Slider-Crank Mechanism - Part 2

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

O A = 0.15 m OA = 0.15 \ m A B = 0.4 m AB = 0.4 \ m A O B = θ \angle AOB = \theta

The center of mass of rods lie at their geometric center. At an instant when: θ = π 3 r a d \theta = \frac{\pi}{3} \: rad θ ˙ = 12 π r a d / s \dot{\theta} = -12\pi \: rad/s

Determine the speed of the sliding body numbered as 3. Enter your answer in meters per second.

Note:

  • The small circles and triangles just indicate joints between bodies.
  • The negative sign of θ ˙ \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 5.8681.

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1 solution

Steven Chase
Sep 6, 2019

The following is true in general:

( O A cos θ B x ) 2 + ( O A sin θ 0 ) 2 = ( A B ) 2 (OA \, \cos \theta - B_x)^2 + (OA \, \sin \theta - 0)^2 = (AB)^2

Solve this equation for B x B_x , given the numbers provided. It is necessary to use a negative root when solving. B x 0.4533 B_x \approx 0.4533

Then differentiate that expression to get the following:

( O A cos θ B x ) ( O A sin θ θ ˙ + B ˙ x ) = ( O A sin θ ) ( O A cos θ θ ˙ ) (OA \, \cos \theta - B_x)(OA \, \sin \theta \, \dot{\theta} + \dot{B}_x ) = (OA \, \sin \theta)(OA \, \cos \theta \, \dot{\theta})

Re-arrange and solve this for B ˙ x \dot{B}_x . B ˙ x 5.868 \dot{B}_x \approx 5.868

This is a nice related-rates problem, which can potentially also become a nice dynamics problem.

Very nice solution, thanks!

As for the dynamics problem, I am thinking of one. A complete dynamic analysis of this system is a doable yet tedious and fairly long exercise.

Karan Chatrath - 1 year, 9 months ago

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I was thinking about the equations for the dynamics. As you say, doable, but not very fun

Steven Chase - 1 year, 9 months ago

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