91Ó°ÊÓ

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y'or the independent variable t. Let F(y) be an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$, that is,$\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity$\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$ is constant; i.e., $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$.

Change of angular momentum:

$\mathrm{m}{\mathrm{â„“}}^2\mathrm{θ}=-\mathrm{â„“}\mathrm{³¾²\mu ²õ¾\pm ²Ôθ}$…… (1)

Newton’s rotational law: The rate of change of angular momentum is equal to torque.

Referring to Problem 7: ${\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{θ}}{{\mathrm{dt}}^2}=-\mathrm{l³¾²\mu ²õ¾\pm ²Ôθ}$…… (2)

To prove: $\frac{{\left(\mathrm{θ}\text{'}\right)}^2}{2}-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{³¦´Ç²õθ}=\mathrm{constant}$

Let us take equation (2) to get,

$$\begin{gathered} {\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{θ}}{{\mathrm{dt}}^2}=-\mathrm{l³¾²\mu ²õ¾\pm ²Ôθ} \\ \mathrm{θ}=-\frac{\mathrm{l³¾²\mu ²õ¾\pm ²Ôθ}}{{\mathrm{ml}}^2} \\ =-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{²õ¾\pm ²Ôθ} \end{gathered}$$

Then,

$$\begin{gathered} \mathrm{f}\left(\mathrm{θ}\right)=\frac{\mathrm{d}}{\mathrm{»åθ}}\mathrm{F}\left(\mathrm{θ}\right) \\ \mathrm{F}\left(\mathrm{θ}\right)=∫\mathrm{f}\left(\mathrm{θ}\right)\mathrm{»åθ}=\frac{\mathrm{g}}{\mathrm{l}}\mathrm{³¦´Ç²õθ} \end{gathered}$$

Substitute the values in E(t).

$$\begin{gathered} \mathrm{E}\left(\mathrm{t}\right)=\frac{1}{2}\mathrm{θ}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{θ}\left(\mathrm{t}\right)\right) \\ =\frac{1}{2}\mathrm{θ}\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{³¦´Ç²õθ} \end{gathered}$$

Hence proved.