Why don't we consider centripetal force while making FBD (free body diagram)? 
Consider a situation where a block is kept on a wedge which is further kept on a surface which rotates with $\omega$ the FBD for the same


My question is why dont we consider centripetal force in fbd? Is it because we are drawing it from a frame such that it cancels?

Edit :My simple definition of fbd tells me that it is a diagram representing all the forces acting on the body when viewed from  a particular frame which ensures that centripetal force must be acting only in a certain frame
 A: From an inertial frame
You are considering the centripetal force. In fact you have already included it if you are using an inertial frame.
The centripetal force is not a new force. It is just a name we give to whichever force that pulls centripetally (towards the centre in circular motion).
In your case there is a component from the normal force which pulls centripetally (horizontally leftwards). So that is called the centripetal force.
From the non-inertial frame
This above explanation is from an inertial frame, as if standing on the ground and watching the scenario. If you want to look at this from a non-inertial frame, by imagining that you are sitting on the block that is in rotation, which might be the case judging from your sketch, then from that frame the block doesn't seem to rotate. Rather it looks like the surroundings are "rotating".
Thus from that frame there seems to be no (resultant) centripetal force acting on the block. This means that the horizontal component of the normal force must be balanced out by something. That "something" is what we invent with the name centrifugal force, pointing opposite in the outwards direction.
This centrifugal force is a so-called pseudo-force in that it doesn't really "exist" - but we invent it mathematically in order to make Newton's laws hold true even in non-inertial frames.
A: 
My question is why dont we consider centripetal force in fbd ? Is it because we are drawing it from a frame such that it cancels?

Centripetal force and centrifugal force never act together. You are working in non-inertial frame and have shown centrifugal force , hence no need to show centripetal force.

I am attaching a picture from my notes to clarify what i meant so that my answer is interpreted correctly. The situation is quite similar to your question.
Consider a case of banking when viewed from 2 different frames :

In ground frame , you don't show any centrifugal force.

From inside car, which is a non inertial frame ,$F_{centrifugal}$ is balanced by $Nsin\theta$.
No doubt $Nsin\theta$ will act in both frame , and is equal to $mv^2/r$ regardless of the frame but since car is at rest in non inertial frame , I wouldn't call it a centripetal force.

Edit 2 :
Why would i call $Nsin\theta$ a centripetal force in the car's frame
According to Wikipedia a centripetal force is defined as :  a force that makes a body follow a curved path.
Since the body is at rest in it's own frame , $Nsin\theta$ is  is not a cause of the centripetal force in the rotating frame.
A: The reason we don't include the "centripetal force" in the FBD is the same reason we don't include the "vertical force" or the "horizontal force" in the FBD. Centripetal is just a direction, just like vertical and horizontal are directions. Just like how forces can be vertical or horizontal, forces can also be centripetal or tangential.
For example, including "centripetal force" on a FBD would just be like including the gravitational force as well as a second "vertical force" on the FBD for an object in free fall. In your case, there is a component of the force of the wedge on the block. This force points in the centripetal direction, so it is a centripetal force. But you wouldn't put an explicit "centripetal force" on the diagram; you already did that.
A: 
why dont we consider centripetal force in fbd ?

Indeed, we should include pseudo forces too, consider this :

Due to different position of wedge and block COM from rotational axis and they difference in mass- there will arise subtle tidal forces (differences in centrifugal forces of that within block and wedge). So in the end normal force upon block will be :
$$ N = mg \cdot \cos \alpha + m\,\omega^2r \cdot \sin\alpha - M\omega^2R 
\cdot \sin\alpha$$
Where $m$ block mass, $M$ - wedge mass, $r$ block distance from rotational axis, $R$ - wedge distance from rotational axis, $\omega$ - angular speed.
A: The descriptions of circular motion often make it sound like centripetal and/or centrifugal force result from circular motion. But it's the opposite: circular motion results from centripetal force. There is nothing specifically exerting a centripetal force on the object. Centripetal force is an "output" force, not an "input" force. Gravity and the normal force are the "input" force; they are distinct forces from distinct physical sources. When we add those two forces together, we get the net force. Since this force is toward the center of the circle, it is a centripetal force. "Centripetal force" is just what we're calling the sum of all the input forces. It's not a force we start with when we're drawing the FBD, it's what we use the FBD to find.
A: We do not consider the centripetal force to only be acting on the block, as it is also acting on the wedge. You have to consider the entire wedge-block system while calculating the centripetal force, so the centripetal force would act on the wedge-block system as a whole, not only on the block.
Edit:
Even if you do consider the centripetal force, then you have to consider a horizontal component of reaction from the wedge which correctly cancels out the centripetal force, so it won't make any difference.
I hope this helps!
