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 ______
Answer: (B) $4$
Concept. At the top of the loop the body is upside down: the normal force from the track and gravity both point down, towards the centre, and together supply the centripetal force. Energy conservation along the smooth track then gives the release height.
Step 1. Speed at the top. The body presses on the track with $3mg$, so by Newton's third law the track pushes it with $N=3mg$, towards the centre:
$$N+mg=\frac{mv^2}{R}\ \Rightarrow\ 4mg=\frac{mv^2}{R}\ \Rightarrow\ v^2=4gR$$
Step 2. Energy from release to the top. The top of the loop is at height $2R$:
$$mgh=mg(2R)+\tfrac12mv^2=2mgR+\tfrac12m(4gR)=4mgR$$
Step 3. Height.
$$h=4R\ \Rightarrow\ \alpha=4$$
Answer. Option B. For the minimum case ($N=0$) the same steps give $h=\tfrac52R$.
Solution by Sreeraj P, M.Sc Physics
