Thursday, June 30, 2011


As we were driving into Boston today... I know I've started a lot of posts this way, so I apologize if you're getting sick of car-related stuff, but it's all I've been doing lately.  Anyway, as we were driving into Boston today, we were behind a van on the Massachusetts Turnpike that suddenly lost one of its hubcaps.  Rather than simply falling to the ground, the hubcap continued to roll alongside the van, slowly losing speed.  It made me start thinking about the stabilizing effects of spinning, and wonder how long it would take to eventually fall over and stop.

Let's start by defining some coordinates:
The hubcap rolls in the x-direction, spinning about the z-axis.  Its angular momentum is
where I is its moment of inertia and ω is its angular velocity.  If we approximate the hubcap as a thin, solid disk, we can write
where m is the mass of the hubcap.  If we also assume it rolls without slipping, we can replace ω with a linear velocity, v:
The hubcap won't be perfectly vertical when it pops off the wheel, so gravity will apply a torque, tipping it over.  The amount of torque will depend on how far the wheel is already tipped.  We'll call the angle off the y-axis that the hubcap has tilted θ.  Then the torque applied by gravity is
Torque and angular momentum are related by
so we'll need to generalize our previous equation for L.  Taking the tilt of the wheel into account, we have
Plugging this into the relationship above, along with our equation for τ gives the differential equation
We're not interested in the z-axis rotation, since this will remain constant, but picking out the x rotation,
Solving this gives
where C is some constant.  The hubcap will be on its side when θ = 90°, so solving for t in this case gives
We can put C in terms of the initial tilt of the hubcap with
Plugging this into the equation for t gives
To get an idea of how this varies with θo, we can make a plot:
Although the time can go to infinity with a perfectly vertical tilt, it quickly drops off with any slight error.  Going at highway speeds, a tilt of just 0.00001° will cause the hubcap to fall over within 23 seconds.

Tuesday, June 28, 2011

Unlikely != Impossible

My sister is currently in law school, and earlier today she sent our father, a former NOAA weather observer, the following question:
I am pondering the nature of black ice. In tort liability cases, a person who slips and falls on business property cannot recover for his injuries if "a reasonable person" should have discovered and realized the danger. For example, if I go to Starbucks and see the person in front of me slip on some ice in front of the door, and then I slip on the ice and injure myself, I "accepted the risk" of slipping, and I can't sue Starbucks, even if Starbucks negligently maintained their property. Knowledge is an essential component of risk acceptance. I cannot accept a risk if I have no knowledge of its existence.

Now to my question: the nature of black ice is that it is nearly impossible to see. A person of ordinary, average intelligence knows that he should "be careful of black ice," but he doesn't have actual knowledge of any specific patch of black ice until it's too late.  Do you agree? Is it possible to "know" black ice?
Here was his response:
Black ice forms from either rain falling on a freezing surface or when moist air comes in contact with the freezing road surface. This can be in the air or can come from car exhaust. I think your question is can one know that there is black ice, right?

I say you can. You can, should know when the temperature is nearing freezing. Most people don't but ignorance is no excuse.
He likes to tease me for my scientific precision, and as we were discussing this he joked that he knew there was no black ice on the road today, with temperatures in the 80s, but that I might allow for the possibility.  I thought it might be fun to figure out exactly what the probability of black ice today would be, to show that there is some finite possibility.

In order to get ice, we'll need temperatures around freezing.  Temperature describes the average speed molecules are moving in a material, but not every molecule will be going at this speed.  We could imagine a small part of a material having molecules that are mostly going a below-average speed.

In an ideal gas (air is pretty close to one), the probability of a molecule going a specific speed is given by the Maxwell Speed Distribution:
where m is the mass of the molecule, k is Boltzmann's constant, and T is the temperature.  It's possible to get an average speed as a function of temperature from this:
Using these two, we can find the probability of finding a patch of freezing air.

According to this, the average speed of nitrogen (the main component of air) at 32°F will be 453 m/s.  Plugging this into the Maxwell Speed Distribution at a temperature of 80°F gives a probability of 0.2%.  This may seem shockingly high, but remember that this is only for a single molecule.  To find the probability of a patch of cool air, we'll need the number of molecules in the patch.

Let's suppose we're looking for a block of freezing air 25 cm by 25 cm by 1 cm.  The density of air is approximately 1.2 kg/m^3.  Multiplying by the volume of our block, and converting to the number of molecules we find 1.6*10^22.  Now, to find the probability of all the molecules moving slowly enough, we raise the single molecule probability to this power.  Unfortunately, there's no easy way to express the tininess of this number, but please note that it is not zero.

