Find Alpha for Motion in Vertical Circular Loop

To find the value of $\alpha$, we analyze the motion of the body using the conservation of mechanical energy and Newton's second law for circular motion.

1. Conservation of Mechanical Energy: Let the lowest point of the circular loop be the reference level for potential energy ($U = 0$). The initial total mechanical energy at height $h$ is entirely potential energy: $E_i = m g h$

At the highest point of the vertical circular loop, the height of the body is $2R$. If $v$ is the velocity of the body at this highest point, its total mechanical energy is: $E_f = \frac{1}{2} m v^2 + m g (2R)$

Since the plane and loop are smooth, mechanical energy is conserved ($E_i = E_f$): $m g h = \frac{1}{2} m v^2 + 2 m g R$ $v^2 = 2 g (h - 2R) \quad \text{--- (Equation 1)}$

1. Dynamics at the Highest Point: At the highest point of the circular loop, the forces acting vertically downwards on the body are:

• The gravitational force (weight), $m g$
• The normal reaction force exerted by the circular track on the body, $N$

By Newton's third law, the force exerted by the body on the track is equal in magnitude to the normal reaction force $N$. Given that this force is three times the body's weight: $N = 3 m g$

The net force acting towards the center of the circle provides the necessary centripetal force: $N + m g = \frac{m v^2}{R}$

Substituting $N = 3 m g$: $3 m g + m g = \frac{m v^2}{R}$ $4 m g = \frac{m v^2}{R}$ $v^2 = 4 g R \quad \text{--- (Equation 2)}$

1. Determining $\alpha$: Equating Equation 1 and Equation 2: $2 g (h - 2R) = 4 g R$ $h - 2R = 2R$ $h = 4 R$

Comparing this with $h = \alpha R$, we get: $\alpha = 4$

Hence, the correct option is B.