Question

Use the energy integral lemma to show that pendulum motion obeys$\frac{{\left(\mathrm{\theta }\text{'}\right)}^2}{2}-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{cos\theta }=\mathrm{constant}$

Short answer

Therefore, the given statement is true. The pendulum motion obeys $\frac{{\left(\mathrm{\theta }\text{'}\right)}^2}{2}-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{cos\theta }=\mathrm{constant}$.

Step-by-step solution

General form 

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:

$\mathrm{m}{\ell }^2\mathrm{\theta }=-\ell \mathrm{mgsin\theta }$…… (1)

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

Prove the given equation

Referring to Problem 7: ${\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\mathrm{lmgsin\theta }$…… (2)

To prove: $\frac{{\left(\mathrm{\theta }\text{'}\right)}^2}{2}-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{cos\theta }=\mathrm{constant}$

Let us take equation (2) to get,

$$\begin{gathered} {\mathrm{ml}}^2\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\mathrm{lmgsin\theta } \\ \mathrm{\theta }=-\frac{\mathrm{lmgsin\theta }}{{\mathrm{ml}}^2} \\ =-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{sin\theta } \end{gathered}$$

So, $\mathrm{f}\left(\mathrm{\theta }\right)=-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{sin\theta }$.

Then,

$$\begin{gathered} \mathrm{f}\left(\mathrm{\theta }\right)=\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{F}\left(\mathrm{\theta }\right) \\ \mathrm{F}\left(\mathrm{\theta }\right)=\int \mathrm{f}\left(\mathrm{\theta }\right)\mathrm{d\theta }=\frac{\mathrm{g}}{\mathrm{l}}\mathrm{cos\theta } \end{gathered}$$

Substitute the values in E(t).

$$\begin{gathered} \mathrm{E}\left(\mathrm{t}\right)=\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{\theta }\left(\mathrm{t}\right)\right) \\ =\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\mathrm{l}}\mathrm{cos\theta } \end{gathered}$$

Hence proved.