Is centripetal acceleration mandatory for circular motion? If we consider a case where a closed circular wall is present, and inside the boundary, just adjacent to the inner wall, two particles are placed touching the surface of inner boundary such that the distance between both particles tends to 0. Particle A and B have identical masses, shapes and physical properties. Particle B applies a constant force on particle A and particle B is in uniform circular motion. Thus, particle A also moves in a circle. And here, centripetal force is not required to make particle A move in a circle. 
What's going on?
 A: Particle A is subject to the centripetal force $F=m\omega r^2$ just the same as particle B. In both cases the wall provides the reaction force to keep both particles in uniform circular motion.
There can be no circular motion without a centripetal force.
A: Given a particle moving uniformly on a circular path $\vec r(t)$, without loss of generality, we can parametrize the motion as:
$$ r(t) = R \big( \vec e_x \cos(\omega t) + \vec e_y \sin(\omega t) \big), $$
where $\omega$ is the angular velocity of the motion.
By using Newton's axiom $\vec F = m \vec a$ we can calculate the force necessary for the particle to move on that path (without even considering what might cause that force):
$$ \vec F(t) = m \partial_t^2 r(t) = - mR\omega^2 \big(\vec e_x \cos(\omega t) + \vec e_y \sin(\omega t) \big). $$
As one can easily see this force is directed towards the centre of the circle and has magnitude $m R \omega^2 = \frac{mv^2}{R}$. This proves that absolutely generally, for an object to move uniformly on a circular path there must be a centripetal force.
We can even generalize this argument further to a general path $\vec r(t)$. The force will still be given $\vec F(t) = m \ddot{\vec r}(t)$. Using a time dependent basis of the form $\vec e_t = \frac{\dot{\vec r}}{\left|\dot{\vec r}\right|}$, $\vec e_n = \frac{\ddot{\vec r} - (\ddot{\vec r} \cdot \vec e_t) \vec e_t}{\left|\ddot{\vec r} - (\ddot{\vec r} \cdot \vec e_t) \vec e_t\right|}$ and $\vec e_b = \vec e_t \times \vec e_n$ (the tangential, normal and binormal vectors) we can write:
$$ \vec F = \vec e_t F_t + \vec e_n F_n. $$
(No term with $\vec e_b$ appears as $\vec F \propto \ddot{\vec r}$ and $\ddot{\vec r} \in \text{span}\{\vec e_t, \vec e_n \}$ by construction).
Propositon $m\partial_t v = F_t$ and $F_n = \frac{mv^2}{R}$ (Notation: $\vec v = \dot{\vec r}$ and $v = \left| \vec v \right|$ as usual, $R$ is the radius of curvature which is independent of $v$).
Proof: 
By direct calculation and using the above definitions, one easily gets:
$$m\partial_t \left| \dot{\vec r} \right| = m\partial_t \sqrt{\dot{\vec r} \cdot \dot{\vec r}} = m\frac{\ddot{\vec r} \cdot \dot{\vec r}}{\sqrt{\dot{\vec r} \cdot \dot{\vec r}}} = \vec F \cdot \vec e_t = F_t.$$
First we simplify the term for $\vec e_n$ using the identity $\vec a \times (\vec b \times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)$:
$$\vec e_n = \frac{\ddot{\vec r} - (\ddot{\vec r} \cdot \vec e_t) \vec e_t}{\left|\ddot{\vec r} - (\ddot{\vec r} \cdot \vec e_t) \vec e_t\right|} = \frac{\ddot{\vec r} - \ddot{\vec r} (\vec e_t \cdot \vec e_t) + \vec e_t \times (\ddot{\vec r} \times \vec e_t)}{\left| \cdots \right|} = \frac{\vec e_t \times (\ddot{\vec r} \times \vec e_t)}{\left|\ddot{\vec r} \times \vec e_t\right|}$$
(Note: $\left|\vec e_t \times (\vec e_t \times \ddot{\vec r})\right| = \left|\vec e_t \times \ddot{\vec r}\right|$ because $\vec e_t \perp \vec e_t \times \ddot{\vec r}$ and $\left|\vec e_t\right| = 1$).
With this we arrive at (using $\vec a \cdot (\vec b \times \vec c) = \vec c \cdot (\vec a \times \vec b)$ in the second step):
\begin{align*}
 F_n &= m \ddot{\vec r} \cdot \frac{\vec e_t \times (\ddot{\vec r} \times \vec e_t)}{\left|\ddot{\vec r} \times \vec e_t\right|} = m \frac{(\ddot{\vec r} \times \vec e_t)^2}{\left|\ddot{\vec r} \times \vec e_t\right|} = mv^2 \left|\frac{\ddot{\vec r} \times \vec e_t}{v^2}\right| = mv^2 \left|\frac{\ddot{\vec r} \times \dot{\vec r}}{v^3}\right|.
\end{align*}
From these terms we identify the radius of curvature $R$ as:
$$ R = \frac{v^3}{\left|\dot{\vec r} \times \ddot{\vec r}\right|}. $$
The proof that $R$ is constant under parameter changes $\vec r'(t) = (\vec r \circ f)(t)$ (with an arbitrary function $f$, that is it is independent of the velocity and only dependent on the curve traced by the path, formally this independence means $R'(t) = R\big(f(t)\big)$) is left as an exercise for the reader.
q.e.d.
In conclusion, this says, that for any movement along a path at any instance we can view the motion as an accelerated circular motion along the osculating circle of the path and only the tangential forces change the speed, while normal forces acting as centripetal force only change the direction of the velocity. So for the motion along any path we get, that if the motion is non-linear we will have a centripetal force whose parameters relate to the osculating circle.
Side note: The analysis of the geometry of the path are closely related to the first steps into the Frenet theory of the differential geometry of curves.
A: Let me try to add a missing intuitive piece in the explanation to the discussion to @Gert's answer of why there must always be radial acceleration for any circular motion.
Remember what acceleration is: Change in velocity. $\vec a=d\vec v/dt$. More mathematically it is change in the velocity vector.
If velocity is changed - either magnetude or direction - then we define this rate of change as acceleration.


*

*If the velocity magnetude is changed, then acceleration is pointing in the same direction as the velocity (and it goes faster or slower). We call this tangential acceleration.

*If the velocity direction is changed, then acceleration is pointing to the side and not parallel to the velocity direction. If it is exactly perpendicular to the direction we call it radial acceleration.


Now imagine what happens if an object has no tangential and only radial acceleration. Then it keeps its magnitude (so the speed doesn't change) and only changes direction. Thinking more about this it should be clear that such a situation is a circular motion. If there both is tangential and radial acceleration, then the we have an elliptical motion.
Naturally, for any path that is not straight, there must be a change of the velocity direction. And we call such a change acceleration - for a circular motion that acceleration happens to be perpendicular and towards the center of the circular path.
Any forces that act on a particle in a circular motion therefor must result in such a sideways acceleration. If they do not, then the motion cannot be circular because of the above explanation.
