= Solution
Treat the carriage as translating along the circular track, neglecting wheel <rotational kinetic energy> and its finite size. At the top, gravity and the inward normal <force> supply the <centripetal acceleration>:
$$
mg+N=\frac{mv_{\rm top}^2}{r}.
$$
For unilateral contact, $N\ge0$, so the limiting <speed> is $v_{\rm top}^2=gr$. <Conservation of energy> from rest at height $h$ then gives
$$
mgh=2mgr+\tfrac12mgr,\qquad \boxed{h_{\min}=\tfrac52r.}
$$
This top contact condition is the most restrictive: at angle $\phi$ measured from the bottom, $N/(mg)=2h/r-2+3\cos\phi$, whose minimum occurs at the top. A mechanically captive carriage able to sustain an outward rail reaction has a different constraint; the result here uses the usual requirement of positive normal contact.
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