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"A grandfather clock has a simple pendulum of length L with a bob of mass m. The effect of air resistance is to produce a force on the bob of magnitude 2cm times its speed, where c > 0 is a positive constant.

(i) Show that, provided the angular displacement θ of the pendulum from the downward vertical is small, θ approximately satisfies the equation,

\[

\ddot{\theta}+2c\dot{\theta}+\frac{g}{L}\theta=0

\]

*[Hint: Use polar coordinates and Taylor’s theorem as we did in lectures for the case without air resistance.]*

(ii) Suppose now that c = $\sqrt{g/L}$. Show that if $\theta(0) = a$ and $\dot{\theta}(0) = 0$, then

$\theta(t) = a(1 + ct)e^{−ct}$

Will the pendulum ever return to the vertical position in this case? Explain

your answer"

so far I have:

The rod has fixed length r=L so

$\vec{r}=L\vec{e}_r$

as r=L is a constant, $\dot{r}=\ddot{r}=0$

so acceleration is

$\ddot{r}=(\dot{r}-r\dot{\theta}^2)e_r+(2\dot{r}\dot{\theta}+r\ddot{\theta})e_\theta$

$=-L(\dot{\theta}^2e_r+L\ddot{\theta}e_\theta$

so the total force is

$(mgcos\theta-T)e_r-(mgsin\theta+2cm\dot{x})e_\theta$

I've completed a very similar question without air resistance without too much problems but not sure if I've added air resistance in right at all. Hope it at least makes some sense :/

Thanks