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Suppose I have two mass rigid pendulum, both of whose masses are equidistant from the pivot point at P. All three points lie on a circle of diamater D and subtend an angle $\alpha$ at the pivot. let the distance of each mass from the pivot point be $\ell$

My questions is:

1. why is it true that at equilibirium, the position of the masses (below the point P) is $$l \cos \left(\frac{1}{2} \alpha\right)=l^{2} / D$$ I understand the leftside of the equation but do not understand why is it equal to right side, in other words why $$ \ cos \left(\frac{1}{2} \alpha\right)=l / D$$

2. why the gravitational potential energy of the system, after being displaced over a small angle $\theta$ is $$U \approx M g \frac{l^{2}}{D} \frac{\theta^{2}}{2}$$ I am having hard time to understand the math behind those two questions

  • $\begingroup$ What do you mean by equilibrium? Are these masses moving in a circle? Are there forces other than tension and gravity? Exactly what is rigid? $\endgroup$ – R.W. Bird Sep 10 '20 at 17:05
  • $\begingroup$ The masses lies on the circle, no other force exept gravity and tension $\endgroup$ – Sagigever Sep 10 '20 at 18:12

Part 1 is simply a consequence of the geometry of the scenario, and has nothing to do with its dynamics. If you draw a line from one mass (at, say, $A$) to the centre of the circle at $O$ then $OAP$ is an isosceles triangle with side $\frac D 2, \frac D 2, l$ and acute angle $\frac \alpha 2$. It follows that

$\displaystyle \cos \left( \frac \alpha 2 \right) = \frac l D$

For part 2, if we assume both masses move through a small angle $\theta$ then the change in PE is

$\displaystyle Mgl \left( \cos \left( \frac \alpha 2 \right) - \cos \left( \frac \alpha 2 + \theta \right) \right) \approx Mgl \cos \left( \frac \alpha 2 \right) \frac {\theta^2}{2} $

  • $\begingroup$ I'm srry mate but I can't understand what you mean by explaining part 1, I tried to draw what you mentioned but AP should be equal to $\ell$ from the given and also I still not understand how the cosine of alpha is $\frac{l}{D}$ if AP is the hyper. About part 2, I still don't understand this approximation $\displaystyle Mgl \left( \cos \left( \frac \alpha 2 \right) - \cos \left( \frac \alpha 2 + \theta \right) \right) \approx Mgl \cos \left( \frac \alpha 2 \right) \frac {\theta^2}{2}$ what the math tool you used? tylor expansion? $\endgroup$ – Sagigever Sep 10 '20 at 14:27
  • $\begingroup$ @Sagigever I have corrected my answer. The short sides of the triangle are length $\frac D 2$, the long side $AP$ is length $l$, and the cosine of $\frac \alpha 2$ is half the long side divided by the short side, so $\frac l D$. For part 2, yes, use Taylor expansion of $\cos \theta$ since $\theta$ is small. $\endgroup$ – gandalf61 Sep 10 '20 at 15:26
  • 1
    $\begingroup$ I do not understand how I should get this approximation in part 2, I know that in small angles I can write $cos(\theta) \approx 1 - \frac{\theta^2}{2}$ but how do I use it here? should I use a trigonometric identities? If you can show me the method I will be really gratitude $\endgroup$ – Sagigever Sep 10 '20 at 15:51
  • $\begingroup$ @Sagigever I wrote you the answer for this $\endgroup$ – Eli Sep 14 '20 at 6:31

I have this answer for you:

enter image description here

Point I:

I use the $\pi/2$ angle those:


Point II:

The total Potential Energy is:

$$U=-M\,g\,l\cos(\varphi+\alpha/2)\tag 1$$

obtain the Taylor series for $\varphi\mapsto \Delta\varphi$

$$U\approx M\,g\,l\,\left(-\cos \left( 1/2\,\alpha \right) +\sin \left( 1/2\,\alpha \right) \Delta \varphi +1/2\,\cos \left( 1/2\,\alpha \right) {\Delta \varphi } ^{2} \right) $$

with $\alpha=2\,\arccos \left( {\frac {l}{D}} \right)$ you obtain:

$$U\approx M\,g\,l\,\left(\underbrace{-{\frac {l}{D}}+{\frac {\sqrt {{D}^{2}-{l}^{2}}\Delta \varphi }{D}}}_{U_\varphi}+\frac 12\,{\frac {l{\Delta \varphi }^{2}}{D}}\right) $$

because the force $F_\varphi=\frac{\partial U_\varphi}{\partial \Delta \varphi}$ is constant we can neglecting this part.

  • $\begingroup$ Thank you for your detailed solution, can you explain why the fatct that $F_\varphi=\frac{\partial U_\varphi}{\partial \Delta \varphi}$ is constant allow us to neglect this part? $\endgroup$ – Sagigever Sep 14 '20 at 16:15
  • $\begingroup$ This is because we are looking for a solution where the pendulum is in static equilibrium state. $\endgroup$ – Eli Sep 14 '20 at 18:56
  • $\begingroup$ I am still not understanding your argue, in equilibiruim $F_{\phi}= 0$ if i understand it correctly $\endgroup$ – Sagigever Sep 14 '20 at 21:50
  • $\begingroup$ If you generate the equation of motion you see that $\phi(0)$ Is the acceleration not zero ,due to a constant force,you don’t have static equilibrium. $\endgroup$ – Eli Sep 15 '20 at 6:24

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