A ball is released from rest from point P of a smooth semi-spherical vessel as shown in figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is α with respect to point P. Which of the following graphs represent the correct relation between A and α when ball goes from Q to R?

Solution

Physics sir · 4 steps

Find speed from energy, then N from the radial equation: both grow as sin α, so their ratio stays fixed.

1. Speed at Q. P is at the level of the centre O. At angle α below OP, the ball has dropped a height R sin α. The vessel is smooth, so energy is conserved and v² = 2gR sin α.
2. Centripetal force. F_c = mv²/R = 2mg sin α, directed towards O.
3. Normal reaction. Along the radius, N points towards O and the gravity component mg sin α points away from O. Their difference supplies the centripetal force, so N - mg sin α = 2mg sin α, which gives N = 3mg sin α.

   A = F_c/N = (2mg sin α)/(3mg sin α) = 2/3

4. So. A = 2/3 for every position from Q to R. The graph is a horizontal straight line, which is option (c).

The idea to keep. In vertical circular motion, write energy conservation for v² and the radial force equation for N together. Here both depend on sin α in the same way, so sin α cancels in the ratio.

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