A track consists of two cylindrical arcs of equal radius R…

A track consists of two cylindrical arcs of equal radius $R$: a concave arc AB and a convex arc BC, joined smoothly at B with a horizontal tangent. A is $h$ above B, and C is $h$ below B. A small body starts from rest at A and slides without friction. What must $h$ be for the body to leave the track exactly at C?

Explanation of Method

Where a body leaves an S-shaped track.

Step 1: Speed at C

From A to C the body drops $2h$: $v_C^2 = 2g \cdot 2h = 4gh$.

Step 2: Leaving the convex arc

B is the top of the convex arc. At C, a height $h$ lower, the radius makes an angle $\varphi$ with the vertical, where $R\cos\varphi = R - h$. The body leaves when the normal force vanishes, i.e. when the radial component of gravity alone provides the centripetal force:

\[ \frac{mv_C^2}{R} = mg\cos\varphi = mg\,\frac{R - h}{R}. \]

Step 3: Condition

\[ 4gh = g\,(R - h) \;\Longrightarrow\; 5h = R. \]

\[h = \frac{R}{5}\]

Check

It does not leave earlier: on the convex arc $N \ge 0$ requires $3\cos\varphi \ge 2 + 2h/R = 2.4$, i.e. $\cos\varphi \ge 0.8$; at C, $\cos\varphi = 1 - h/R = 0.8$, exactly the limit $\checkmark$

$A$ $B$ $C$ $h$ $h$ $O_2$ $R$ $4gh = g(R - h) \Rightarrow h = \tfrac{R}{5}$

Assumed shape (drawn to scale for $h = R/5$): the concave arc AB and the convex arc BC meet at B with a horizontal tangent; A is $h$ above B, C is $h$ below it. The body leaves the convex arc at C (red dashed path).

Result

\[\boxed{\,h = \frac{R}{5}\,}\]

Remark

About the original solution. The result $h = R/5$ is correct, but the derivation does not lead to it: it used only one drop of $h$ instead of $2h$ from A to C, and then simply stated $R/5$ ("plugging this back gives"). The shape of the track is taken as the usual S-curve, since the figure is not available.