This talk of exceedingly unlikely events reminded me of a design I made one year for Swarthmore's annual Physics Department T-Shirt Contest.  I forget whether I ever submitted it (it didn't win in any case), but here it is (click to enlarge):

Monday, June 27, 2011

On Further Reflection

As we were driving into Boston today, Steve was marveling at the appearance of the clouds through his polarizing sunglasses.  I figured it was a good excuse to talk a bit about polarization, and why it's useful in sunglasses.

Light is made up of oscillating electric and magnetic fields (hence its alternate name, electromagnetic wave).  Polarization refers to the direction of the electric field, which can be fixed in one direction (linear polarization), or can rotate about the axis the light travels along (circular/elliptical polarization).  Typical light from the sun or other mundane sources is unpolarized, meaning it consists of a random assortment of polarizations.  However, under certain conditions, reflected light can become polarized.

When light hits a reflective surface at a specific angle (called Brewster's angle), the reflected light will be completely polarized parallel to the reflective surface.  For angles close to Brewster's angle, the reflected light will be mostly polarized in the same direction.  By cutting out that particular polarization, the reflected light will be blocked.  Here's a picture demonstrating the effect (from Wikipedia):
The photo on the left was taken without a polarizing filter, while the one on the right was taken with a filter adjusted to cut out the light reflected from the window.  This can be useful when driving by eliminating reflections from puddles on the road, or glare on glass.

Polarizing lenses work by having electrons that are free to move along a particular axis.  When the light hits the lens, if its electric field is aligned with the axis the electrons can move along, the field causes the electrons to start oscillating with the field.  This absorbs the light and re-emits it in the opposite direction, in essence reflecting it.  Therefore, any light emerging from the lens has polarization perpendicular to the direction the electrons move in.

In addition to its usefulness in driving, polarization can also create cool effects like the clouds I mentioned in the opening.  Here's a photo of a similar situation (from Wikipedia):

Saturday, June 25, 2011

Rain, Rain, Go Away

I've been trying to get this post done for the past couple days, but I've been feeling some of the fatigue I was warned about with the radiation, so I was a bit delayed.

It was another rainy trip in and out of Boston Thursday.  Steve's car has the interesting feature of wipers that automatically adjust their speed, using a sensor along the top of the windshield.  Watching them speed and slow with the rain got me thinking: the faster you drive, the more rain you'll hit, but the extra speed also drives rain up the windshield and into the sensor.  Do the two balance each other out, or will the wipers end up going too fast or too slow?

The most difficult part of this will be figuring out how fast the rain slides up the windshield, so we'll handle that first.  Once the rain hits the windshield, it's going the same velocity as the car, and it's the oncoming wind that will push it up.  The force of the wind drag is given by
where ρ is the density of air, v is the velocity of the water drop relative to the wind (basically the car's velocity), C is the drag coefficient, and A is the cross-sectional area of the drop.  To simplify things, we'll assume gravity is much weaker than this and take it to be the only force on the drop.  Then the distance it moves up the windshield after a time t is
where m is the mass of the drop.  If the drops are uniformly spread over the windshield, then on average they'll need to travel up half of it to get to the sensor.  This gives a rate of
where d is the length of the windshield.

Suppose the number of drops that have entered the sensor after a time t is n(t).  Then the rate of drops is
where rr is the rate of rainfall directly into the sensor.  The density of rain in the air (drops per volume) is
where vr is the velocity of the falling rain and As is the area of the sensor.  Using this, we can find the rate drops fall on the windshield
where θ is the tilt of the windshield.

Comparing to the rate the drops are entering the sensor, we can see that both are linear in v with a constant offset, meaning that for sufficiently high velocities, the relationship should work out evenly.  Clearly those Volvo engineers knew what they were doing...

Tuesday, June 21, 2011


On our daily trips into Boston, we pass a lot of trucks on the highway, and a fair portion of them are designed for carrying liquids.  I was struck by the variety of tank shapes, given that the regular dry containers all look basically the same.  I assumed this had something to do with distributing the pressure such that the tank wouldn't rupture, and I decided to take a look at how the pressures would compare between three different designs: a cylindrical tube, a triangular tube, and a standard box.  I chose the dimensions such that they all had the same width (they have to fit on the road after all) and all the same volume.  The pressure exerted by standing water is given by
where ρ is the density of water, g the acceleration due to gravity, and d the depth of the water.  Applying this to each of the shapes, we find

where red is the highest pressure.  We can get a more quantitative assessment of the shapes by finding the total force on the tank walls.
Evaluating this for each shape gives
where w is the width of the tanks.  Throwing in some arbitrary values for our variables, we have
Circle 4
Triangle 3.14
Rectangle 5.6
suggesting that the triangular tube will have the least pressure.  However, the triangle requires a much taller container to hold the same volume, so the circular tube offers a nice compromise.

