Phys 239
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Quantitative Physics
Problem Set 3 Solutions
Capillary Action
Use the Buckingham Pi Theorem (and the usual dose of physical reasoning) to arrive at an expression for
the height of the water column in a thin (1 mm diameter) tube via capillary action. Among the relevant
parameters is surface tension, which for water is about γ = 0.07 N/m. Don’t look up capillary action or seek
formulas elsewhere: Use the Pi, Luke. That’s what it’s for. At the end, put in some numbers and see if it
makes any sense.
Our table:
i
1
2
3
4
5
vi
h
R
g
ρ
γ
units
m
m
m/s2
kg/m3
kg/s2
notes
sought
tube radius
fighting gravity
mass loading of fluid
surface tension
We have n = 5, r = 3, so seek two Pi variables. Let’s start with the easy one (containing the thing we seek):
Π1 = h/R. Next, we incorporate surface tension, noting that γ and ρ are the only two that employ kg, and
therefore must be divided. γ/ρ leaves m3 /s2 , which we can get from R2 g. Thus we have our set:
Π1 =
h
γ
, Π2 =
.
R
ρgR2
And now we set these into the usual Buckingham formulation:
γ
Π1 = f (Π2 ) → h = Rf
.
ρgR2
So how would we expect this to scale? A single power of Π2 looks attractive: we get a larger capillary rise
if the surface tension is high, gravity and/or density are small, and tube radius is small. I’m sold:
h=
γ
.
ρgR
Does this match approximate observation? A very thin glass tube with a 1 mm diameter (R = 0.0005 m)
should sport a sizable capillary effect in water:
h≈
0.07
0.07
=
≈ 0.014 m.
1000 · 10 · 5 × 10−4
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So the prediction is 14 mm, which seems utterly reasonable. A 1 cm diameter tube would realize a much
less noticeable 1.4 mm rise. The height would equal the diameter for a diameter of 3.7 mm.
The “real” formula has a 2 cos θ in the numerator, where θ is the contact angle between fluid and vessel.
Water on glass has a contact angle of zero (clings), so the cosine term is effectively gone (unity), leaving a
factor of two we missed.
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Deep Water Waves
Use Buckingham Pi and physical reasoning to estimate wave speed in deep water for large waves where surface
tension is unimportant. After you have a relation, evaluate the speed for some familiar-scale wavelength and
compute the (time) period of the waves. If this seems off, figure out a place to throw a factor of 2π to make
things about right (if you think you need it at all: depends on your choice of variables). Note: You can
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also use this technique to get at small capillary waves, where surface tension is important (and gravity is
not)—but I am not asking you to do this here/now.
We will follow the hint that surface tension is not important for large waves, but presumably gravity is.
Let’s start our table:
i
1
2
3
4
5
6
vi
v
λ
g
ρ
h
A
units
m/s
m
m/s2
kg/m3
m
m
notes
sought
wavelength
gravity important
density of water
depth of water
amplitude of wave
It’s fine to over-build your table and then pare down, which is what we do here. For deep water, we may
assume that the surface waves know nothing about the bottom, so let’s chop it from the table. Next, even
though it seems density ought to be important, it is the only unit with kilograms. Rather than strain
ourselves to concoct another variable containing kilograms, we might recognize that the period/speed of
a pendulum does not depend on the mass, nor does its amplitude of oscillation impact the speed/period.
Likewise, the free-fall time of objects in gravity are independent of mass. So it is with water waves: the
inertia (resistance to motion) exactly compensates the forcing from gravity: both linearly dependent on
mass. It may be harder to reconcile the independence of velocity on wave amplitude, but think about the
following: would very small amplitude waves get essentially stuck? No more so than a small-amplitude
pendulum (absent friction) would fail to oscillate—and at the normal frequency, I might add. You can also
think about superposition: a wave may be thought of as a superposition of two waves at half the amplitude,
yet the pair travel together at the speed of the larger wave.
After all the trimming, we are left with only the top three variables, using two fundamental units, and
therefore have a single Pi variable: v 2 λ/g. This must be a constant at best, so:
p
v 2 ∝ gλ, or v ∝ gλ.
