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I was wondering whether maybe you could help me with this problem. I have a double pendulum. I have set the origin of cartesian coordinates to be the "head" of the first arm, which is fixed. The end of the second arm is attached to a block that slides along the x-axis. What I want to do is to derive the equations relating the pendulum's angles with the distance from the origin to the block.

Now, I know how I could go about deriving the equations without the constraint.

$$x_1 = L_1cos(a_1)$$ $$y_1 = L_1sin(a_1)$$

Where $x_1$ and $y_1$ is where the first arm joins the second arm and $a_1$ is the angle between the horizontal and the first arm.

Similarly, I can derive the equations for the end of the second arm $x_2 = x_1 + L_2 cos(a_2)$ and $y_2 = y_1 - L_2 sin(a_2)$

Now then, if I attach a sliding block to the end of my second arm, I don't know whether my equation for $x_2$ would change at all. I don't think it would but would I have to somehow restrict the swing angles so that the block only moves along the x direction?

Well, basically the problem is finding the equation of $x_2$ if it's attached to a block that only moves along the x- direction.

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  • $\begingroup$ If you provided an image your question would be a lot easier to understand. $\endgroup$ – Bending Unit 22 Aug 18 '15 at 14:04
  • $\begingroup$ Yes, I figure I should have included a diagram, I just wasn't sure how to draw one in here. Yours is a good one and it does show what happens in my problem. Thank you. $\endgroup$ – Emilio Botero Aug 18 '15 at 14:46
  • $\begingroup$ Check this article in the help center on how to use markdown to include images. There's also a little icon in the editor that you can click which inserts the necessary formatting code for you and also allows you to upload images to imgur. If you still have problems (or any other question regarding the use of the site itself) go right ahead and ask on the meta site $\endgroup$ – Bending Unit 22 Aug 18 '15 at 14:52
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What you are describing is a crankshaft.

if I attach a sliding block to the end of my second arm, I don't know whether my equation for $x_2$ would change at all.

No it wouldn't, because your restraint:

the second arm is attached to a block that slides along the x-axis

doesn't restrain $x_2$ but $y_2$

"Being on the x axis" basically means $y = 0$, which in your case means $y_2 = 0$, which is where you "lose" the second degree of freedom.

Take a look at the following image:

crankshaft

The formula for the relation between $x$ and $\varphi$ in the image is:

$$x=l_1\cos \varphi + l_2 \sqrt{1-\left(\frac{l_1}{l_2}\right)^2 \sin^2 \varphi}$$

To get there:

  1. use the Pythagorean theorem for the second summand in the formula for $x_2$ instead of trigonometry
  2. replace $y_1$ with its definition

$$ \begin{align} x_2 &= x_1 + \sqrt{l_2^2 - \underset{\overbrace{l_1\sin\varphi}}{y_1^2}}\\ \end{align}$$

The relationship between the two angles can be derived from their shared opposite side in their respective triangles:

$$l_1\sin a_1 = l_2\sin a_2$$ $$ a_1 = \arcsin \left(\frac{l_2}{l_1}\sin a_2\right)$$

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  • $\begingroup$ Thank you for your answer. I would like to derive that same relationship but in terms of both angles. If you could make a suggestion as to how I might approach this, I'd be very grateful. $\endgroup$ – Emilio Botero Aug 18 '15 at 14:56
  • $\begingroup$ @EmilioBotero I added the relationship between the two angles to my answer, please take a look $\endgroup$ – Bending Unit 22 Aug 18 '15 at 15:05
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    $\begingroup$ Ah I misunderstood the orientation. I'd like to caution the calculation for $a_1$ in the answer here because it's still not clear to me what OP is using to define that angle. Assuming it's the most acute of the angles in Bender's drawing, the acute angle between $L_2$ and horizontal, the equation is correct. $\endgroup$ – Chuck♦ Aug 18 '15 at 15:39
  • $\begingroup$ @Chuck: Indeed, I consider $a_2$ to be the most acute angle (the alternate angle thereof) as indicated by "shared opposite side", which appears to be too subtle in hindsight. It appeared to me that $a_2$ is defined in a clockwise orientation (as opposed to the ccw definition of $\varphi$ in the image). I came to this conclusion by looking at the definition of $y_1$ in the question, which is given as $y_2 = y_1 - L_2 \sin(a_2)$. The minus sign indicates a different orientation of the angles. $\endgroup$ – Bending Unit 22 Aug 18 '15 at 16:46

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