Download On the Russell Hall case Andrei Galiautdinov

On the Russell Hall case
Andrei Galiautdinov
Department of Physics and Astronomy, University of Georgia, Athens, GA 30602, USA
(Dated: November 27, 2015)
A conservative estimate for the stopping force acting on a falling human body on impact with the
ground is provided. Several hydrodynamical paradoxes relevant to this problem are also discussed.
Contents
I. The meaning of physical estimation
II. The statement of the problem
III.
1
1
III. Solution (ignoring air resistance)
1
IV. Comments
2
V. Effect of air resistance
A. On the importance of the Reynolds number
B. A quick justification of Newton’s formula
VI. Paradoxes
Acknowledgments
References
I.
2
2
3
3
SOLUTION (IGNORING AIR RESISTANCE)
The following quick estimate takes a few seconds to
perform. We think this is the most conservative (minimal) estimate for the impact force one can make under
typical conditions. [NOTE: “Typical”, as usual, is in
the eyes of the beholder.] The estimate does not distinguish between landing on concrete and landing on compacted soil. It cannot predict with any level of confidence
whether the subject will be able to survive. For further
comments on the results obtained see Sec. IV.
Ignoring air resistance, the speed on impact is
v =
p
2gh
p
2 · 9.81 · (85 · 0.3048)
=
≈ 22.5 [m/s],
(1)
4
4
where g = 9.81 [m/s2 ] is the gravitational acceleration
and 0.3048 [m/ft] is the conversion factor. [NOTE: in
physics, there is no such thing as the exact value. Everything is approximate.]
THE MEANING OF PHYSICAL
ESTIMATION
In physics, to estimate something means to perform a
quick order of magnitude calculation using simple formulas describing the basic laws of Nature, which, in physicist’s personal opinion, are most relevant to the situation
at hand. It’s like saying that
82 × 114 ≈ 100 × 100 = 10, 000 (exact value is 9, 348)
in mathematics. Physicists don’t usually care when the
estimate is off by a factor of less than 10. For example, to
a physicist, a 60- and a 100-watt light bulbs are equally
bright, while a 10- and 100-watt are not.
II.
THE STATEMENT OF THE PROBLEM
On November 15, 2015, a student weighing about
mg = 155 [lbs] (the value quoted on the “Red & Black”
website, see [1]) fell from the 9th floor (h = 85 [ft]) of
Russell on the ground below. Estimate the stopping force
acting on the body on impact.
FIG. 1: The Russell Hall case diagram.
Assuming the body fell flat on the ground (simultaneous impact by entire body, see Fig. 1) and using the
most conservative estimate for the stopping distance to
be about s = 1 [ft] (roughly, the “thickness” of the human
body), we first find the “stopping” average acceleration
2
to be
a=
v2
h
= g
2s
s
(= 85 g, in our estimate),
(2)
from which the average stopping force is, by Newton’s
Second Law,
h
Fminimal = ma = mg ,
s
(3)
(this formula works for any mg, h, and s, so plug in
whatever they were in actual landing, just make sure you
use the right units). Since we chose s = 1 [ft], we get
bone structure and elasticity, all must be taken into account.
People doing karate can exert 100’s lbs per square inch
and yet damage to fighter’s hand is minimal. Intuitively,
damage comes from tear, which comes from relative displacements of various body parts, which, if the impact is
brief, can be rather small. That may suffice to break
a brick (think of a rigid atomic structure that crumbles when a single atomic layer is dislocated) but not
the fighter’s hand (which is made of stretchable organic
molecules embedded in an almost incompressible fluid).
V.
Fminimal
85[f t]
= 155[lbs] ×
≈ 13, 175 [lbs].
1[f t]
(4)
A.
If we model the impact area as a rectangle
A = 1[f t] × 6[f t],
(5)
(typical dimensions of the human body), we find the average minimal pressure exerted on the body to be
Pminimal =
Fminimal
mg h
=
A
A s
(6)
or,
Pminimal =
13, 175[lbs]
2
6[ft]
lbs
lbs
= 2, 200
= 15
. (7)
f t2
in2
REMARK: The “stopping distance”, s, is the vertical distance traveled by object’s center of mass during
stopping. Thus for a water balloon falling on a concrete
floor the stopping distance could be as large as balloon’s
radius due to splattering! That would give a relatively
small stopping force (see Sec. VI).
