$s$ is the length of the curve from $O$ to $P$. The tangential force at $P$ is $F(s)$. Does $$m\frac{d^2s}{dt^2}=F(s)$$ hold?

Question

I am reading "An Introduction to Mechanics" by Daniel Kleppner and Robert Kolenkow.

The sketch shows a simple pendulum of length $l$, with mass $M$, and corresponding weight $W=Mg$.
The mass moves in a circular arc in a vertical plane. Denoting the angle from the vertical by $\theta$, we see that the velocity is $l\frac{d\theta}{dt}$ and the acceleration is $l\frac{d^2\theta}{dt^2}$. The tangential force is $-W\sin\theta$. Thus the equation of motion is $$Ml\frac{d^2\theta}{dt^2}=-Mg\sin\theta$$ or $$\frac{d^2\theta}{dt^2}+\frac{g}{l}\sin\theta=0.$$

I don't understand the above solution.
The mass doesn't move in a straight line. (The mass moves in a circular arc.)
But the authors derived the differential equation like the above.

Let $s$ be the length of the curve from the vertical to the mass.
Then $s=l\theta$.
Then $$M\frac{d^2s}{dt^2}=Ml\frac{d^2\theta}{dt^2}=-Mg\sin\theta=-Mg\sin\left(\frac{s}{l}\right)$$ holds.

My question is the following:
Please see the above figure.
The mass $m$ is at the point $P$.
Suppose $s$ is the length of the curve from the point $O$ to the point $P$.
Suppose the tangential force at the point $P$ is $F(s)$.
Does $$m\frac{d^2s}{dt^2}=F(s)$$ hold?

Answer 1

Your force should explicitly be a function a $(\theta)$ only because it is independent of length of pendulum. So $$m\frac{d^2s}{dt^2}=F(\theta)$$ where s=$l\theta$ and $F=-mgsin(\theta)$

Answer 2

If a mass is moving on a 2D curve , the mass position (inertial system) is:

\begin{align*} &\vec{P}=\begin{bmatrix} x(~q(\tau)~) \\ y(~q(\tau)~) \\ \end{bmatrix} \end{align*} where q is the generalized coordinate and $~\tau~$ is the time

from here you obtain the velocity $~\vec{v}~$

\begin{align*} &\vec{v}=\begin{bmatrix} x'(q) \\ y'(q) \\ \end{bmatrix}\,\dot{q} \end{align*} where $~'=\frac{\partial (..)}{\partial q}~$

now the equation of motion , Newton low ,is: \begin{align*} &m\,\frac{d}{d\tau}\left(\vec v\cdot\,\hat{{t}}\right)=\vec{F}\cdot\,\hat{{t}}\tag 1 \end{align*} where $~\vec{F}~$ is the external force components and $~\hat{t}~$ is the tangent vector at point $~q(\tau)~$ \begin{align*} &\hat{t}=\frac{1}{\sqrt{x'^2+y'^2}}\begin{bmatrix} x'(q) \\ y'(q) \\ \end{bmatrix} \end{align*}

Example: Pendulum

the generalize coordinate is $~s~$ (not $~\theta~$) thus

\begin{align*} &\vec{P}=l\,\begin{bmatrix} \sin\left(\frac{s}{l}\right) \\ -\cos\left(\frac{s}{l}\right) \\ \end{bmatrix} \quad, \vec{v}= \begin{bmatrix} \cos\left(\frac{s}{l}\right) \\ \sin\left(\frac{s}{l}\right) \\ \end{bmatrix}\,\dot{s} \quad, \hat t=\begin{bmatrix} \cos\left(\frac{s}{l}\right) \\ \sin\left(\frac{s}{l}\right) \end{bmatrix} \quad\text{and}\\ &\vec{F}=\begin{bmatrix} 0 \\ -m\,g \\ \end{bmatrix} \end{align*}

with equation (1)

\begin{align*} & \ddot{s}=-m\,g\,\sin\left(\frac{s}{l}\right) \end{align*}

Example: $~\sin~$

the generalized coordinate is $~x~$ (it doesn't make sense to choose the line element s to be the generalized coordinate ) thus

\begin{align*} &\vec{P}=l\,\begin{bmatrix} x\\ \sin(x)\\ \end{bmatrix} \quad, \vec{v}= \begin{bmatrix} 1\\ \cos(x)\\ \end{bmatrix}\,\dot{x} \quad, \hat t=\frac{1}{\sqrt{1+\cos^2(x)}}\begin{bmatrix} 1\\ \cos(x)\\ \end{bmatrix} \quad\text{and}\\ &\vec{F}=\begin{bmatrix} 0 \\ -m\,g \\ \end{bmatrix} \end{align*}

the equation of motion is

\begin{align*} &\frac{d}{d\tau}\left(\sqrt{1+\cos^2(x)}\,\dot{x}\right)=-m\,g\frac{\cos(x)}{\sqrt{1+\cos^2(x)}} \end{align*}

this is not what you obtained

Conclusion

only if the magnitude of the tangent vector equal one $~|\vec t|=\sqrt{x'^2(s)+y'^2(s)}=1~$ you obtain $$\ddot s=x'\,F_x+y'\,F_y$$