Simple pendulum with air resistance

In summary, Carla1985 was getting stuck on a few questions for her mechanics homework, and was helped by another user by modifying the equation and providing an approximation for sin theta.
  • #1
Carla1985
94
0
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 I am 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
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
 
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  • #2
Carla1985 said:
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?
 
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  • #3
Actually your few alterations put me back on track, surprising how silly mistakes can confuse the whole thing. Thank you very much :D
 
  • #4
I like Serena said:
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?
 
  • #5
Ackbach said:
Did you mean Newton's 2nd Law?

Yep.
Thanks for spotting that.
 

FAQ: Simple pendulum with air resistance

What is a simple pendulum with air resistance?

A simple pendulum with air resistance is a physical system that consists of a mass attached to a pivot point by a string or rod and is subject to the forces of gravity and air resistance. The air resistance acts as a damping force that slows down the motion of the pendulum.

How does air resistance affect the motion of a simple pendulum?

Air resistance creates a drag force on the pendulum, which opposes its motion. This means that the pendulum takes longer to swing back and forth, decreasing its amplitude and increasing the time period of each swing.

Can air resistance be ignored in a simple pendulum?

No, air resistance cannot be ignored in a simple pendulum because it significantly affects the motion of the pendulum. Ignoring air resistance would result in inaccurate predictions of the pendulum's behavior.

How does the mass of the pendulum affect the impact of air resistance?

The mass of the pendulum has a direct impact on the effect of air resistance. A heavier mass will experience a stronger air resistance force, resulting in a greater decrease in amplitude and increase in time period compared to a lighter mass.

Is there a way to reduce the impact of air resistance on a simple pendulum?

Yes, the impact of air resistance can be reduced by using a more streamlined shape for the pendulum and increasing the length of the string or rod. Additionally, conducting experiments in a vacuum can eliminate the effects of air resistance altogether.

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