I have that the energy $E$ in a system is

$$E = \frac{1}{2}L^2\dot{\theta}^2(m_1+4m_2) - Lg(m_1+2m_2)cos(\theta) + c$$

where $c\in\mathbb{R}$.

I have used the conservation of energy over time to show that

$$L^2(m_1+4m_2)\dot{\theta}\ddot{\theta} = -Lg(m_1+2m_2)sin(\theta)\dot{\theta} \implies L(m_1+4m_2)\ddot{\theta} = -g(m_1+2m_2)sin(\theta)$$

but I don't understand where to go from here.


closed as off-topic by Bob D, G. Smith, Gert, ZeroTheHero, tpg2114 May 27 at 3:32

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  • 1
    $\begingroup$ On a style note, when typesetting trig functions use \sin and \cos. The result is visibly better $$ \sin(\theta) \;{\rm vs.}\; sin(\theta) $$ $$ \cos(\theta) \;{\rm vs.}\; cos(\theta) $$ $\endgroup$ – ja72 May 27 at 0:08

Typically one would use a small angle approximation, letting $\sin\theta\approx\theta$. The resulting differential equation is that of a simple harmonic oscillator. Otherwise you will need to deal with elliptic functions.


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