Polynomial Wire Dynamics

Here's my approach which avoids tedious differentiation:

Since the beam is forced to move along the wire, we have: $\begin{aligned} \dot{\vec{v}} &= \dot{v}\hat{T} + v\dot{\hat{T}} \\ &= \dot{v}\hat{T} + v\frac{d\hat{T}}{ds}\frac{ds}{dt} \\ &= \dot{v}\hat{T} + v^{2}\frac{d\hat{T}}{ds}\end{aligned}$ Here, $v$ is speed, $\dot{v}$ is the tangential acceleration and $\frac{d\hat{T}}{ds}$ is the curvature vector, which is normal to tangent line. If we denote $\theta$ to be the angle between the tangent line and x-axis, we have: $\begin{aligned} \dot{v} &= g\sin(\theta) = g\frac{\tan{\theta}}{\sqrt{1+\tan^{2}{\theta}}} = g\frac{y'}{\sqrt{1+y'^{2}}} \\ \frac{d\hat{T}}{ds} &= \kappa\hat{N} = \frac{y''}{(1+y'^{2})^{\frac{3}{2}}}\hat{N}\end{aligned}$ It follows that the squared magnitude of acceleration is, by Pythagorean theorem: $\begin{aligned} \left \| \vec{a} \right \|^{2} &= g^{2}\frac{y'^{2}}{1+y'^{2}} + v^{4}\frac{y''^{2}}{(1+y'^{2})^{3}}\end{aligned}$ We can write this whole expression as a function of $x$ , which then becomes: $\begin{aligned} \left \| \vec{a} \right \|^{2} &= g^{2}\frac{(4x^{3}+9x^{2})^{2}}{1+(4x^{3}+9x^{2})^{2}} + 4g^{2}(4-x^{4}-3x^{3})^{2}\frac{(12x^{2}+18x)^{2}}{(1+(4x^{3}+9x^{2})^{2})^{3}} \\ &= g^{2}x^{2}\left[\frac{(4x+9)^{2}}{1+x^{4}(4x+9)^{2}}+144(4-x^{4}-3x^{3})^{2}\frac{(2x+3)^{2}}{(1+x^{4}(4x+9)^{2})^{3}} \right ]\end{aligned}$ I used the same arguments as you in the finish, but I get a slightly different result - 195.85619 - which is probably due to my imperfect method of numerical integration.