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 _________.
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Apply conservation of mechanical energy between the release point and the top of the circle. Use the given normal force condition at the top to find the required velocity.
We can also solve this problem using the work-energy theorem, which states that the work done by all forces equals the change in kinetic energy ($W_{net} = \Delta K$).
In this case, the only force doing work is gravity, because the normal force is always perpendicular to the displacement and the plane is smooth (no friction). When the body moves from the release height $h$ to the top of the loop (height $2R$), the change in height is $(h - 2R)$. Work done by gravity, $W_g = mg(h - 2R)$. The change in kinetic energy is $\Delta K = \frac{1}{2}mv^2 - 0$, where $v$ is the velocity at the top.
$$mg(h - 2R) = \frac{1}{2}mv^2$$
... (Equation A)
Now, we examine the centripetal acceleration at the highest point of the loop. The net force toward the center of the circle must satisfy:
$$\sum F_{radial} = \frac{mv^2}{R}$$
At the top, both gravity and the normal force $N$ point toward the center:
$$mg + N = \frac{mv^2}{R}$$
We are given that the body exerts a force on the track equal to $3mg$. This is the magnitude of the normal force, $N = 3mg$. Substituting this into our dynamic equation:
$$mg + 3mg = \frac{mv^2}{R}$$
$$4mg = \frac{mv^2}{R}$$
$$mv^2 = 4mgR$$
Now, plug the value of $mv^2$ into Equation A:
$$mg(h - 2R) = \frac{1}{2}(4mgR)$$
$$mg(h - 2R) = 2mgR$$
Dividing through by $mg$: $$h - 2R = 2R$$ $$h = 4R$$ Thus, $h = 4R$. By comparing this with $h = \alpha R$, we obtain $\alpha = 4$.
