Rotational Dynamics Problem Involving Multiple Rods 
In the figure shown, the rod $OA$ of length $1 \ \rm m$ is rotating
about the fixed point $O$ and the rod $AB$ of length $1\ \rm m$ is
rotating about $A$, in a vertical plane. At a certain moment, when
$OA$ is making an angle  $60^\circ$ with the horizontal, the rod $AB$
is horizontal. At the same moment angular speed and angular
acceleration of $OA$ and $AB$ are $2 \ \rm rad/s$, $2\sqrt 3 \rm \
 rad/s^2$ and $1 \ \rm rad/s$, $3\sqrt3 \rm \ rad/s^2$ respectively.
Calculate the acceleration of point $B$ in $\rm m/s^2$  at that
moment.


My approach:
I assumed the acceleration
of centre of mass of rod AB to have x and y components and via constraints at point A(using angular acceleration of rod OA as it is hinged to the ground and equating it to acceleration due to rod AB caused by its x,y acceleration and angular acceleration and also ensuring that there are no components of acceleration perpendicular to acceleration caused by hinge(O) at A.). The net acceleration at B can by found by using x and y components of acceleration of rod Ab, its angular acceleration and centripetal acceleration. However I am not getting the answer. Am i going conceptually wrong somewhere?. It would be great if someone could help me out with this one
 A: This is a standard kinematics problem.
The steps to solve these is to first designate the degrees of freedom of the system. Use the two orientation angles $\theta_1$ and $\theta_2$ is conventionally what is used for pin joints.

Then track the points of interest (since dynamics are not involved, each center of mass isn't important here) in terms of position, velocity and acceleration. This is done by taking the time derivative using the chain rule.
$$ \begin{aligned} \vec{\rm pos}_A &=  \pmatrix{ \ell \cos \theta_1 \\ \ell \sin \theta_1} & \vec{\rm pos}_B & = \vec{\rm pos}_A + \pmatrix{ \ell \cos \theta_2 \\ \ell \sin \theta_2 } \end{aligned} $$
$$ \begin{aligned} \vec{\rm vel}_A &=  \pmatrix{ \ell \omega_1 \sin \theta_1 \\ -\ell \omega_1 \cos \theta_1} & \vec{\rm vel}_B & = \vec{\rm vel}_A + \pmatrix{ \ell \omega_2 \sin \theta_2 \\- \ell \omega_2 \cos \theta_2 } \end{aligned} $$
where $\dot{\theta}_1 = -\omega_1$ and $\dot{\theta}_2 = -\omega_2$ from the rotational sense of the sketch above.
and one more time
$$ \begin{aligned} \vec{\rm acc}_A &= \pmatrix{ \ell \alpha_1 \sin \theta_1 - \ell \omega^2 \cos \theta_1 \\ -\ell \alpha_1 \cos \theta_1 - \ell \omega_1^2 \sin \theta_1} & \vec{\rm acc}_B & =\vec{acc}_A + \pmatrix{ \ell \alpha_2 \sin \theta_2 - \ell \omega_2^2 \cos \theta_2 \\ -\ell \alpha_2 \cos \theta_2 - \ell \omega_2^2 \sin \theta_2  } \end{aligned} $$
where $\dot{\omega}_1 = \alpha_1$ and $\dot{\omega}_2 = \alpha_2$.
Now you have symbolic expressions of the acceleration of each point, which you use in the situation you want to examine.

You could also solve the above using the velocity and acceleration transformation equations
$$ \begin{aligned}
  \vec{\rm vel}_A & = \vec{\omega}_1 \times \vec{\rm pos}_1 \\
  \vec{\rm vel}_B & = \vec{\rm vel}_A + \vec{\omega}_2 \times ( \vec{\rm pos}_2 - \vec{\rm pos}_1 ) \\
  \vec{\rm acc}_A & = \vec{\alpha}_1 \times \vec{\rm pos}_1 + \vec{\omega}_1 \times \vec{\rm vel}_1 \\
  \vec{\rm acc}_B & = \vec{\rm acc}_A + \vec{\alpha}_2 \times( \vec{\rm pos}_2 - \vec{\rm pos}_1) + \vec{\omega}_2 \times ( \vec{\rm vel}_2 - \vec{\rm vel}_1) \\ 
\end{aligned}$$
but I think the above is far more complex and tedious compared to the direct differentiation method presented above at first.
