Rate at which a pendulum bob slows due to air resistance?

Question

I know that "perfect" pendulums would be able to swing forever, unperturbed by air resistance. However, since there is air resistance around us, pendulums (swinging bobs) slow down and move closer and closer to a halt. Say we have a metal sphere of mass m and radius r as the bob, suspended at length l from a point. The bob is allowed to rise to a height $h_{0}$ above its equilibrium position (with the string remaining fully stretched), and is then released. After one swing, the bob reaches a new height $h_{1}$ above equilibrium, and so on, until after swing n, it reaches height $h_{n}$ above equilibrium. At what rate will this damping take place (i.e. how can one theoretically calculate $h_{n}$)? What are the factors that affect it?

Answer 1

Drag for a sphere is roughly proportional with velocity squared over a wide range of velocities (as long as the Reynolds number is reasonably large) and given by

$$F= \frac12 \rho v^2 A C_D$$

Where $\rho$ is th density of the medium (about 1.2 kg/m$^3$ for air), $v$ is the velocity, A the cross sectional area ($\pi r^2$) and $C_D$ the drag coefficient which varies with Reynolds number but which can be approximated to 0.5 for a wide range of velocities.

Since velocity squared is proportional to the height from the top of the swing, this suggests that the work done by the force of drag is roughly proportional to the height of the swing multiplied by the arc of the swing.

Answer 2

You will note from the answers and comments already given that this is a complicated system to analyse.

Two frictional force regimes have been suggested.

One is $F_D= \frac12 \rho v^2 A C_D$ the frictional force being proportional to the velocity squared and the other is $F_D = 6 \pi r v \eta$ where the frictional force is proportional to the velocity (Stokes’ law).

$r$ is the radius of the bob, $A=\pi r^2$, $\rho$ is the density of the air, $C_D$ is the drag coefficient which depends on Reynolds number (see below) and $\eta$ which is the viscosity of the air.

An important dimensionless parameter in fluid dynamics is Reynolds number which in this example can be written as $R_e = \dfrac {r v \rho}{\eta}$.

If the Reynolds number is large $(>1000)$ then the velocity squared regime predominates whereas if Reynolds number is small $(<1)$ then the velocity regime predominates.

Note that for low Reynolds numbers the drag coefficient is inversely proportional to Reynolds number and hence inversely proportional to the velocity. So the drag is again proportional to the velocity in this regime.

There is also a massive complication as to whether the flow of air passing the bob is laminar or turbulent and also the surface condition of the bob will contribute to the complication.

Now the experiment that you have done is one which has been done by many students and there are variations in terms of recording the data.

You variation is measuring a height from the equilibrium position which experimentally might be difficult to do accurately.

Other variations are to measure the angular amplitude of the swing $\theta$ as it varies with with either time $t$ or number of swings $n$.
Another variant is to measure the amplitude of the swing by using a horizontal ruler. With this variation you can count the number of swings as the amplitude drops progressively by say, a centimetre.

There are complications in that the period of the simple pendulum varies with amplitude etc. This means that the time $t$ for a given number of swings is not proportional to the number of swings $n$.

The theoretical analysis with the assumption that the friction force is proportional to the velocity is much easier (as described in one of the comments) than using the assumption of frictional force is proportional to the velocity squared.

Given a number of assumptions the velocity regime predicts a relationship of the form

$A_n=A_o e^{-k n}$ or $A(t) = A(0) = e^{-k' t}$

for the variation of amplitude. with number of swings or time.

So $\ln(A_n) = -k\; n + \ln(A_n)$ might be a reasonable relationship if you plotted a graph of $\ln(A_n)$ against $n$ and obtained a straight line graph.

As an aside you could then see if the Reynolds number is low in your experiment by substituting in values of $v, r \rho$ and $\eta$?

Answer 3

There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph:

http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf ,

and then calculated the Reynolds number to find the drag force using the formula given by the first answerer (Floris). You ask for the rate, that's the force (found above) times the speed. In the region of small angle (defined by another answerer) and if the loss (dissipation) is small i.e. a high Q, then one may approximate by differentiating the position; also given above. I suspect you want the total loss for each cycle (period). For that one must integrate the rate over time. At least two engineering texts do that: "Vibration Theory and Applications (Thomson) and "fundamentals of mechanical vibrations" (KELLY), and I've done the "whole shebang" in my draft for the Horological Science Newsletter. It may be found here:

http://www.cleyet.org/Pendula,%20Horological%20and%20Otherwise/HSN/Drafts/

when I "get around to" putting it there.

bc, recommends studying Eisberg's (and Lerner) Physics/ foundations and applications. That text deals at the simplest possible level.

Answer 4

The pendulum motion is described by $\ddot{\theta} + 2 \gamma \dot{\theta} + \omega^2 \sin(\theta)= 0$, where $\theta$ is the angle from the vertical position, $\gamma$ is the dissipation coefficient, and $\omega^2 = g/l$. For small angles ($\theta < \sim \pi/6)$, this equation can be approximated as $\ddot{\theta} + 2 \gamma \dot{\theta} + \omega^2 \theta= 0$. The last equation has solution $$ \theta = e^{-\gamma t} [\theta_0 \cos(\Omega t) + (v_0 + \gamma \theta_0) \sin(\Omega t)/ \Omega], $$ where $\Omega = \sqrt{\omega^2 - \gamma^2}$, $\theta(0) = \theta_0$ and $\dot{\theta}(0) = v_0$. A height from equilibrium position is found as $h(t) = l [1-\cos(\theta)] \approx l(1-\theta^2/2)$. If you plot $h(t)$, then every even (or odd) maximum gives the height after one swing, i.e. $h_0$, $h_1$, $h_2$, $\dots$ If the angles are large, one has to solve the nonlinear equation. It is also possible to write an explicit solution in this case, but it is much more involved.