How can a particle which is not moving have an acceleration? Suppose a rod is rotating around a fixed point located at an extreme point of it and there are two points on it. One, somewhere in the middle and the other at the other extreme.Call them  $A$ and $B$ respectively.
The Question is relatively simple, but it is confusing me a lot! For $A$ , $B$ is at rest. How can it have an acceleration? Yeah, for $A$ there exists a centripetal force on $B$. But the point is that there is no relative motion! How can there be an acceleration as acceleration is the rate of change of velocity ( here relative velocity ) which is 0?` 
EDIT - This was a question in my book and it asked to calculate the acceleration of $B$ relative to $A$ and the answer was not 0 . 
 A: The phrasing "moving", "velocity", "acceleration" or whatsoever else always depend on the choice of a reference frame. 
Choose a reference frame $S$ and, in that reference frame, calculate $\mathbf{r}, \dot{\mathbf{r}}, \ddot{\mathbf{r}}$.
Now choose another reference frame $S'$ and calculate $\mathbf{r'}, \dot{\mathbf{r'}}, \ddot{\mathbf{r'}}$.
The latter quantities are in general different from the former and it can as well happen that either set vanish (in particular the velocity of a particle is always zero in its own reference frame, by definition of reference frame integral with the particle).
A: You are asking "what is the acceleration of B relative to A?" The confusion is that you (or the book) are not specifying exactly what reference frame is being used. Are we using A's rotating reference frame which is fixed in the rod, with A always facing O? Or are we using a frame in which A is always facing in a fixed external direction - North, for example?
In the 1st case B is stationary in A's rotating frame of reference and does not accelerate, as you state. In the 2nd case B is rotating around A with constant speed, hence B has a relative velocity and a relative acceleration which are both constant in magnitude but varying in direction. The constant relative speed is $r\omega$ where $r$ is the fixed distance between A and B, and $\omega$ is the angular velocity of the rod. The magnitude of the relative acceleration is $a=r\omega^2$. The direction is always towards A.
A: I'm not quite sure what you're asking but i think your getting confused by only accounting for change in magnitude component of momentum when change in direction is also a change in momentum and therefore technically acceleration. The velocity doesn't change but the momentum does.
If you're question is about how something has a momentum in a theoretical instant of time thats a much deepet and probably a more philosophical matter.
A: If I stand and reach my hands into the air holding a rod with both hands.  One end of the rod's end is in my hand while the other end of the rod extends past my other hand let's say a foot.  My hands are reaching up in such a way that my body and arms stretched out makes me resemble the letter Y.  The end of the rod that extends past my one hand will be point B. The other point of the rod that is directly above my head will be point A. Point A is center and has no centripetal motion relative to me once my feet start doing their thing.  Now I swirl around using my feet but holding my body rigid.  Point B will have the centripetal force as it is at the end of the rod and not center.  No point is moving in relation to me and are at rest relative to me. I speed up my foot work and still no motion relative to me but we have acceleration.  I only think this is what you are after but let's just see what else pops up here.
A: If $A$ is the pivot, $B$ is not at rest relative to $A$.  $B$ experiences angular displacement. In polar coordinates, $B$ can be described by
\begin{align*}
  \mathbf{r} &= r \, \hat{r} \\
  \mathbf{v} &= \dot{r} \, \hat{r}+r\dot{\theta} \, \hat{\theta} \\
  \mathbf{a} &= (\ddot{r}-r\dot{\theta}^2) \, \hat{r}+
                (2\dot{r}\dot{\theta}+r\ddot{\theta}) \, \hat{\theta}
\end{align*}
In particular, $\dot{r}=0$ and $\dot{\theta}=\omega$
\begin{align*}
  \mathbf{v} &= r\omega \,\hat{\theta} \\
  \mathbf{a} &= -r\omega^2 \hat{r}
\end{align*}
A: 
How can a particle which is not moving have an acceleration?

Using the Taylor series expansion for the trajectory of the particle $x(t)$ about $t = 0$, we have
$$x(t) = x(0) + \frac{dx}{dt}t + \frac{1}{2}\frac{d^2x}{dt^2}t^2 + \cdots = x(0) + v(0)t + \frac{1}{2}a(0)t^2 + \cdots$$
This is a perfectly general result.  In the case that the initial velocity is zero, $v(0) = 0$, there is no reason to constrain the initial acceleration (or jerk or higher order derivatives) to be zero.
Physically, for a particle with zero velocity but non-zero acceleration, this means that $x(0 + dt) = x(0)$, i.e., the object isn't moving at $t = 0$ but it is beginning to move.
