The crank of a slider-crank mechanism rotates counter-clockwise (CCW) with a constant angular velocity w, as shown. Assume the length of the crank to be r.

Using exact analysis, the acceleration of the slider in the y-direction, at the instant shown, where the crank is parallel to x-axis, is given by

To find the acceleration of the slider in the y-direction using exact analysis, we define the coordinate system with the origin at the crank pin pivot $O$.

Let $\theta$ be the angle of the crank relative to the positive x-axis. The coordinates of the crank pin $A$ are given by:
$x_A=r\cos \theta$
$y_A=r\sin \theta$

The slider $B$ is constrained to move along the vertical direction. From the diagram, when the crank is horizontal ($\theta =0$), the horizontal distance between $A$ and $B$ is $r$. Therefore, the constant x-coordinate of the slider is:
$x_B=2r$

The length of the connecting rod $AB$ is given as $L=\sqrt{2}r$. The distance formula between $A$ and $B$ gives:
${(x_B-x_A)}^2+{(y_B-y_A)}^2=L^2$

Substituting $x_B=2r$, $x_A=r\cos \theta$, $y_A=r\sin \theta$, and $L=\sqrt{2}r$ into the equation:
${(2r-r\cos \theta )}^2+{(y_B-r\sin \theta )}^2=2r^2$

Differentiating this equation with respect to time $t$, keeping in mind that $\frac{d\theta }{dt}=\omega$ is constant:
$2(2r-r\cos \theta )(r\sin \theta ·\omega )+2(y_B-r\sin \theta )(v_B-r\cos \theta ·\omega )=0$

Simplifying by dividing by 2:
$(2r-r\cos \theta )r\sin \theta ·\omega +(y_B-r\sin \theta )(v_B-r\cos \theta ·\omega )=0$

At the instant shown in the diagram, the crank is parallel to the x-axis, meaning $\theta =0$, $\cos \theta =1$, and $\sin \theta =0$. Also, the vertical position of the slider B is $y_B=-r$.
Substituting these values to find the slider velocity $v_B$:
$(2r-r)\left(0\right)+(-r-0)(v_B-r\omega )=0$
$-r(v_B-r\omega )=0\Rightarrow v_B=r\omega$

To find the acceleration, we differentiate the simplified first derivative equation with respect to time once more:
$\frac{d}{dt}\left[\right(2r-r\cos \theta )r\sin \theta ·\omega ]+\frac{d}{dt}\left[\right(y_B-r\sin \theta \left)\right(v_B-r\cos \theta ·\omega \left)\right]=0$

Using the product rule, the first term differentiates to:
$(r\sin \theta ·\omega )(r\sin \theta ·\omega )+(2r-r\cos \theta )(r\cos \theta ·{\omega }^2)$
At $\theta =0$, this becomes:
$0+(2r-r)\left(r{\omega }^2\right)=r^2{\omega }^2$

The second term differentiates to:
${(v_B-r\cos \theta ·\omega )}^2+(y_B-r\sin \theta )(a_y+r\sin \theta ·{\omega }^2)$
At $\theta =0$ and substituting $v_B=r\omega$:
${(r\omega -r\omega )}^2+(-r-0)(a_y+0)=-r{a}_y$

Combining the results of both differentiated parts:
$r^2{\omega }^2-r{a}_y=0$
$a_y={\omega }^{2}r$

Thus, the acceleration of the slider in the y-direction at this instant is ${\omega }^{2}r$.