# Simple pendulum with air resistance

#### Carla1985

##### Active member
Im getting stuck on a couple of the questions for my mechanics homework. Il put what I've done so far and hopefully someone can check im doing it right n push me into the right direction for the rest.

"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

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
Hi Carla1985! Actually, if I'm nitpicking a bit, your formula for acceleration should be:
$$\ddot{\mathbf{\vec r}}=(\ddot{r}-r\dot{\theta}^2)\mathbf{\vec e_r} + (2\dot{r}\dot{\theta}+r\ddot{\theta})\mathbf{\vec e_\theta} = -L\dot{\theta}^2\mathbf{\vec e_r} + L\ddot{\theta}\mathbf{\vec e_\theta}$$
And your total force should be:
$$\mathbf{\vec F} = (mg\cos\theta-T)\mathbf{\vec e_r}-(mg\sin\theta+2cm \cdot L\dot{\theta})\mathbf{\vec e_\theta}$$

Since each of the components of $\mathbf{\vec F}$ must be equal to each of the components of $m\ddot{\mathbf{\vec r}}$ (Newton's 3rd law), you get:
$$\left\{\begin{array}{lll} mg\cos\theta-T &=& -mL\dot{\theta}^2\\ mg\sin\theta+2cm \cdot L\dot{\theta} &=& mL\ddot{\theta} \end{array}\right.$$
Furthermore, you can use the approximation $\sin \theta \approx \theta$.

So where are you stuck?

Last edited:

#### Carla1985

##### Active member
Actually your few alterations put me back on track, surprising how silly mistakes can confuse the whole thing. Thank you very much #### Ackbach

##### Indicium Physicus
Staff member
Since each of the components of $\mathbf{\vec F}$ must be equal to each of the components of $m\ddot{\mathbf{\vec r}}$ (Newton's 3rd law), ...
Did you mean Newton's 2nd Law?

Staff member