That boy should have been careful

Let $F_{N}$ is the normal force of the ice on the boy and $m$ is his mass.The net inward force is $mg \cos\theta-F_{N}$ and according to newton’s second law,this must be equal to $\frac{mv^{2}}{R}$ ,we wish to find his speed If the gravitational potential energy is taken to be zero when he is at the top of the ice mound, then his potential energy at the time shown is $-U=mgR(1-\cos\theta)$ He starts from rest and his kinetic energy at the time shown is $\frac{1}{2} \times mv^{2}$ Thus conservation of energy gives $0= \frac{1}{2} \times mv^{2} - mgR(1-\cos\theta)$ Or $v^{2}=2gR(1-\cos\theta)$ We substitute this expression into the equation developed from the second law to obtain $g \cos\theta=2g(1-\cos\theta)$ .This gives $\cos\theta=\frac{2}{3}$ .The height of the boy above the bottom of the mound is $h=R \cos\theta=\frac{2}{3}R$ .Hence $a+b=\boxed{5}$

Let us assume that angle the radius makes with the vertical when the boy loses contact is A and the point is L m below the top.
Boy lose contact when...... cent. acc = radial component of g.
Hence, V^2/R = g * Cos(A)..<......> Loss of P.E. = Gain in K.E. =====>
(1/2) * m * V^2 = m * g * L
Hence, ( 2 * g * L)/R = g * Cos(A) = g * (R - L)/R..<..> L = R/3.

Height h = R - L =R - R/3 = (2/3) * R =(a/b) * R....... a+b= 5.