The problem statement suggested that perhaps a 2π could be missing. So let’s imagine a familiar wave
scenario and see if we can figure out where it might go. I’ll do a couple, for good measure. First, your friend
jumps into deep water (at a pool, maybe), and you see waves approximately 2 m apart leave the scene at
the
above without a proportionality constant would suggest
√ rate of a brisk walk. Maybe 2 m/s. Our relation
√
20 = 4.5 m/s, which is a factor of 2.25 ≈ 2π off from our guess. Waves arriving on the beach have a
wavelength closer to 10 m (between crests), and travel at jogging speed, maybe 3 m/s. But the unscaled
relation would have v ≈ 10 m/s, which is again about 2–3 times the guessed speed. We can also evaluate
the period: 10 m wavelength at 10 m/s translates to a one-second period, or a wave every second—which is
obviously off. So perhaps we should replace λ with λ̄ = λ/2π (such that λ̄k = 1, where k is the wavevector,
or wavenumber). Having developed some confidence, we have:
r
p
gλ
v = gλ̄ =
.
2π
Trying this out on a larger scale, waves in the “roaring 50’s” in the southern hemisphere have giant wavelengths
in the 100 m range. Their speed would be about 13 m/s, which is no joke. But this is substantially less than
the terrifying 30 m/s (freeway speeds) that would result had we not inserted the 2π factor.
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Biking Speeds
How fast might a person expect to go on a bicycle in the following scenarios, assuming drag dominates over
rolling resistance?
a. On a flat road, no wind, keeping up a steady (and not shabby) power output of 200 W.
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b. Coasting down a long hill. You make up a reasonable grade (think percent) for the problem.
c. Climbing the same hill, again pumping out 200 W.
Part a (flat): We are fighting drag here, so Fd = 12 cD ρAv 2 , and the power exerted is Fd · v = 12 cD ρAv 3 =
200 W. If we pick cD ∼ 1 for our generic blob, a cross-sectional area of 0.5 m2 (approx. 0.4 m wide by 1.2 m
high (somewhat crouched/folded), and use sea-level density of air of 1.25 kg/m3 , we calculate v ≈ 8.5 m/s,
or about 19 m.p.h.—seeming imminently reasonable. This would not be considered easy to maintain—which
is one way of saying that 200 W output (one-quarter horsepower) is not easy to sustain. But well-fit people
can.
Part b (downhill): A steep grade for a road is 10%, meaning an angle of tan−1 0.1 ∼ 6◦ . For a long hill,
I’ll use the more moderate grade of 5%, which we would still call a significant slope (typical of roads into or
out of mountains). Here, the rate at which potential energy is lost is equivalent to the power feeding drag.
In other words, Fd · v = 12 cD ρAv 3 = mgvθ, where θ ∼ 0.05 is the slope. In this case, we get v ≈ 11 m/s,
or 24 m.p.h. for an 80 kg bike plus rider with the cross section used above. The rate of power dissipation
works out to 440 W. A 10% grade would produce 16.6 m/s, or 35 m.p.h.
Part c (uphill): Now we go up the same 5% slope under power, and P = Fd · v + mgvθ. This cubic relation
in velocity has no direct solution. We have tools to find the solution, but let’s punt in the spirit of the class.
If we guess the solution to be something in the neighborhood of 5 m/s, we can ask what the contribution
from drag and climbing will be. For climbing, mgvθ ∼ 200 W for 80 kg of mass, while for air resistance,
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2 cD ρAv ∼ 45 W. So climbing dominates air resistance roughly 4:1, but we overshot the budget, at 246 W.
So picking 4 m/s, or 9 m.p.h. results in 160 W and 25 W, respectively, for a much closer value of 185 W.
But wait a minute! Where did the 200 W number come from? Why should we work so hard to recover it
exactly? It suffices to say that a fit individual can sustain 4–5 m/s up a 5% slope hill. Once we get to 10%,
we’re talking 2.5 m/s with only 6 W fighting the air–completely negligible by this point.
