A fixed hollow sphere with centre O has a smooth inner surface of radius a - CIE - A-Level Further Maths - Question 3 - 2011 - Paper 1

To solve for $\cos \theta$, we will use the conservation of energy and the equations of motion for the particle.

1. Using Conservation of Energy:

   At the lowest point, the total energy is the kinetic energy of the particle, given by:

   $E_1 = \frac{1}{2} m (2\sqrt{4g})^2 = 16mg$

   At the point of losing contact with the surface, the potential energy gain is:

   $E_2 = mgR(1 + \cos \theta)$

   where $R = a$ is the radius of the sphere. The total energy at this point can be expressed as:

   $E_2 = mgR + \frac{1}{2} mv^2$

   Setting the total energy before and after equal:

   $16mg = mgR(1 + \cos \theta) + \frac{1}{2} mv^2$

2. Equating Radial Forces: At the point of losing contact, we also need the radial forces to balance. The force due to gravity component acting towards the center is:

   $F_{gravity} = mg \cos \theta$

   The centripetal force needed is given by:

   $F_{centripetal} = \frac{mv^2}{a}$

   Setting these equal at the point of contact loss yields:

   $mg \cos \theta = \frac{mv^2}{a}$

   Substituting $v^2$ from the conservation of energy equation, we can eliminate $v$ and express everything in terms of $\cos \theta$.

3. Eliminating Variables:

   From the first equation:

   $v^2 = 16g - 2g(1 + \cos \theta)$

   Solve for $\cos \theta$:

   $16g = mg(a(1 + \cos \theta) + \frac{1}{2} m(16g - 2g(1 + \cos \theta))$

   Simplifying leads to:

   $2g(1 - \cos \theta) = 2g$

   Finally, we arrive at:

   $\cos \theta = \frac{2}{3}$