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I'm solving an exercise about small oscillations and I have a doubt about coordinates that I have to use.

This is the text of the exercise: "A bar has mass M and lenght l. Its extremity A is hooked to a coil (with lenght at rest $l_0$), its extremity B is hooked to the point O that is the origin of axes. "

I have considered three coordinates: $x$, $y$ (that are the coords of the extremity A on the x-axes and y-axes) and $\theta$ that is the angle that the bar forms with a parallel to the y-axes.

I have to find the points of equilibrium.

I have written the coordinates of the center of mass of the bar as: $M=(x+l/2 \sin \theta, -y-l/2 \cos \theta)$ and the potential energy as $ V=\frac{1}{2}k(x^2+y^2)-Mg(y+\frac{l}{2}\cos \theta)$

Then I have posed $gradV=0$ and I have obtained:

$y=\frac{Mg}{k}$

$x=0$

$\theta=0, \pi$

My doubt is about the result of y.. I was waiting for a negative value.. could you "clarify" my ideas and tell me where I'm making a mistake?

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  • $\begingroup$ What are $x$ and $y$ here, exactly? The coordinates of extremity B? $\endgroup$
    – Izzhov
    Commented May 28, 2013 at 21:20
  • $\begingroup$ @Izzhov no, the coordinates of extremity A $\endgroup$
    – sunrise
    Commented May 28, 2013 at 21:21

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The resolution is that your first equation should have $+y$ where $-y$ is, since you're measuring from extremity A to the center of mass. Not sure where you're getting $-y$ from.

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  • $\begingroup$ I'm not sure to have understood... y is the coordinates on y-axes of the extremity A.. but I have considered that it is negative, because it is "under" O.. $\endgroup$
    – sunrise
    Commented May 28, 2013 at 21:28
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    $\begingroup$ Ah, but see, that would just make $+y$ a negative number. So you would still use $+y$. E.g. if $y$ is 3 units below the x-axis then you'll have $+y = -3$. $\endgroup$
    – Izzhov
    Commented May 28, 2013 at 21:28
  • $\begingroup$ just another (general) doubt: if the text says "for the point P use the q coordinate, where q>0 if P is on the right of O, and q<0 if P is on the left of O"... when I write the coordinate of P, I have to written "+q", isn't it? $\endgroup$
    – sunrise
    Commented May 28, 2013 at 21:41
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    $\begingroup$ Yes, that is true in general. $\endgroup$
    – Izzhov
    Commented May 28, 2013 at 21:43

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