A note on rolling resistance: for cars the rolling resistance is typically 0.01 times the weight of the car (think
of this as a small coefficient of friction). For bikes, the number is half this (or less for racing road bikes
under high pressure). So in the three cases above (at speeds of 8.5, 11, and 4.5 m/s), the associated power
in rolling resistance for 80 kg is about 34, 44, and 18 W, respectively, and is therefore not large compared
to the 200, 440, and 200 W relevant to each section.
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Drag Regimes
a. How long will it take a small round pebble to reach the bottom of the deep end of a swimming pool?
Justify your assumptions.
b. If we adopt a single form for the drag force for a sphere: Fd = 12 cD ρAv 2 , but have cD change from
roughly constant (∼ 0.5) at high Reynolds number to inversely proportional to Reynolds number (cD =
kν/Rv) for some constant k) at low Reynolds number so that the Stokes drag, 6πρνRv is recovered,
what is the constant, k, and at what Reynolds number does the viscous regime version of cD meet the
inertial regime (constant) version of cD ?
c. What is the largest size of a pebble that would fall in a viscous regime in water? Would we call this a
pebble?
Part a: We’ll start by considering a 1 cm diameter pebble. We can guess its velocity will be in the neighborhood of 1 m/s so that we can check the Reynolds number and verify the flow regime. Re ∼Rv/ν ∼
0.005 · 1/10−6 = 5000. So we are safely in the non-laminar inertial drag regime. At terminal velocity,
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mg = 43 ρrock πR3 g = 12 cD ρwater Av 2 , so that v = ( 38 ρrock Rg/cD ρwater ) 2 . If we pick cD ∼ 0.5 for an almost
sphere-like, but likely rough pebble, and ρrock /ρwater ∼ 2.5, we find that v ≈ 0.8 m/s, very nicely in line
with our prediction.
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But let’s not be utterly complacent. If we had a rock with ρrock = ρwater , the formulation above would
still produce a velocity (about 0.5 m/s), but we know this “rock” to be neutrally buoyant, so no downward
velocity. We should therefore replace ρrock /ρwater with (ρrock − ρwater )/ρwater in the relation above. Doing
so, we get v ∼ 0.6 m/s. Okay, so this is order-of-magnitude, and the modification in question is inside a
square root, so less influential. But we do what we can.
The deep-end of a pool is typically about 3 meters, so we are looking at about 5 seconds for the pebble
to reach the bottom. The Reynolds number is now about 3000. This range of Reynolds number will likely
result in alternating vortices being shed, making the rock wobble back and forth in its decent. I’ve always
seen objects of this size take wavy paths when falling in water—perhaps this is partly why.
Part b: If we replace cD with kν/Rv, our inertial drag equation—noting that A = πR2 for the sphere—
becomes Fd = 21 kνρπRv. We want to associate this with Stokes drag: 6πνρRv, for which we can immediately
see k = 12. Since cD = 12/Re in the viscous regime, then we would find it “crossing” the inertial value of
cD ≈ 0.5 at a Reynolds number of 24. This is certainly a sensible transition point. If we wanted one regime
smoothly transitioning to the other, we might say that cD = 0.5 + 12ν/Rv, or something like this.
Part c: We can approach the question of the smallest pebble that would fall in a viscous regime in several
ways. We can presume to know the Reynolds number at which this happens (we could use the result Re ∼ 24
from part b, for instance), we could guess at the crossover Reynolds number and check, or we could pick the
size for which the inertial terminal velocity and the viscous terminal velocity are equal.
Starting first with the target Reynolds number, and using the viscous drag force: Fd = 6πνρwater Rv = meff g,
we can produce (by substituting νRv = ν 2 Re and including the buoyant effect that meff = (ρrock −ρwater )·V ):
R3 =
9
ρwater
ν 2 Re
,
2 ρrock − ρwater g
which for Re ∼ 24 (and ν = 10−6 m2 /s) makes R ∼ 0.2 mm—a small grain of sand. The velocity would be
0.12 m/s under viscous drag, and 0.12 m/s under inertial drag. That these two are the same indicates the
correct balance (by the fruits of part b) and choice in Reynolds number crossover.
If we did not already know the crossover Reynolds number, we might guess 10 or 100 and try it out.
Setting terminal velocities equal produces exactly the same result as the first method, but the route is slightly
different.
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