Slider-Crank Mechanism - Part 3

Classical Mechanics Level 3

Consider the mechanism shown in the figure. The system comprises two rigid rods named $1$ and $2$ respectively. The point mass numbered as $3$ can only translate along the X-axis. The bodies are linked by using appropriate hinges and joints.

$OA = 0.15 \ m$ $AB = 0.4 \ m$ $\angle AOB = \theta$

The centers of mass of rods lie at their geometric center. At the following instant: $\theta = \frac{\pi}{3} \: rad$ $\dot{\theta} = -12\pi \: rad/s$

The solver is required to find the point on body $2$ which has minimum speed. Enter your answer as the distance from point $A$ along the length of the rod $AB$ .

Note:

The small circles and triangles just indicate joints between bodies.
The negative sign of $\dot{\theta}$ indicates a clockwise rotation of body 1 about O.
Body 3, even though depicted as a rectangle is just a point mass.
The motion of the system is confined to the X-Y plane.

The answer is 0.1450.

1 solution

Steven Chase
Sep 10, 2019

Let $(x,y)$ be a point on rod $2$ . The following relation generally holds:

$x = x_A + \alpha(x_B - x_A) \\ y = y_A + \alpha(y_B - y_A) \\ 0 \leq \alpha \leq 1$

Time-differentiating gives:

$\dot{x} = \dot{x}_A + \alpha(\dot{x}_B - \dot{x}_A) \\ \dot{y} = \dot{y}_A + \alpha(\dot{y}_B - \dot{y}_A) \\ 0 \leq \alpha \leq 1$

The speed is:

$v = \sqrt{\dot{x}^2 + \dot{y}^2}$

For point $A$ :

$x_A = OA \, \cos \theta \\ y_A = OA \, \sin \theta \\ \dot{x}_A = -OA \, \sin \theta \, \dot{\theta} \\ \dot{y}_A = OA \, \cos \theta \, \dot{\theta}$

For point $B$ (from the previous problem):

$\dot{x}_B \approx 5.868 \\ \dot{y}_B = 0$

So we know the velocities of the two ends points. All that remains is to sweep the parameter $\alpha$ and see which value yields the smallest speed. As it turns out, $\alpha \approx 0.3625$ yields the smallest speed, corresponding to a length of $\approx 0.145$ from point $A$ .

Thanks for the solution. Another way in which this can be approached is by computing the square of the speed as a function of $\alpha$ and equating its derivative (with respect to $\alpha$ ) to zero. This would be an analytical method.

Karan Chatrath - 1 year, 9 months ago

Indeed, I considered that. But laziness prevailed, as it often does. This has turned into a nice problem series

Steven Chase - 1 year, 9 months ago

I understand, and thanks! I am still contemplating whether I should post a dynamics variant of this. Thinking of something insightful and yet not too tedious in terms of calculations. I'll post one when I do have an idea.

Karan Chatrath - 1 year, 9 months ago

Slider-Crank Mechanism - Part 3

The answer is 0.1450.

1 solution

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