A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $$h$$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $$h = \alpha R$$. The value of $$\alpha$$ is _______.

Solution :

At the highest point of the vertical circle :

Centripetal force is provided by weight and normal reaction.

Given body exerts force three times its weight on the track.

Therefore, reaction by track on body is :

$$N = 3mg$$

At highest point :

$$mg + N = \frac{mv^2}{R}$$

$$mg + 3mg = \frac{mv^2}{R}$$

$$4mg = \frac{mv^2}{R}$$

$$v^2 = 4gR$$

Using conservation of mechanical energy :

Initial energy at height $$h$$ :

$$E_i = mgh$$

Energy at highest point of loop :

Height of highest point above ground :

$$2R$$

Therefore,

$$E_f = mg(2R) + \frac{1}{2}m(4gR)$$

$$= 2mgR + 2mgR$$

$$= 4mgR$$

Equating energies :

$$mgh = 4mgR$$

$$h = 4R$$

Comparing with :

$$h = \alpha R$$

Therefore,

$$\alpha = 4$$

Final Answer :

$$4$$