$A$ block of mass $m$ is sliding on a fixed frictionless concave surface of radius $R$. It is released from rest at point $P$ which is at a height of $H \ll R$ from the lowest point $Q$.
$(a)$ What is the potential energy as a function of $\theta$,taking the lowest point $Q$ as the reference level for potential energy?
$(b)$ What is the kinetic energy as a function of $\theta$?
$(c)$ What is the time taken for the particle to reach from point $P$ to the lowest point $Q$?
$(d)$ How much force is exerted by the block on the concave surface at the point $Q$?

(D) The mass $m$ is at height $H$ from point $Q$,where potential energy is taken as zero. From the geometry of the figure,if at some angle $\theta$,the height of mass $m$ above the lowest point $Q$ is $h$,then from $\triangle ABC$,$\cos \theta = \frac{R-h}{R} \Rightarrow h = R(1 - \cos \theta)$. Hence,the potential energy $U(\theta)$ is $U(\theta) = mgh = mgR(1 - \cos \theta)$.
$(b)$ Kinetic energy $K(\theta)$ at position $\theta$ is the loss of potential energy from the initial position $P$ to position $\theta$. Thus,$K(\theta) = mgH - U(\theta) = mgH - mgR(1 - \cos \theta) = mg(H - R(1 - \cos \theta))$.
$(c)$ For $H \ll R$,the motion is simple harmonic with time period $T = 2\pi \sqrt{\frac{R}{g}}$. The time taken to travel from $P$ to $Q$ is one-fourth of this time period. Thus,$t = \frac{T}{4} = \frac{2\pi}{4} \sqrt{\frac{R}{g}} = \frac{\pi}{2} \sqrt{\frac{R}{g}}$.
$(d)$ From energy conservation at the lowest point $Q$,if $m$ has velocity $v$,then $\frac{1}{2}mv^2 = mgH \Rightarrow mv^2 = 2mgH$. The centripetal force is $F_c = \frac{mv^2}{R} = \frac{2mgH}{R}$. The normal reaction $N$ at $Q$ satisfies $N - mg = \frac{mv^2}{R} \Rightarrow N = mg + \frac{2mgH}{R} = mg(1 + \frac{2H}{R})$. This is the force exerted by the block on the surface.