Hello I have a problem how to write down equations for the pendulum correctly. Say I would use Cartesian coordinates $x, y$ representing the position of the mass. Then the velocities would be usually written as $\dot{x}, \dot{y}$, but here the problem appears.

Of course, the pendulum may move in two different directions at every position. So I think there should be two signs $\pm \dot{x}, \pm \dot{y}$, but $x, y$ have only a single sign. If polar coordinates are used this is realized with the angular velocity that is a pseudo quantity. Is it also possible using Cartesian coordinates to incorporate this somehow?

  • 1
    $\begingroup$ Why do $x$ and $y$ only have a single sign? $\endgroup$
    – noah
    Commented Apr 23, 2019 at 22:16
  • $\begingroup$ The velocity at any $(x(t),y(t))$ point is just $(\dot{x}(t), \dot{y}(t))$, simple as that. Any positive or negative signs can contained in $\dot{x}(t)/\dot{y}(t)$ without any need to assign an explicited sign to them. $\endgroup$
    – Triatticus
    Commented Apr 24, 2019 at 0:06
  • $\begingroup$ Oh you are right, I mean a single sign for each of them. Not necessarily the same of course. $\endgroup$
    – Q.stion
    Commented Apr 24, 2019 at 10:39
  • $\begingroup$ I don't understand what your problem is. Please give a concrete example where the math gives an incorrect answer unless the signs are flipped manually. $\endgroup$ Commented Apr 24, 2019 at 12:52

1 Answer 1


I think you are overcomplicating the problem. If the mass is located at

$$ \begin{matrix} x = \ell \sin \theta \\ y = -\ell \cos \theta \end{matrix} $$

relative to the pivot point then, the velocity kinematics are

$$ \begin{matrix} \dot x = \ell \dot{\theta} \cos \theta \\ \dot{y} = \ell \dot{\theta} \sin \theta \end{matrix} $$

The values of $\dot{x}$ and $\dot{y}$ may be positive meaning motion to the right and up or negative menaing motion left or down. The same sign convention applies for the coordinates $(x,y)$ as for their derivatives $(\dot{x}, \dot{y})$.

  • $\begingroup$ Okay, but I don't see it. You are right, but the problem is still there. Say for example in your equations that we are in the fourth quadrant (x > 0, y < 0). Then $x = +x$, but $\dot{x} = \pm \dot{x}$ (sign is in $\dot{\theta}$). Now, if you do not have $\dot{\theta}$ because you use Cartesian coordinates, then I would have $x = +x$, but $\frac{d}{dt} x = \pm \dot{x}$. But this is strange, since taking a derivative usually is not adding a second sign. $\endgroup$
    – Q.stion
    Commented Apr 24, 2019 at 10:45
  • $\begingroup$ You know it is ok to have $x>0$ and $\dot{x}<0$. The mechanics of the problem are such that the velocity vector is always perpendicular to the position vector. So if the position is on the 4th quadrant, then the velocity will point towards the 1st quadrant. Go outside and observe a swing, and visualize the velocty vector. You will see how the math works out. $\endgroup$ Commented Apr 24, 2019 at 12:50

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