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.
To find the value of $\alpha$, we utilize the principles of conservation of mechanical energy and circular motion dynamics.
First, let's consider the energy of the body. Since the plane and the loop are smooth, there is no work done by friction. The total mechanical energy is conserved throughout the motion. Let the reference level for potential energy ($U=0$) be the lowest point of the loop. At the initial point on the inclined plane, the body is released from height $h$ at rest ($v_i = 0$). Initial Mechanical Energy, $E_i = U_i + K_i = mgh + \frac{1}{2}m(0)^2 = mgh$
At the highest point of the vertical circle, the height is $2R$ (the diameter of the circle). Let the velocity of the body at this point be $v$. Final Mechanical Energy, $E_f = U_f + K_f = mg(2R) + \frac{1}{2}mv^2$ By the Law of Conservation of Energy, $E_i = E_f$:
$$mgh = 2mgR + \frac{1}{2}mv^2$$
This can be simplified to:
$$gh = 2gR + \frac{v^2}{2}$$
... (Equation 1)
Next, we analyze the dynamics at the highest point of the vertical loop. The forces acting on the body are the gravitational force $mg$ (downward) and the normal reaction force $N$ from the track (downward). Together, these provide the required centripetal force for circular motion:
$$N + mg = \frac{mv^2}{R}$$
The problem states that at the highest point, the body exerts a force of three times its weight ($3mg$) on the track. According to Newton's Third Law, the track exerts an equal and opposite normal force $N = 3mg$ on the body. Substituting $N = 3mg$ into the force equation:
$$3mg + mg = \frac{mv^2}{R}$$
$$4mg = \frac{mv^2}{R}$$
$$v^2 = 4gR$$
Now, substitute this expression for $v^2$ back into Equation 1:
$$gh = 2gR + \frac{4gR}{2}$$
$$gh = 2gR + 2gR$$
$$gh = 4gR$$
Dividing by $g$, we find: $$h = 4R$$ Comparing this with the given form $h = \alpha R$, we get $\alpha = 4$. This corresponds to option 2.
