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 the angular position of point $Q$ is $\alpha$ with respect to point $P$. Which of the following graphs represents the correct relation between $A$ and $\alpha$ when the ball goes from $Q$ to $R$?
Answer: (C) see figure
At angle $\alpha$ below the rim the ball has fallen $R\sin\alpha$, so $mv^2 = 2mgR\sin\alpha$.
Radially (towards $O$): $N - mg\sin\alpha = \dfrac{mv^2}{R} = 2mg\sin\alpha\Rightarrow N = 3mg\sin\alpha$.
$$A = \frac{mv^2/R}{N} = \frac{2mg\sin\alpha}{3mg\sin\alpha} = \frac23$$
$A$ is constant, so the graph is a horizontal line.
Solution by Sreeraj P, M.Sc Physics