Finding accelerations using Instantaneous axis of rotation Disclaimer: I present a question which is homework-like, However it is simply to demonstrate an example. I only wish to clarify one thing, that is the validity of the IAR in determining the accelerations.
IAR: stands for the Instantaneous Axis of Rotation
Consider a Massless rod (initially vertical) with two Balls A and B of the same mass $m$ attached at the two ends. A slides along a frictionless wall, B slides along a frictionless floor.

My interpretation of the IAR: at the "instant", the system can be thought of as rotating purely about the IAR.'
I have used this interpretation of the IAR a lot: and have successfully managed to tackle quite a few problems by  representing the entire kinetic energy of the system as $1/2I_{iar}\omega^2$, or even (sometimes) use $\tau _{iar}=I_{iar}\alpha$. However, accelerations aren't modelled property by this interpretation.
For instance, in the diagram above, the interpretation suggests that:
$\vec{A_a}$=$(\omega ^2(l\cos(\theta)))\hat{i}$-$(\alpha(l\cos(\theta)))\hat{j}$.
However, $x_a$=0 always. So $\ddot{x_a}$=$0$, and thus we have a contradiction. Likewise for B.
Things get more bizarre when you consider the instant when A leaves the wall.
My observations: This happens at $\sin(\theta)=2/3$, and at this instant,$\ddot{x_b}$=$0$ ,$\ddot{y_b}$=$0$, $\ddot{x_a}$=$0$ and $\ddot{y_a}$=$-g\hat{j}$. This cannot be explained by the IAR at all which leads to my question: When is it appropriate to say that  $\vec{A}$=-$(\omega ^2(r)\hat{n}$+$(\alpha(r))\hat{\theta}$., where r is the distance from IAR?
 A: This is a good question, and highlights a common misconception in Newtonian mechanics. For concreteness, let's work in two dimensions.
It's true that at every moment, you can write the velocity as a rotation about an instantaneous point $\mathbf{r}_0$ of rotation, which means that the velocity $\mathbf{v}$ of any point in the body satisfies
$$\mathbf{v} = \mathbf{\omega} \times (\mathbf{r} - \mathbf{r}_0).$$
The derivative is the acceleration, so there are three terms by the product rule, because of the three places the derivative can act,
$$\mathbf{a} = \mathbf{\alpha} \times (\mathbf{r} - \mathbf{r}_0) + \omega \times \mathbf{v} - \omega \times \mathbf{v}_0.$$
The first term represents the angular acceleration. The second term represents the usual centripetal acceleration due to rotation. And the third term is due to the rate of change $\mathbf{v}_0 = d\mathbf{r}_0/dt$ of the instantaneous point of rotation $\mathbf{r}_0$ itself. That's the term you're missing. Your statement in bold is correct exactly when this term vanishes.
Almost all introductory physics textbooks ignore this term, because they work with only simple situations where $\mathbf{v}_0 = 0$, but it is there. For example, I wrote question B3 on the 2019 US Physics Olympiad with the sole purpose of illustrating this point; in that problem the instantaneous point of rotation is changing, and without accounting for this, one gets incorrect physical answers.
A: If you work out the problem, you will find that the acceleration on the end is just tangent to the walls. This is true for a solid body, or for two masses attached to a rigid link between them.
$$ \begin{aligned} \vec{a}_{\rm K} & = \pmatrix{ 0 \\ -L \omega^2 \cos \varphi - L \dot \omega \sin \varphi } \\
\vec{a}_{\rm H} & = \pmatrix{L \dot \omega \cos \varphi - L \omega^2 \sin \varphi \\ 0} \end{aligned}  \tag{1}$$

