(q)A particle, mass m , slides down the smooth track, Figure below, from a height H un- der gravity, It is to complete a circular trajectory of radius R when reaching its lowest point. Determine the smallest value of \( \mathrm{H},[3] \)

To determine the smallest height \( H \) from which a particle of mass \( m \) must slide down a smooth track to complete a circular trajectory of radius \( R \) at the lowest point, we can use principles of energy conservation and circular motion. 1. **Energy Conservation**: As the particle slides down from height \( H \), its potential energy at the top is converted into kinetic energy at the lowest point. The potential energy at height \( H \) is given by: \[ PE = mgh \] where \( g \) is the acceleration due to gravity. At the lowest point, the potential energy is zero, and all the energy is converted into kinetic energy: \[ KE = \frac{1}{2} mv^2 \] By conservation of energy: \[ mgh = \frac{1}{2} mv^2 \] Simplifying this gives: \[ v^2 = 2gH \] 2. **Centripetal Force Requirement**: For the particle to complete the circular motion at the lowest point, it must have sufficient speed to provide the necessary centripetal force. The centripetal force required to keep the particle moving in a circle of radius \( R \) is given by: \[ F_c = \frac{mv^2}{R} \] At the lowest point of the circular path, the gravitational force acting on the particle also contributes to the centripetal force. The total force acting on the particle at the lowest point is: \[ mg = F_c + N \] where \( N \) is the normal force. For the minimum speed required to maintain circular motion, we set \( N = 0 \) (the particle is just at the verge of losing contact with the track): \[ mg = \frac{mv^2}{R} \] Canceling \( m \) from both sides (assuming \( m \neq 0 \)) gives: \[ g = \frac{v^2}{R} \] Substituting \( v^2 \) from the energy conservation equation: \[ g = \frac{2gH}{R} \] Rearranging this gives: \[ H = \frac{gR}{2g} = \frac{R}{2} \] Thus, the smallest height \( H \) from which the particle must slide down to complete the circular trajectory of radius \( R \) is: \[ \boxed{\frac{R}{2}} \]