On the contrary, if s is very small, the force can be
much larger. To give an example, a brick weighing mg =
10 [lbs] with face area A = 0.5 [ft2 ] falling on solid floor
from just h = 3 [ft] may experience a great stopping force
(as well as pressure) and break apart due to extremely
small s. Let’s say the floor caved by s = 0.1 [in] =
0.008 [ft], then the stopping pressure acting on the brick
is
mg h
lbs
10
3
lbs
P =
=
×
= 7, 500
=
52
,
A s
0.5 0.008
f t2
in2
(8)
which is already 3.5 times greater than the pressure found
in (7).
IV.
EFFECT OF AIR RESISTANCE
COMMENTS
The moral is: The numbers given in (4) and (7) can’t
tell us much about the actual effect on the human body.
The system is just too complicated. Various other parameters, such as membrane strengths, properties of blood,
On the importance of the Reynolds number
In order to estimate the effect of air resistance on the
motion of a falling body we have to decide which formula
to use: either the Stokes formula for a solid sphere of
radius r (as corrected by Oseen),
3
Stokes
Fdrag
= 6πηrv 1 + NR ,
(9)
8
representing drag due to friction (viscosity), where η is
the viscosity coefficient and NR is the Reynolds number,
or Newton’s formula,
Newton
Fdrag
= cx
ρv 2
A,
2
(10)
representing drag due to inertial forces (including turbulence). Here, ρ is the density of air, and cx is the
dimensionless drag coefficient which is a function of NR ,
cx = f (NR ),
NR ≡
`v
,
ν
(11)
with ` being object’s characteristic linear size (so that
A ≈ `2 ) and η, and
ν≡
η
ρ
(12)
being the so-called kinematic viscosity. By writing
NR =
ρ`v
ρv 2 A
=
,
η
η(v/`)A
(13)
we uncover the physical meaning of the Reynolds number: it represents the ratio of inertial to viscous forces
acting on the moving object. Thus, when NR 1 we
use (9), when NR 1 we use (10).
For air under normal conditions, ρ = 1.2 [kg/m3 ] and
η = 1.9×10−5 [kg · s/m], and thus ν ≈ 1.6×10−5 [m2 /s],
which for ` ∼ 1 [m] and v ≈ 10 [m/s] gives
NR =
`v
≈ 600, 000 1,
ν
(14)
3
so we have to use Newton’s formula (10).
Now, typically, cx ∼ 1, so we can estimate the drag
force to be roughly
Newton
Fdrag
= cx
1.2 · 102
ρv 2
A=1·
· 1 = 60 [N ].
2
2
is proportional in strength to v. Therefore, the velocity
gradients near the object, and hence the viscous forces
caused by those gradients and acting on the object, are
also proportional to v. Thus, at low speeds,
(15)
Fdrag ∝ v.
(21)
In comparison, the weight of the body is
mg = 155[lbs] · (1/2.2)[kg/lbs] · 9.81[m/s2 ] ≈ 690[N ],
(16)
which is about 10 times greater than the drag force, making the drag ignorable for the estimation purposes.
B.
VI.
PARADOXES
If you ever watched YouTube videos depicting exploding balloons you might have noticed that bursting always
starts at a single location. Let’s call it the weakest link.
A quick justification of Newton’s formula
A simple Force-Work-Energy argument gives (10) as
follows.
Move the object at (high) speed v by distance dx
through air initially at rest. Doing so causes the air to
be accelerated to the same speed, v, in the volume dV
that the object sweeps through,
dV = Adx,
(17)
with A being the frontal area of the object (facing ”into
the wind”). The mass of air in dV is
dm = ρdV,
(18)
and hence the kinetic energy imparted by the object onto
that accelerated air volume is
dK = (1/2)dmv 2 ,
(19)
which is also the work, dW , done by the object on the
air volume. Hence, the force Fdrag that the air exerts on
the object is
Fdrag = dW/dx = dK/dx = (1/2)ρAv 2 ∝ v 2 .
