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[parent] penny falling from Empire State Building (Example)

A Penny's Journey

As an example of constant acceleration, we will analyze a penny falling from the roof of the Empire State Building. Hopefully, any worries one might have of being killed by a falling penny as they stroll by tall buildings in New York will be quelled. After all, people are usually not injured by hail falling from the clouds.

We will apply Newton's laws of motion to two cases. At first we will neglect air resistance (drag) so we can solve the problem analytically. Then we will use a simple model of drag to show how the penny reaches a terminal velocity and how this dramatically affects the impact velocity of the penny on the ground.

Free Fall Acceleration

When the only force acting on a body as it falls toward another body, similar to an apple falling to the Earth, is gravity then we say the body is in free fall. The only way we can do this is if we neglet things like drag caused by the atmosphere, wind or variations in The Gravitational Field. This situation is depicted in Figure 1, where the force due to gravity is

$\displaystyle F = -mg$
\includegraphics[scale=1]{FreeFallAcceleration.eps}
Figure 1: Penny falling from Empire State Building

The two things we want to calculate is the velocity of the penny when it impacts the ground and the time it takes the penny to fall. Applying Newton's second law to the penny

$\displaystyle F = ma$

yields the acceleration of the penny in the y direction

$\displaystyle a(t) = \frac{F}{m} = -\frac{mg}{m} = -g \, .$

Notice that the minus sign comes from the coordinate system choice with the y axis pointing up. Next, integrate the acceleration equation to get the velocity of the penny at a given time. Remember that the acceleration is defined as

$\displaystyle a = \frac{dv}{dt}.$

This leads to the integration setup of

$\displaystyle \int dv = - \int g dt.$

Integrate to get

$\displaystyle v(t) = -gt + C.$

The constant is found using initial conditions, so at $t = 0$, the velocity of the penny is zero

$\displaystyle v(0) = 0 = -g(0) + C.$

This is only true if C is also zero, therefore the velocity equation is

$\displaystyle v(t) = -gt$ (1)

Well, this is great, we can find the velocity at any given time. However, we do not know the time of the impact. For this we must integrate (1) to get the position of the penny. Remember, that velocity is defined as

$\displaystyle v = \frac{dx}{dt}$

This leads to the integration setup of

$\displaystyle \int dx = -\int gt dt.$

Integrate to get

$\displaystyle x(t) = -\frac{1}{2} g t^2 + C .$

Once again, the constant is found from the initial condition, at $t = 0$, $x(0) = h$. Therefore, the position equation is

$\displaystyle x(t) = -\frac{1}{2} g t^2 + h$ (2)

What we want is the time of impact. This happens when $x = 0$. So solve (2) for $t$.

$\displaystyle 0 = -\frac{1}{2} g t_f^2 + h .$

Rearrange to get

$\displaystyle t_f = \sqrt{ \frac{2h}{g} }$ (3)

Evaluate (3), using the value $g = 9.8$, to get

$\displaystyle t_f = \sqrt{ \frac{(2)*(381)}{9.8} }$

$\displaystyle t_f = 8.82 \,\, [s].$

Finally, we get the velocity at impact by substituting (3) into (1)

$\displaystyle v(t_f) = -g \sqrt{ \frac{2h}{g} }.$

Bringing $g$ under the square root to get

$\displaystyle v(t_f) = -\sqrt{2hg}.$ (4)

This evaluates to

$\displaystyle v(t_f) = -\sqrt{2*381*9.8} = -86.42 \,\, [m/s] .$

In english units, the velocity is

$\displaystyle v(t_f) = -193.3 \,\, [mi/hr] .$

This is fairly fast for a penny, however this does not come very close to the actual speed because for such a large height, air resistance will become a big factor. In the next section we will model air resistance to capture this effect.

Resist the Penny

Since the penny is traveling through the atmosphere (a fluid), we need to use some basic results from fluid mechanics to model the drag force acting on the penny. A simple model combined with experimental data results in the drag expression

$\displaystyle F_d = \frac{1}{2} C_d \rho A v^2$ (5)

where $C_d$ is the drag coefficient of the penny. It is an experimantly derived value and here we will use the value given by [1],

$\displaystyle C_d = 1.17 .$

The drag force is in the opposite direction of the velocity vector as shown in the free body diagram in Figure 2.

\includegraphics[scale=1]{FreeBodyDiagram.eps}
Figure 2: Free Body Diagram

The density we will use is the standard day value at sea level given in the tables from [2],

$\displaystyle \rho = 1.225 \,\, [kg/m^2] $

Next $A$, is the effective cross-sectional area of the penny, From Wikipedia [3], the diameter of the US penny is

$\displaystyle D = 19.05 \,\, [mm] = 1.905 \times 10^{-2} \,\,[m] $

So the cross-sectional area is then

$\displaystyle A = \pi r^2 = \pi \frac{D^2}{4} = 0.000286 m^2$

Combining all these constants gives us the drag force that is proportional to the square of the velocity

$\displaystyle F_d = 0.0002 v^2.$

Now we have 2 forces to use with Newton's second law, this gives the acceleration equation as