Saturday, June 18, 2011

I Am Not a Crooke

At my parents' new condo where I've been staying, we have an interesting scientific trinket that once resided at my grandparents' house: a Crookes radiometer.  For those unfamiliar, they generally look like this (from Wikipedia):
When exposed to a light source, the vanes turn.  For a long time, the process that caused this was misunderstood.  Initially, it was believed that the momentum carried by the light was being transferred to the vanes.  The momentum carried by a photon is given by
where h is Planck's constant, and λ is the wavelength of the photon.  The majority of photons from the sun have a wavelength around 500 nm, giving a momentum of 1.3 x 10^-27 kg m/s.  Assuming the vanes are about 4 cm^2, there is about 1/4 watt worth of photons hitting each one at 500 nm.  Converting this into the actual number of photons gives about 6.3 x 10^17 photons/second.  We can find the force on the vanes by multiplying this by the single-photon momentum we found above.  This gives the miniscule force of 8.2 x 10^-10 Newtons, hardly enough to move anything macroscopic.

Aside from the lack of sufficient force, this explanation would also predict that the vanes would turn with the white side trailing.  The black sides would absorb the photons' momentum, but the white sides would reflect it, gaining twice as much.  However, as you can see from the animation above, the black sides trail.

It turns out the real explanation has to do with the heat the vanes gain from absorbing light.  The black sides heat more quickly than the white, creating a pocket of hotter air in front of them.  Near the edges of the vanes, the cooler air from the white side flows into the hotter air, pushing the vanes forward.

Thursday, June 16, 2011

Prismatic Spray

It was raining as we were driving into Boston the other day, and the cars in front of us would send up blinding sprays of water droplets on the highway.  It got me thinking about the fact that, by themselves, water and air are each basically transparent, but mixing them into a mist makes them nearly opaque.  This is caused by refraction – when light crosses the boundary between two materials, it bends, so sending it through thousands of small droplets bends it many times, and spreads it out.

I tried to figure out an exact form for the degree of bending for a small drop, but that ended up being too difficult, so I used a geometry program called GeoGebra to make the construction I was looking for.  I collected a couple points and fit a parabola:
The x-axis is the distance off-center that the light hits the drop, and the y-axis is the degree of bending as a fraction of pi.  The equation for the parabola is
so for a large beam of parallel light entering a drop, the exit path looks like this:
Averaging over all y values gives a bend of 0.11π.  

If we assume the air/water mixture was about 10% water, then we have one 1 mm^3 drop in every 10 mm^3 of space, or approximately one drop every 4 mm.  If some light enters a drop, given our average bending it will spread out over the 4 mm to cover approximately 3 times the original space.

To get an idea of what this looks like, I took an image,
and divided it into a 5x5 grid.  Each grid square I divided into 3x3 and expanded the middle square to fill the larger square.  Mixing this with the original gives
This is a pretty coarse approximation of what's really going on, but it shows how quickly this process can make an image indistinct.

Sunday, June 12, 2011

The Pressure's On

For the summer, I'm staying at my parents' new condo in downtown Northampton.  It's a nice place, but as usual, it's the physics examples that catch my eye.  The shower head is unusually large, and I've noticed that when I turn off the water, some stays trapped inside the head and slowly drips out.  What's interesting is the way it drips out – since the head is tilted slightly, the drips start off coming from the whole head, but gradually stop in the front as the trapped water diminishes.

Dripping water, as opposed to a stream, is caused by surface tension, the extra binding force at the surface of a liquid.  For a drip clinging to a tube, the force of the surface tension is
where d is the diameter of the tube, and γ is a constant that depends on the specific liquid.  To find when the drip falls, we can set this force equal to the forces acting against it – gravity and the pressure in the tube.
where m is the mass of the drip and p is the pressure in the tube.  Notice that these two are the only real variables in the equation; everything else is a constant.

First, we'll consider the pressure, p.  Suppose the shower head is tilted at an angle θ:
We'll assume it's a cylinder with radius r.  Then we can relate the volume of water V to the depth of the water h at a specific point x by integrating over the area of rectangular slices:
It can be a little difficult to see what exactly h(x) looks like, but I found this diagram helpful:
We can think of h(x) as a segment of a right triangle such that
where xc is the length of the clipped portion of the triangle.  Then the volume is
or evaluating,
Using this, we can eliminate xc from our equation for h(x):
or putting it in terms of the pressure at a point x
where ρ is the density of water, and g is the acceleration due to gravity.  Simplifying things a bit, we can write
where M is the mass of the trapped water.  Substituting into the force equation found above, we have
Note that
so by integrating over x, we can turn this into a differential equation.
Solving this gives
where C is some constant.  Interestingly, this says that you'll never lose all the trapped water, since the pressure will eventually drop to zero, but a few drops will stay stuck from the surface tension.  Substituting back into the force equation and solving for m gives
Here's what this looks like (y-axis represents drip size, x-axis represents position):
I admit, that was a lot of work for an only mildly interesting result, but they can't all be winners...