Question

Verify equation (18.3) i.e.$\mathbf{l}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\theta }}{{\mathbf{dl}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{d\theta }}{\mathbf{dl}}\mathbf{+}\frac{\mathbf{g}}{{\mathbf{V}}^{\mathbf{2}}}\mathbf{\theta }\mathbf{=}\mathbf{0}$

Short answer

The resultant answer $\mathrm{l}\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}+2\frac{\mathrm{d\theta }}{\mathrm{dl}}+\frac{\mathrm{g}}{{\mathrm{V}}^2}\mathrm{\theta }=0$is verified.

Step-by-step solution

Concept of Equation of motion:

The equation of motion:

$\frac{\mathbf{d}}{\mathbf{dt}}\left({\mathrm{ml}}^2\theta \right)\mathbf{=}\mathbf{mglsin\theta }$

Consider jn(x)=∑r=0∞(-1)ΓΓ(r+1)Γ(r+1+n)(x2)2Γ+n and simplify it:

Considering the equation of motion:

$\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{ml}}^2\mathrm{\theta }\right)=\mathrm{mgl}\mathrm{sin\theta }$

For the small value of $\theta$:

$\sin =\theta$

Hence,

${\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}+2\mathrm{lm}\frac{\mathrm{dl}}{\mathrm{dt}}\times \frac{\mathrm{d\theta }}{\mathrm{dt}}+\mathrm{mgl\theta }=0$ ….. (1)

Consider the following formula.

$\mathrm{l}={\mathrm{l}}_0+\mathrm{vt}$

Take a derivation with respect to t .

role="math" localid="1659272604614" $$\begin{gathered} \frac{\mathrm{dl}}{\mathrm{dt}}=\mathrm{v} \\ \mathrm{dl}=\mathrm{vdt} \end{gathered}$$

Then,

$$\begin{gathered} \frac{\mathrm{d\theta }}{\mathrm{dt}}=\frac{\mathrm{d\theta }}{\mathrm{dl}}\times \frac{\mathrm{dl}}{\mathrm{dt}} \\ =\mathrm{v}\frac{\mathrm{d\theta }}{\mathrm{dt}} \end{gathered}$$.

And,

$\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}={\mathrm{V}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}$.

Put the values into equation (1):

$$\begin{gathered} {\mathrm{ml}}^2{\mathrm{v}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}+2\mathrm{lmv}\times \mathrm{v}\frac{\mathrm{d\theta }}{\mathrm{dt}}+\mathrm{mgl\theta }\frac{{\mathrm{v}}^2}{{\mathrm{v}}^2}=0 \\ {\mathrm{mlv}}^2\left[\mathrm{l}\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}+2\frac{\mathrm{d\theta }}{\mathrm{dt}}+\frac{\mathrm{g}}{{\mathrm{v}}^2}\mathrm{\theta }\right]=0 \\ \mathrm{l}\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}+2\frac{\mathrm{d\theta }}{\mathrm{dt}}+\frac{\mathrm{g}}{{\mathrm{v}}^2}\mathrm{\theta }=0 \end{gathered}$$

Hence, $\mathrm{l}\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dl}}^2}+2\frac{\mathrm{d\theta }}{\mathrm{dl}}+\frac{\mathrm{g}}{{\mathrm{V}}^2}\mathrm{\theta }=0$ is verified.