Question 6.Q.2: Does P leave the surface of the sphere? For the particular c...

When u=(3 g b)^{1 / 2}, the formulae (6.17), (6.18) for v² and N become

v^{2}=u^{2}-2 g b(1-\cos \theta)                       (6.17)

N=\frac{m u^{2}}{b}+m g(3 \cos \theta-2)                    (6.18)

v^{2}=g b(1+2 \cos \theta), \quad N=m g(1+3 \cos \theta).

If P remains in contact with the sphere, then it comes to rest when v = 0, that is, when cos θ = −1/2. This first happens when θ = 120° (a point on the upper half of the sphere, higher than O). If P were threaded on a circular wire (from which it could not fall off) this is exactly what would happen; P would perform periodic oscillations in the range −120° ≤ θ ≤ 120°. However, in the present case, the reaction N is restricted to be positive and this condition will be violated when θ > \cos ^{-1} (−1/3) ≈ 109°. Since this angle is less than 120°, the conclusion is that P loses contact with the sphere when θ = \cos ^{-1} (−1/3); at this instant, the speed of P is (g b / 3)^{1 / 2}. P then moves as a free projectile until it strikes the sphere.