$\displaystyle a(t) = \frac{0.0002 v^2}{m} - g$

The integral setup is then

$\displaystyle \int dv = \int \frac{0.0002 v^2}{m} dt - \int g dt .$

Unlike the free fall example, the analytical solution is a litte more difficult. However, all is not lost. We will first turn to numerical methods such as Euler's method or numerical programs such as Matlab, Octave or Maple and then later compare with analytical solution (coming soon). Here we will solve the differential equation using Octave [4]. First a little setup must be done to use Octave's lsode solver. The differential equation of motion that we will solve with the mass of the penny, $m = 2.5 \times 10^{-3} \,\, [kg]$ [3]

$\displaystyle \frac{d^2x}{dt^2} = 0.0817 \left( \frac{dx}{dt} \right)^2 - 9.8$ (6)

We cannot use a 2nd order differential equation in lsode, so we must use two 1st order equations. This is done by using two states, one for position and another for velocity. Then our equations of motion become

$\displaystyle \dot{y} = \dot{y}$

$\displaystyle \ddot{y} = 0.0817 \dot{y}^2 - 9.8$

The commands in octave are

t = linespace(0, 100, 10000);
y = lsode(“penny”, [381 0], t);

The function “penny” passed to lsode is here. The results are displayed in Figure 3, Figure 4 and Figure 5.

\includegraphics[scale=1]{figure1.eps}
Figure 3: Penny's Velocity

Note how the max velocity is now a measly $-10.952 \,\, [m/s]$ and that it stays constant for quite awhile up until impact. This is because the penny reaches what is called terminal velocity where the upward drag force at the terminal velocity cancels out the gravitational force,

$\displaystyle 0.0817 v_t^2 = g $

$\displaystyle v_t = \sqrt{\frac{g}{0.0817}} = 10.952 \,\, [m/s] .$

Figure 2 is a plot of the position and upon further inspection we see impact occurs at

$\displaystyle t_f = 35.36 \, \, [s]$

and the impact velocity is the terminal velocity

$\displaystyle v(t_f) = -10.952 \,\, [m/s] .$
\includegraphics[scale=1]{figure2.eps}
Figure 4: Penny's Position

Finally, we will zoom in on the first few seconds of the fall to see the position before terminal velocity is reached.

\includegraphics[scale=1]{figure3.eps}
Figure 5: First Few Seconds

So to wrap things up, we see that air resistance makes a dramatic difference in the maximum velocity of the penny and the time of impact. In reality, the motion is a little more complicated due to winds and the fact that the penny will tumble, but understanding how a falling object reaches terminal velocity is something to take away. For a more detailed analysis of chaotic dynamics of disks, see [6]

Bibliography

1
Yoon, Joe, Scott, Jeff, Gravity, Acceleration & Speed. Aerisoaceweb
2
Andersonn, John, Fundamentals of Aerodynamics. 4th Edition. McGraw-Hill, New York, 2007.
3
Wikipedia, Cent (United States coin)
4
Octave, GNU Octave
5
Halliday, D., Resnick, R., Walker, J. Fundamentals of Physics. 5th Edition. John Wiley & Sons, New York, 1997.
6
Stuart B. Field, M. Klaus, M. G. Moore, Franco Nori Chaotic dynamics of falling disks. Nature 388, 252-254, 17 July 1997.



"penny falling from Empire State Building" is owned by bloftin.

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Other names:  free fall acceleration example, drag on a penny, constant acceleration example
Keywords:  penny, Empire State Building, free fall, constant acceleration

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Cross-references: dynamics, object, function, mass, motion, differential equation, programs, diagram, vector, drag force, mechanics, section, speed, square, position, system, The Gravitational Field, velocity, resistance, Newton's laws of motion, acceleration

This is version 7 of penny falling from Empire State Building, born on 2006-12-18, modified 2009-06-21.
Object id is 239, canonical name is PennyFallingFromEmpireStateBuilding.
Accessed 33450 times total.

Classification:
Physics Classification: 45.50.Dd (General motion)
 45.50.Pk (Celestial mechanics )
 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
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Discussion
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How does a penny fall? by arkolbus on 2009-06-21 14:23:38
I stumbled across this article and am not familiar with the website.  Forgive me if this is the wrong location for this post.

This is a wonderful article on free-fall and the significance of air resistance and the concept of terminal velocity.  I noticed that in this article that the area of the penny is taken as the area of the penny's face.  The assumption is made that the penny would fall with its face perpendicular to the direction of motion.

Is this assumption valid?  It seems to me that the penny is more likely to fall on its edge, which presents a smaller cross-section and would result in a higher terminal velocity.  In the "flat" orientation presented in the article, the drag forces on the penny would be in equilibrium over the face of the penny, but this would not be a stable equilibrium.  It seems that the "edge" orientation would be a more stable equilibrium, and hence be favored.

The article does acknowledge that the situation is more complicated and the penny would tumble, but I'm curious as to the most likely orientation during fall.  Would the penny fall flat, on edge, or is there so much tumbling that there is no favored orientation?
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