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I tried to get the approximation for small angle of a simple pendulum using only $\sum \mathbf F = m\mathbf a$ and cartesian coordinates (that means only $x$'s and $y$'s, without $\theta$). After some derivatives I got the expression:

$$a_x = -\left(\frac{x}{L}\right)\left(\frac{v_x^2}{L} + g\right)$$

By energy conservation, it is easy to realize that $\frac{v_x^2}{L} << g$ for small angle, leading to the familiar harmonic oscillator equation.

Is it the only way to reach that conclusion?

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If you work in terms of the torque $N$ instead of the force, then the only approximation needed is literally the small angle approximation.

Call $\theta$ the angle that the pendulum The rotational form of Newton's Second Law, for rotational motion around the attachment point of the pendulum is $$I\ddot{\theta}=N.$$ If the pendulum is a point mass located a distance $l$ away from the center of rotation, then the moment of inertia $I$ is $ml^{2}$. Even if the moment of inertia is more general, the torque about the rotation axis is $N=-mgl\sin\theta$, where $l$ is the distance to the center of mass; the minus sign indicates that the torque (which is, more generally vector) is, when limited to its action in the plane of motion, a restoring torque that pushes the pendulum back toward $\theta=0$.

Thus the equation of motion for $\theta$ is $$I\ddot{\theta}=-mgl\sin\theta.$$ The small-$\theta$ approximation for the sine function is $\sin\theta\approx\theta$, so the approximate equation of motion of the pendulum is $$I\ddot{\theta}=-mgl\theta,$$ which is a simple harmonic oscillator with angular frequency $\omega_{0}=\sqrt{mgl/I}$; if the pendulum bob is a point mass, this becomes $\omega_{0}=\sqrt{g/l}$. Notice that no mention of energy is involved. Indeed, there is no need for energy to be conserved; we could add an additional viscous damping force acting against the pendulum's motion and get the standard equation of motion for a damped harmonic oscillator, once we approximate $\sin\theta\approx\theta$.

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  • $\begingroup$ OP doesn't want to use angles. While technically accurate, this doesn't address the question in the terms the OP specified. (That doesn't mean the question is good.) $\endgroup$ – Bill N May 9 at 1:17
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That energy balance is not the only way, but any way will probably require using some position coordinate other than $x$, in addition to whatever simplifications you make on the potential. To see this probably the best way is the Lagrangian formalism, where postulating some $y(x)$ constraint will create a Lagrangian$$ \mathcal L(x,\dot x)= \frac12 \dot x^2\left[1 + \big(y'(x)\big)^2\right]-g~y(x),$$ so the equations of motion are going to be $$ \ddot x \left[1 + \big(y'(x)\big)^2\right] = -y'(x)\big[g +\dot x^2 y''(x)\big] $$ and approximating the bracketed part on the right as $g$ is exactly the step that you are trying to avoid.

This term comes in the Lagrangian formalism from an incomplete cancellation of two components stemming from this cross term $\frac12(y')^2 \dot x^2$, the $y$-component of kinetic energy, and so while I would struggle to prove that it’s necessary to eliminate such cross terms, it’s certainly sufficient to do so. In which case we might define $$\frac{\mathrm ds}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}\sqrt{1 + \left(y'(x)\right)^2},\\ s(x) = \int_0^x\mathrm dz~ \sqrt{1 + \big(y'(z)\big)^2}, $$and so you want an arclength coordinate to phrase the problem in, and then the problem will go away.

For circles the arclength is just the angle times the radius. So you have $y=L-L\cos(s/L)$ and $x=L\sin(s/L)$ and the equations of motion become$$\ddot s = -g\frac{\mathrm dy}{\mathrm ds} = -g\sin(s/L),$$ and now just one approximation, $\sin(s/L)\approx s/L$, gives both $\ddot s\approx-(g/L) s$ and $s\approx x$, which is the approach showed in the other answer.

The approach in an earlier version of this answer first approximated the circle with a parabola, $y\approx x^2/(2L),$ to try to get a simpler result, but the arclength of a parabola is not all that simple—it can be written with a logarithm or a hyperbolic arcsine, but neither looks nice and invertible to get an $x(s)$ or $y'(s).$

If one wanted to go down this road an interesting approach might instead approximate the bottom of the circle as a hyperbolic cosine, $y=L\cosh(x/L)-L\approx x^2/(2L).$ In that case one gets $s=L \sinh(x/L)$ and $$y(s)=-L+L\sqrt{1+(s/L)^2},\\\ddot s=-\frac gL\frac{s}{\sqrt{1+(s/L)^2},$$which has somewhat of a similar feel to that answer without so much arclength pain, because hyperbolic functions have convenient arclengths. But I don't think you're going to get anything more elegant than the other answer because it’s just so direct:“here is the arclength on the circle, let's get the equation for $\ddot\theta.$

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  • $\begingroup$ The implicit assumption is also that a (small) change in the differential equation implies a (small) change in the solution. $\endgroup$ – lalala May 7 at 15:25
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    $\begingroup$ I thought that your answer was right. But you made the assumption that the tension in the rod equals the projection of $mg$ in the radial direction, as it was a static situation, isn't it?. The tension is a little greater because there is a centripetal component. $\endgroup$ – Claudio Saspinski May 7 at 19:26
  • $\begingroup$ Yes that is true, the full equations of motion on a parabola actually say,$$\ddot x= -\frac{x}{1+(x/L)^2}\left(\frac gL + \frac{\dot x^2}{L^2}\right)$$and this is the same approximation you had! Hm. Then I think what makes the angular approach work is that you get a decoupling between the kinetic energy and the potential energy in the Lagrangian? There should be an analogue for parabolas using something other than sines and cosines but I would have to look at it a while... $\endgroup$ – CR Drost May 8 at 16:21
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I realized that it is possible to get rid of the "centripetal" term $\frac{v_x^2}{L}$ using D'Alembert approach:

$\mathbf{F.\delta r} = m\mathbf{a.\delta r}$

If 2 vector are equal, so do their dot product with the same vector. So, notions of energy or work are not necessary, but comes as a consequence.

enter image description here

$\mathbf{\delta r} = \left(\frac{y}{L},\frac{x}{L}\right)\delta r $

Where $L$ is the length of the pendulum. Because the tension $T$ is normal to the allowable displacement, the dot product reduces to: $$\mathbf{F.\delta r} = (0,-mg)\mathbf.\left(\frac{y}{L},\frac{x}{L}\right)\delta r= -\frac{mgx}{L}\frac{\partial r}{\partial t}\delta t = -\frac{mgx}{L}v\delta t$$

$$m\mathbf{a.\delta r} = m\frac{\partial \mathbf v}{\partial t}.\frac{\partial \mathbf r}{\partial t}\delta t = m\frac{\partial \mathbf v}{\partial t}.\mathbf v\delta t = m\frac{1}{2}\frac{\partial v^2}{\partial t}\delta t$$

Equalizing the expressions: $$-\frac{gx}{L}v = \frac{1}{2}\frac{\partial v^2}{\partial t}$$

Because $v^2 = v_x^2 + v_y^2$, and the velocity is normal to the vector radius from the center of the circle:

$$(v_x,v_y)\mathbf.(x,y)=0 \implies v_x = \frac{-y}{x}v_y \implies v^2 = \frac{v_x^2 L^2}{y^2}$$

For small angles, $L^2 \approx y^2 $

$$ -\frac{gx}{L}v_x \approx \frac{1}{2}\frac{\partial v_x^2}{\partial t} = v_x \frac{\partial v_x}{\partial t} \implies a_x \approx -\frac{g}{L}x$$

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