# Physics of Walking While Holding a Plastic Bag

When we are walking in a straight line, while holding a plastic bag on our hand, with an acceleration much smaller than $g$. There are two possible cases that may happen….

Sometimes the plastic bag would swing so peacefully that it makes almost no difference whether it is there or not….

But sometimes it may swing with a large amplitude, annoying enough to force us to make a conscious attempt to stop it.

Then how should we walk in order to avoid the latter case?

Let’s take a look at the plastic bag:

Yes, that is how it looks like in the physics world!

We can neglect the mass of the plastic bag relative to the total mass of the stuffs inside the bag. Due to centrifugal force, the stuffs inside will distribute themselves as far as possible from the pivot(finger holding the bag), so it is quite likely that they will end up having moment of inertia close to $ml^2$ relative to the pivot. And since the plastic bag has a large cross sectional area, we cannot underestimate the power of air drag. Let’s say it is proportional to the velocity of the bag, so we can write the torque due to the drag force as $\tau_{drag}=-k l^2 \frac{d\theta}{dt}$. Further I will assume that the mass of the plastic and the stuffs inside is small enough that we can neglect the transient effect. Yet another assumption, we may say that while walking human’s center of mass translates in approximately sinusoidal motion so that we can write the acceleration of center of mass as $a=a_0sin\omega t$. Since we are working in the hand’s frame, actually the drag force will change the equilibrium position so that it is not vertical anymore, but we will assume that this effect is small.

Okay now let’s write down the equation of motion, here is the torque about pivot equation:

$ml^2\frac{d^2\theta}{dt^2}+kl^2\frac{d\theta}{dt}+mgl\theta=ma_0 lsin\omega t$

Forget about the homogeneous solution, the steady state solution for this differential equation is:

The amplitude of oscillation

$\theta_0= \frac{a_0}{l\sqrt{(\gamma^2\omega^2+((g/l)^2-\omega^2)^2}}$

where $\gamma=k/2m$

So the high amplitude motion will occurs at resonance, when the $\omega$ of walking pace close to $\sqrt{\frac{g}{l}}$

But it is not that simple.. I measured my own standard walking pace using my ipod, the average time it takes for me to do ten steps is $4.91 seconds$, hence the average time between steps is $0.491 seconds$ or the frequency of pace is $2.04 Hz$. And the natural frequency of oscillation of a random plastic bag filled with snacks and drinks can be calculated using $f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}$ , and the result is $0.760 Hz$ for $l=43cm$. Thus resonance won’t occur at around the standard walking pace.

But we may pass through the resonance frequency during the process of building up speed. During the process, the frequency pace is increased from zero to $2.04 Hz$, at some point it will hit $0.760 Hz$. Now the thing is during the process how long will it stay inside the “near resonance region” which is roughly $\omega_r -\gamma<\omega<\omega_r+\gamma$? Where the amplitude of the oscillation is at least half the maximum amplitude. The longer it stay in that region the more effective the resonance will be. We know that the characteristic time required for the oscillations to die out is $1/\gamma$. Therefore our walking pace must pass through this “near resonance region” in time more than $1/\gamma$ to get significant excitation of the resonance. The value of gamma can be calculated by comparing the natural frequency of oscillation with the experimental frequency of oscillation which turns out to be $f_1=0.735Hz$. We get $\gamma=1.45 Hz$

The average walking velocity is related to the pace frequency as follows

$v=d\frac{\omega}{2\pi}$

Where $d$ is the pace length. It can be measured by wetting my sandals before walking on the dry road, and then I came back and measure the distance between two  footsteps using a ruler.

The average distance turns out to be $d=110cm$ . Assuming constant pace length, the change in velocity corresponds to the change in $\omega$ as follows:

$\delta v=\frac{d}{2\pi}\delta\omega$

Thus the time required to pass through the resonance width $\delta\omega=2\gamma$ is

$\delta t=\frac{d\gamma}{a\pi}$

Which must be larger than the damping time $1/\gamma$ in order to excite an effective resonance. At the end we get

$a<\frac{d\gamma^2}{\pi}=0.736 m/s^2$

Therefore to avoid an annoying high amplitude swinging plastic bag, we must accelerate faster than $0.736 m/s^2$, at least for this plastic bag