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With air resistance

For the sake of simplicity, we will only consider the air resistance experienced by the pendulum bob itself, not by the rod. The air resistance force vector for the pendulum bob would be:

FD=−ρCDA∣v∣v2.\begin{aligned} \textbf{F}_D &= -\dfrac{\rho C_D A|v|\textbf{v}}{2}. \end{aligned}

Where v\textbf{v} is the velocity vector of the bob. It is equal to (x˙,y˙)(\dot{x}, \dot{y}). Hence:

FD=−ρCDAl∣θ˙∣2(x˙,y˙)=−ρCDAl∣θ˙∣2(−lθ˙sin⁡θ,lθ˙cos⁡θ)=ρCDAl2∣θ˙∣θ˙2(sin⁡θ,−cos⁡θ).\begin{aligned} \textbf{F}_D &= -\dfrac{\rho C_D Al|\dot{\theta}|}{2}(\dot{x}, \dot{y}) \\ &= -\dfrac{\rho C_D Al|\dot{\theta}|}{2} (-l\dot{\theta}\sin{\theta}, l\dot{\theta}\cos{\theta}) \\ &= \dfrac{\rho C_D Al^2|\dot{\theta}|\dot{\theta}}{2} (\sin{\theta}, -\cos{\theta}). \end{aligned}

To find the generalized force term, we need to find ∂r∂θ\dfrac{\partial \textbf{r}}{\partial \theta}:

∂r∂θ=(∂x∂θ,∂y∂θ)=(−lsin⁡θ,lcos⁡θ).\begin{aligned} \dfrac{\partial \textbf{r}}{\partial \theta} &= \left(\dfrac{\partial x}{\partial \theta}, \dfrac{\partial y}{\partial \theta}\right) \\ &= (-l\sin{\theta}, l\cos{\theta}). \end{aligned}

Hence:

Qθ=FD⋅∂r∂θ=ρCDAl2∣θ˙∣θ˙2(sin⁡θ,−cos⁡θ)⋅(−lsin⁡θ,lcos⁡θ)=ρCDAl2∣θ˙∣θ˙2(−lsin⁡2θ−lcos⁡2θ)=−ρCDAl3∣θ˙∣θ˙2.\begin{aligned} Q_{\theta} &= \textbf{F}_D \cdot \dfrac{\partial \textbf{r}}{\partial \theta} \\ &= \dfrac{\rho C_D Al^2|\dot{\theta}|\dot{\theta}}{2} (\sin{\theta}, -\cos{\theta}) \cdot (-l\sin{\theta}, l\cos{\theta}) \\ &= \dfrac{\rho C_D Al^2|\dot{\theta}|\dot{\theta}}{2}(-l\sin^2{\theta}-l\cos^2{\theta}) \\ &= -\dfrac{\rho C_D Al^3|\dot{\theta}|\dot{\theta}}{2}. \end{aligned}

Hence Equations (??) become:

ml2θ¨+mglcos⁡θ=−ρCDAl3∣θ˙∣θ˙2.\begin{aligned} ml^2 \ddot{\theta} + mgl\cos{\theta} &= -\dfrac{\rho C_D Al^3|\dot{\theta}|\dot{\theta}}{2}. \end{aligned}

Dividing by ml2ml^2 yields:

θ¨+glcos⁡θ=−ρCDAl∣θ˙∣θ˙2m.\begin{aligned} \ddot{\theta} + \dfrac{g}{l}\cos{\theta} &= -\dfrac{\rho C_D Al|\dot{\theta}|\dot{\theta}}{2m}. \end{aligned}

Subtracting glcos⁡θ\dfrac{g}{l}\cos{\theta} from both sides yields:

θ¨=−glcos⁡θ−ρCDAl∣θ˙∣θ˙2m.\begin{aligned} \ddot{\theta} &= -\dfrac{g}{l}\cos{\theta} -\dfrac{\rho C_D Al|\dot{\theta}|\dot{\theta}}{2m}. \end{aligned}

This equation cannot be analytically solved, so numerical solutions are necessary. I numerically solve an even more complex version of the single pendulum equation here.

If you would like to see my derivation of the double rigid pendulum equations, it is here.

© Brenton Horne. I do not claim to be infallible, so feel free to open issues or pull requests on GitHub if you notice flaws or things that can be improved.
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Last modified: October 31, 2026.