A rod $AB$ of length $3m$ which remains in the same vertical plane has its ends $A$and $B$ constrained to remain in contact with a horizontal floor and a vertical wall respectively. Determine the acceleration of the end B at the position, if the point $A$has a velocity of $2 m/s$ and an acceleration of $1.6 m/s^2$ rightward.

Acceleration of end B :

The acceleration of the end B has three components.

$\left(1\right)$ Due to translation of end $A$, $a_A = 1.6 m/s^2$ .

$\left(2\right)$ Due to rotation of end $B$about $A$, i.e., tangential acceleration

$a_t=\alpha r_{AB}=\alpha \times 3=3\alpha m/s^2$

$\left(3\right)$ Radial acceleration (from $B$towards $A$)

Also $v_{BA}=\frac{v_A}{\sin 60°}=\frac{2}{\sin 60°}=2.3 m/s$

$$\begin{gathered} a_n=\frac{v_{BA}^2}{r_{AB}} \\ =\frac{{\left(2.3\right)}^2}{3}=1.76 m/s^2 \end{gathered}$$

The component of the acceleration of the end $B$ are shown in Figure. As the net acceleration of the end is vertically downward, so net horizontal component of the accelerations must be zero. i.e., 

${\alpha }_x=0=1.6+1.76\cos 60°-3\alpha \cos 30°$

or $\alpha =0.95 rad / s^2$

Considering vertical components, we get

$$\begin{gathered} a_B=a_y=1.76\sin 60°+3\alpha \sin 30° \\ =2.98 m/s^2 \end{gathered}$$