There is no horizontal component of $\vec{a}_{\rm K}$ and no vertical component of $\vec{a}_{\rm H}$. And this makes sense since points K and H are each kinematically linked to their sliding action.
When you find the acceleration vector for each point in space, you can find the location B where the acceleration is zero. That is the instant axis of acceleration (IAA) and according to my calculations, it is always down and to the left of IAR. When the body isn't moving (initially) then the IAA is on top of the IAR.
The center of rotation is found to be at
$$ \begin{aligned} x_{\rm A} & = L  \sin \varphi & y_{\rm A} & = L \cos \varphi \end{aligned} \tag{2}$$
The center of acceleration is found to be at
$$ \begin{aligned} x_{\rm B} & = L \left( \frac{6 \sin \varphi}{5-3 \cos \varphi} - \sin \varphi \right) & y_{\rm B} & = \left( \frac{9 \sin^2 \varphi}{2(5-3 \cos \varphi)} - \frac{\sin^2 \varphi}{2(\cos \varphi-1)} \right) \end{aligned} \tag{3}$$
This is using the following solution of two separated masses
$$ \begin{aligned} \dot \omega & = \frac{g}{L} \sin \varphi & \omega^2 = \frac{2 g ( 1-\cos \varphi)}{L} \end{aligned} \tag{4} $$
But the result is the same as using a thin rod and
$$ \begin{aligned} \dot \omega & = \frac{3 g}{2 L} \sin \varphi & \omega^2 = \frac{3 g ( 1-\cos \varphi)}{L} \end{aligned} \tag{5} $$
Interesting things happen at $\varphi \approx 48.19^\circ$ making $\vec{a}_{\rm H}=0$.
Why isn't there an acceleration towards the IAR on the endpoints?
Although the body is rotating about IAR, it is also accelerating downwards and to the right. This means that the IAR is moving with time and the acceleration of the body @IAR isn't zero it is $$ \vec{a}_{\rm IAR} = \pmatrix{ 2 g \sin\varphi \cos \varphi -2 g \sin \varphi \\ 2 g \cos^2 \varphi -2 g \cos \varphi} \tag{6}$$
Note the above isn't the acceleration of the IAR as that would be meaningless, but the acceleration of the body at the IAR.
It is exactly such as so the body does not violate the kinematics (sliding joints) in the future.
A: your system has one generalized coordinate $\varphi$
thus:
$$x=L\sin \left( \varphi  \right)$$
$$y=L\cos \left( \varphi  \right) $$
with:
$$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$
and
$$U=m\,g\,y$$
you get:
$$\ddot{\varphi}=\frac{g}{L}\,\sin(\varphi)\tag 1$$
the velocity vector is :
$$\vec{v}= \begin{bmatrix}
\dot{ \varphi}\,y \\
  -\dot{\varphi}\,x \\
\end{bmatrix}=\dot{\varphi}\,\left[ \begin {array}{c} L\cos \left( \varphi  \right)  
\\ -L\sin \left( \varphi  \right) \
\end {array} \right] 
\tag 2$$
the time derivative of equation (2) is the acceleration vector:
$$\vec{a}=\left[ \begin {array}{c} -L\sin \left( \varphi  \right) \dot{\varphi }^{
2}+\ddot{\varphi}\,L\cos \left( \varphi  \right) 
\\ -L\cos \left( \varphi  \right) \dot{\varphi }^{2}-
\ddot{\varphi} \,L\sin \left( \varphi  \right) \end {array} \right] 
$$
and with equation (1)
$$\vec{a}=\left[ \begin {array}{c} -L\sin \left( \varphi  \right) \dot{\varphi }^{
2}+[\frac{g}{L}\,\sin(\varphi)]\,L\cos \left( \varphi  \right) 
\\ -L\cos \left( \varphi  \right) \dot{\varphi }^{2}-
[\frac{g}{L}\,\sin(\varphi)] \,L\sin \left( \varphi  \right) \end {array} \right] 
$$
edit

