must equal mv²/R. From the figure, cos θ = h/R. The block leaves the surface when N = 0.

   mg h/R = (m · 2g(R-h))/R

3. Solve for h. Cancel mg/R to get h = 2R - 2h, so 3h = 2R and h = 2R/3.
4. So. h = (2 × 3)/3 = 2 m above the base.

The common slip: 1. Reporting the vertical drop from the top, R - h = R/3, gives 1 m instead of the height h above the base.

The idea to keep. For a body sliding off a smooth sphere, combine energy conservation with N = 0 in the radial equation. It always leaves at cos θ = 2/3, that is, at height 2R/3 above the centre.

Answer from the NTA final answer key. Solution by PrepWiser.