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# Let m, g, and L be positive constants. (a) Show that $V(\theta, \omega)=\frac{1}{2} m L^{2} \omega^{2}+m g L(1-\cos \theta)$ is positive definite in an open neighborhood containing $(\theta, \omega)=(0,0)$. (b) Show that $\dot{V}(\theta, \omega)$ is negative semidefinite on solution trajectories of$\begin{array}{l} \dot{\theta}=\omega \\ \dot{\omega}=-\frac{\mu}{m} \omega-\frac{g}{L} \sin \theta \end{array}$on an open neighborhood containing $(\theta, \omega)=(0,0)$.

Solution

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(a) We have the following function

\begin{align*} V(\theta, \omega)&=\dfrac{1}{2}mL^2\omega ^2+mgL(1-\cos(\theta))\\ \end{align*}

We know that m, g and L are positive constants. and cos ($\theta$) is varying between -1 to 1.

\begin{align*} -1\le \cos(\theta)le 1\\ 1\ge -\cos(\theta)ge -1\tag{\textcolor{#4257b2}{inequality sign changes after multiplying with -1}}\\ 1+1\ge 1-\cos(\theta)ge -1\\ 2\ge -\cos(\theta)ge 0 \end{align*}

Now we will check the function:

\begin{align*} V(\theta, \omega)&=\dfrac{1}{2}mL^2\omega ^2+mgL(1-\cos(\theta))\\ &=\dfrac{1}{2}\underbrace{mL^2\omega ^2}_{\text{ Positive}}+\underbrace{mgL}_{\text{ Positive}}\underbrace{(1-\cos(\theta))}_{\text{ Positive}}\\ \end{align*}

Hence, the function is strictly positive for all $(\theta,\omega)\in \bf{R}^2$ except (0,0). So the function $V(\theta, \omega)=\dfrac{1}{2}mL^2\omega ^2+mgL(1-\cos(\theta))$ is positive definite.

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