Riffing on the question in Simple Pendulum Why Generalized Coordinate Always Angle? , I'm trying to write down Newton's law for a simple pendulum in Cartesian coordinates. (I'm doing this as an exercise for myself. I know that using the displacement angle $\theta$ is traditional and likely much easier. I just want to see what it looks like, in full glory, done the naive way.)

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To make things concrete, let the origin be at the pivot point, and define $x$ to be positive going to the right and $y$ to be positive going down. Then $(x,y) = (r\sin\theta, r\cos\theta)$.

The force on the bob is $W+T$, where $W =Mg \widehat{y}$ is the weight of the bob (of mass $M$) due to the force of gravity (positive b/c $y$ is directed downwards), and $T = - Mg\cos\theta \widehat{r}$ is the (radially inward) tension force from the string.

To write down the force in $(x,y)$ coordinates, we need to convert the tension $T$ into that coordinate system. When I do that I get \begin{align*} T &= -Mg \cos\theta \widehat{r}\\ &= -Mg \cos\theta ( \sin\theta \widehat{x} + \cos\theta \widehat{y})\\ &= -Mg ( \cos\theta\sin\theta \widehat{x} + \cos^2\theta \widehat{y}). \end{align*} Using that $\sin\theta = \frac{x}{\sqrt{x^2+y^2}}$ and $\cos\theta = \frac{y}{\sqrt{x^2+y^2}}$, the above gives \begin{eqnarray} T = -Mg \left( \frac{xy}{x^2+y^2} \widehat{x} + \frac{y^2}{x^2+y^2} \widehat{y} \right). \end{eqnarray} But $x^2+y^2 = L^2$ in this problem, so \begin{eqnarray} T = -\frac{Mg}{L^2} \left( xy \widehat{x} + y^2 \widehat{y} \right). \end{eqnarray} and the total force (tension plus weight due to gravity) is \begin{eqnarray} F = \frac{Mg}{L^2} \left( -xy \widehat{x} + (L^2 - y^2) \widehat{y} \right). \end{eqnarray}

I'm not sure how to check if the above is correct. The resulting 2nd order ODEs are \begin{align*} M\ddot x &= -\frac{Mg}{L^2} xy \\ M\ddot y &= -\frac{Mg}{L^2} (L^2 - y^2), \end{align*}

which don't quite feel right. For example, linearizing the above about the bottom of the pendulum $(x,y) = (0, L)$ gives the 1st order linear equation \begin{eqnarray*} \left[ \begin{matrix} \ddot x \\ \ddot y\end{matrix} \right] &= \left[ \begin{matrix} -\frac{g}{L} & 0 \\ 0 & -2\frac{g}{L} \end{matrix} \right] \left[\begin{matrix} x \\ y \end{matrix}\right]. \end{eqnarray*} I'm not sure that's obviously wrong, but I'm not sure how to explain the factor of 2, for example.

So...help? What, if anything, is missing, how could I have checked the above equations, and how do they relate to the treatment of the simple pendulum in angular coordinates?

EDIT: Fixed a typo.

  • 1
    $\begingroup$ Final equation is missing double dots on the left. $\endgroup$ – G. Smith Jul 4 at 3:13
  • $\begingroup$ The tension is not equal to the radial component of the weight. It's an unknown force, whose value you will get from the equations of motion. $\endgroup$ – Javier Jul 4 at 18:34

I don't think you're equations for the 2nd order ODEs are correct. It helps to look at limiting cases:

Edit: Considering $\ddot\theta\propto\sin\theta$, this limiting case doesn't invalidate your solution.

First, let's look at the case where $x\to 0$, setting $\theta = 0$, so we're at the bottom of the pendulum swing with $y=L$. This yields the expression

$$\ddot{\vec r} = \left[ \begin{matrix} \ddot{x} \\ \ddot{y}\end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0\end{matrix} \right]$$ which does not seem correct to me.

But I still think the tension is incorrect

I believe the issue might start with your expression for the tension; I disagree with $T=mg\cos{\theta}$. Assuming circular motion, the key fact to remember is that we must have $\sum{Forces} = F_{centripetal}$ from Newton's second law, as we only have the centripetal acceleration. The non-uniform motion of the pendulum only serves to change the centripetal force's magnitude depending on where the pendulum is in its swing. For more details, this answer may be a good reference.

Continuing, we can write more explicitly

$$T + F_g = \frac{mv^2}{L} \implies T = \frac{m}{L}(\dot{x}^2 + \dot{y}^2) +mg\cos\theta$$

The tension is indeed in the radial direction, so you can get the full vector using $\vec T = T\hat r$ and the components using the trigonometric functions. Both terms are positive because the centripetal force term is pointed toward the origin, and the second term compensates for the outward facing gravitational force $\vec F_g = -mg\cos\theta\hat r$.

I'm not sure how to go about including this into the full solution, but it might lead you on the right track.

  • $\begingroup$ I believe the acceleration is correct at the bottom. The typical angle based acceleration equation gives $\ddot\theta = \frac{g}{L}\sin\theta$, so the acceleration is 0 when $\theta=0$. $\endgroup$ – cjordan1 Jul 4 at 11:45
  • $\begingroup$ The motion of a pendulum follows the arc of a circle, but is not uniform circular motion. So I'm not sure $\sum Forces = F_{centripetal}$ applies. $\endgroup$ – cjordan1 Jul 4 at 11:48
  • $\begingroup$ I agree with your first comment because your reference to the standard differential equation makes sense. However, I disagree with your concerns about uniform circular motion. I'll edit my original answer accordingly and maybe we'll move toward agreement. $\endgroup$ – spanishinquisitor Jul 4 at 18:08
  • $\begingroup$ This answer may be helpful for the second point. $\endgroup$ – spanishinquisitor Jul 4 at 18:25

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