I) acceleration point A and B
the velocity of $m_1$ is:
$$\vec{v}_A=v_A\vec{\hat{e}}_y=-\omega\,X\,\vec{\hat{e}}_y$$
the velocity of $m_2$ is:
$$\vec{v}_B=v_B\vec{\hat{e}}_x=\omega\,Y\,\vec{\hat{e}}_x$$
with :
$X=L\sin(\varphi)\quad,Y=L\cos(\varphi)\quad$ and $\omega=\dot{\varphi}$
$\Rightarrow$
$$\vec{v}=\begin{bmatrix}
  v_A \\
  v_B \\
0\\
\end{bmatrix}=
\vec{\omega}\times \vec{R}_p\tag A$$
where :
$$\vec{R}_p=\begin{bmatrix}
  X \\
  Y \\
  0\\
\end{bmatrix}\quad,\vec{\omega}=\dot{\varphi}\,\begin{bmatrix}
  0 \\
  0 \\
  1\\
\end{bmatrix}$$
acceleration
$$\vec{a}=\begin{bmatrix}
  a_A \\
  a_B \\
  0   \\
\end{bmatrix}=\vec{\dot{\omega}}\times \vec{R}_p+
\vec{\omega}\times \vec{\dot{R}}_p\tag B$$
where
$$\vec{\dot{R}}_p=\begin{bmatrix}
  \dot{X} \\
  \dot{Y} \\
  0   \\
\end{bmatrix}$$
II) acceleration arbitrary point $u$ of the rod
velocity of point u is:
$$\vec{v}_u=\vec{\omega}\,\times \vec{u}$$
where:
$$\vec{u}=\vec{R}_u-\vec{R}_p$$
and
$$\vec{R}_u=\begin{bmatrix}
  x \\
 y(x) \\
  0   \\
\end{bmatrix}=\begin{bmatrix}
  x \\
 -\frac{Y}{X}\,x+Y \\
  0   \\
\end{bmatrix}\quad,0\le x\le X
$$
to "move" the point u on the rod you substitute $x\mapsto a_u\,X$ to $\vec{R}_u$
thus
$$\vec{R}_u=\begin{bmatrix}
  a_u\,X \\
 -\frac{Y}{X}\,a\,X+Y \\
  0   \\
\end{bmatrix}=\begin{bmatrix}
  a_u\,X \\
 -a_u\,Y+Y \\
  0   \\
\end{bmatrix}\quad,0\le a_u\le 1$$
$\Rightarrow$
$$\vec{u}=\begin{bmatrix}
  a_u\,X \\
 -a_u\,Y+Y \\
  0   \\
\end{bmatrix}-\begin{bmatrix}
  X \\
  Y \\
  0   \\
\end{bmatrix}=\begin{bmatrix}
  X\,(a_u-1) \\
  -a_u\,Y \\
  0   \\
\end{bmatrix}\quad,0\le a_u\le 1$$
Acceleration:
$$\vec{A}_u=\vec{\dot{\omega}}\times \vec{u}+
\vec{\omega}\times \vec{\dot{u}}=\ddot{\varphi}\,L\,\left[ \begin {array}{c} \cos \left( \varphi  \right) a_u
\\\sin \left( \varphi  \right)  \left( a_u-1 \right) 
\\0\end {array} \right] +\dot{\varphi}^2\,L\,\left[ \begin {array}{c} -a_u\,\sin \left( \varphi  \right) 
\\ \cos \left( \varphi  \right)  \left( a_u-1 \right) 
\\ 0\end {array} \right] 
\tag C$$
Simulation:
With equation (1) you get $\ddot{\varphi}$ and the solution of the differential equation $\varphi(t)$
the position vector to the IAR point is $$\vec{R}_p=\begin{bmatrix}
  X \\
 Y \\
\end{bmatrix}=L\,\begin{bmatrix}
  \sin(\varphi(t) \\
\cos(\varphi(t) \\
\end{bmatrix}$$
I stop the simulation when the Y coordinate is zero.

Figure  1
The  trajectory of the IAR points (blue points) and the position of the rod.

Figure 2
The trajectory of the  acceleration components $\vec{A}_u$ for three different rod points
A: The weight of mass, A , produces a torque about the center of rotation: (mgL)cosθ = Iα  solving gives α = (g/L)cosθ (counter-clockwise).  Multiplying by the corresponding radii gives the negative y acceleration for mass, A, = -g (cosθ)^2  and x acceleration for mass, B, = g(sinθ)(cosθ).
The x acceleration for the center of mass is half that of mass, B;  (H/2m) = (1/2)g(sinθ)(cosθ)  Solving gives the horizontal force from the left hand wall; H = mg(sinθ)(cosθ)  A similar calculation gives the upward vertical force from the lower surface; V = 2mg – mg(cosθ)^2.  If these calculations are correct, mass, A, does not leave the wall before striking the lower surface.
A: instantaneous rotation point is a point where velocity of that point at that point is zero but while you are write acceleration in iaor point you try to see that acceleration work in ground frame which is incorrect you have add acceleration of pseudo force and then you see acceleration of A in ground frame in y direction is zero so in conclusion iaor is point where velocity at that instant is zero not acceleration. $\alpha$ and $\omega$ remain same in all frames and don't change.
Let me know if this is incorrect.
