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

The correct answer is:
ω²r

Step-by-Step Derivation and Analysis:

1. Define the Coordinate System:
Let us establish a Cartesian coordinate system with the crank's pivot O at the origin $(0,0)$.
From the given diagram:

• The crank rotates counter-clockwise with a constant angular velocity $\omega$. The angle of the crank with the positive $x$-axis is $\theta$.
• The coordinates of the crank pin $A$ at any instant are:

  $$x_A=r\cos \theta$$

  $$y_A=r\sin \theta$$

• At the instant shown in the image, the crank is parallel to the $x$-axis (pointing in the positive $x$-direction), so $\theta =0$.
• The slider is constrained to move along a vertical guide parallel to the $y$-axis. The horizontal distance from the crank pin $A$ to the vertical line of stroke is labeled as $1$ (which represents $r$). Since $x_A=r$ at this instant, the constant horizontal coordinate of the slider $B$ is:

  $$x_B=2r$$

• The length of the connecting rod is given as $L=\sqrt{2}r$.

2. Mathematical Formulation:
Using the distance formula between the crank pin $A$ and the slider $B(2r,y_B)$:

$${(x_B-x_A)}^2+{(y_B-y_A)}^2=L^2$$

Substituting the expressions:

$${(2r-r\cos \theta )}^2+{(y_B-r\sin \theta )}^2=2r^2$$

At the instant shown ($\theta =0$):

$${(2r-r)}^2+{(y_B-0)}^2=2r^2\Rightarrow r^2+y_B^2=2r^2\Rightarrow y_B^2=r^2$$

Since the slider is located below the horizontal line of the crank pin, we select the negative root:

$$y_B=-r$$

3. First Derivative (Velocity Analysis):
Differentiating the position relation with respect to time $t$:

$$2(2r-r\cos \theta )(r\sin \theta \stackrel{·}{\theta })+2(y_B-r\sin \theta )({\stackrel{·}{y}}_B-r\cos \theta \stackrel{·}{\theta })=0$$

Dividing by 2 and substituting $\stackrel{·}{\theta }=\omega$:

$$(2r-r\cos \theta )r\omega \sin \theta +(y_B-r\sin \theta )({\stackrel{·}{y}}_B-r\omega \cos \theta )=0$$

At $\theta =0$ and $y_B=-r$:

$$0+(-r-0)({\stackrel{·}{y}}_B-r\omega )=0\Rightarrow {\stackrel{·}{y}}_B=r\omega$$

4. Second Derivative (Acceleration Analysis):
We differentiate the velocity equation with respect to time $t$ once more. Note that $\stackrel{·}{\theta }=\omega$ is constant, so $\ddot{\theta }=0$:
Differentiating the first term $(2r-r\cos \theta )r\omega \sin \theta$:

$$\frac{d}{dt}[(2r-r\cos \theta )r\omega \sin \theta ]=(r\omega \sin \theta )r\omega \sin \theta +(2r-r\cos \theta )r{\omega }^2\cos \theta$$

At $\theta =0$, this derivative simplifies to:

$$0+(2r-r)r{\omega }^2\left(1\right)={\omega }^2{r}^2$$

Differentiating the second term $(y_B-r\sin \theta )({\stackrel{·}{y}}_B-r\omega \cos \theta )$:

$$\frac{d}{dt}\left[(y_B-r\sin \theta )({\stackrel{·}{y}}_B-r\omega \cos \theta )\right]={({\stackrel{·}{y}}_B-r\omega \cos \theta )}^2+(y_B-r\sin \theta )({\ddot{y}}_B+r{\omega }^2\sin \theta )$$

At $\theta =0$, $y_B=-r$, and ${\stackrel{·}{y}}_B=r\omega$, this simplifies to:

$${(r\omega -r\omega )}^2+(-r-0)({\ddot{y}}_B+0)=-r{\ddot{y}}_B$$

Combining the terms of the differentiated equation:

$${\omega }^2{r}^2-r{\ddot{y}}_B=0$$

$${\ddot{y}}_B={\omega }^{2}r$$

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