PARADOX 1: An ideal spherical balloon falling
vertically on the ideal horizontal floor will never
burst.
JUSTIFICATION: Follows from axial symmetry.
In an ideal universe, the weakest link is a single mathematical point. Since a single link has finite strength
and none of the uncountable infinity of links on balloon’s
equator is special, the net equatorial strength is infinite.
RESOLUTION: None. (However, in the real world,
the resolution should have something to do with atomic
structure and inherent randomness present in Nature in
the form of thermal and quantum mechanical motion —
thus, in the real Universe some molecules are more unequal than others, thus there is always the weakest link!)
FURTHER QUESTIONS:
1. Ideal balloon falls at an angle. Where’s its weakest
link? In the front, in the back, or elsewhere?
2. Does friction against the floor play a role? Does it
affect the location of the weakest link? (Compare friction
to no friction situation.)
(20)
Aside from a dimensionless factor cx of order 1, this is
Eq. (10).
The reason why the air in dV gets accelerated to full
object speed v is due to the fact the object moves ”too
fast”, that is NR 1. Hence, the air’s inertia prevents if
from ”getting out of the way” (by flowing gently around
the object). Thus, a sort of ”stagnation zone” in front
of the object is formed, where the air is roughly at rest
relative to the object, i. e., moving at speed v relative
to air at rest further away. (It’s easier to intuit this in
the rest frame of the object, instead of the rest frame of
distant air).
By contrast, at low speeds, when NR < 1, the air has
enough time to move sideways and around the object.
Hence, only a much smaller air volume picks up the full
speed v. In that case, the energy dissipated by the object into the air is governed by the stationary (in the comoving frame of the object) laminar velocity field, which
PARADOX 2: The stopping pressure acting on
a bursting water balloon (or just a droplet) filled
with an ideal fluid approaches zero, thus violating
the impulse-momentum theorem.
JUSTIFICATION: Should follow from A → ∞ when
used in (6) due to splattering down to a single atomic
layer.
RESOLUTION: The A used in Eq. (6) is not the area
of a single atomic layer, but the much smaller area occupied by the fluid at the moment when the “last” fluid
particle loses its vertical speed.
PARADOX 3 (d’Alambert): For incompressible and potential (that is inviscid and irrotational) flow, the drag force on a body moving with
constant velocity is zero.
4
JUSTIFICATION: See [2]. Non-zero drag at constant
speed means that some external agent is doing work.
This work must either (a) thermally dissipate into the
fluid, or (b) by transforming into fluid’s kinetic energy,
produce an outgoing to infinity energy flux. However, (a)
in an ideal fluid there is no dissipation (by definition), and
(b) the fluid speed drops too fast with the distance from
the object (typically, ∼ 1/r3 ) for the outgoing energy flux
to exist.
REMARKS: The above only applies to motion in
boundless fluid. If, for example, the fluid has free surface
then the object moving parallel to the surface will expe-
[1] N. Harris and A. Vander Heuvel, Alcohol, age likely play
role in student’s survival after nine-story fall, Red & Black,
Nov. 19 (2015); online at www.redandblack.com.
[2] L. D. Landau, E. M. Lifshitz, Fluid Mechanics, Course of
rience the so-called wave drag, in which case the energy
will be flowing to infinity in the form of surface waves.
MATHEMATICAL DETAILS: E. g., [3].
RESOLUTION: The paradox is believed to be solved
by Prandtl using his boundary-layer theory, in which the
effects of viscosity are taken into account.
Acknowledgments
The author thanks Todd Baker, J. P. Caillault, and
Bernd Schuttler for helpful discussions.
Theoretical Physics 6, 2nd ed., Pergamon Press (1987).
[3] G. Batchelor, An introduction to fluid dynamics, 2nd ed.,
Cambridge University Press (2000).