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Hertz Theory of Elasticity of Steel Balls
By Free-Fall and Applied Pressure
Antonio T. Bill
Advisor: Tracy Hodge
Monday April 22nd, 2013
Introduction
The deformation that a ball undergoes when it strikes a surface seems to have some affect
on how high the ball bounces back. The energy of a ball that undergoes free-fall can be
calculated using the conservation of energy where kinetic energy is equal to gravitational
potential energy. In practice, if a ball that is dropped on a surface where no additional force is
given to the ball, it will never reach its initial height. This is interesting, because if the ball does
not bounce back to its initial height, then that means energy is being transferred or dissipated.
There are several reasons why a ball does not bounce
Transverse
Force
Figure 1
back to its initial height after being dropped. To list a
few: the deformation of the ball and internal friction of
Ball
Longitudinal
Force
the ball during deformation (Cross, Rod 1999). The
degree of deformation that the ball experience is related
to the elasticity of the ball.
Figure 1 displays a ball during
its decompressed state (the
solid line) and the ball again in
its compressed state (dotted
line).
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Usually a ball’s elasticity is determined by measuring its Poisson’s ratio. Poisson’s ratio is the
ratio of contraction along the line of the applied force to the amount of stretch orthogonal to the
line of the applied force. (Figure 1) offers a visualization of the effect of Poisson’s Ratio.
Poisson’s Ratio:
However, I decided to take a different approach to measure the elasticity of steel balls
which is deemed by Gugan as a more “efficient” way, which I will discuss later. Steel was
chosen because it has the ability to withstand a great amount of force and also because I find it to
be an interesting material. Steel has a Poisson ratio of .3; the lower the ratio, the greater the
durability of the material.
Using five smooth and steel balls of different sizes and diameters an oscilloscope, a
voltage generator and a steel plate, I was able to measure the contact time of the steel balls when
dropped from known heights. The voltage generator was used to apply a voltage (10V) across
the steel plate. I attached the positive probe from the voltage
generator to the plate and the negative probed connected to the
probe on the oscilloscope and then from there it connected to
Figure 2
the steel ball creating an open circuit with the ball and plate
acting as a switch location. Thus, when ball is dropped onto the
plate, the circuit is closed. Prior to the ball’s collision, the
oscilloscope reads 0 V, then once collision occurs, the value
Steel Plate
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jumps up to the 10 V marker on the oscilloscope screen and stays here for a very brief time, then
goes back down to zero. The time that it spent reading 10 V is the contact time, which is what I
am after. (Figure 2) shows a simple schematic of how the ball was dropped. The data showed
trends where the smaller steel balls spent less time in contact with the steel plate than the larger
steel balls. I also captured video footage of each ball drop to determine how much energy was
left in the ball in order to precisely declare how much energy the ball sustained once it struck the
plate. In addition to using the oscilloscope to measure contact time, I compressed each steel ball
in a hydraulic press placing the balls under a few thousand pounds of pressure to measure the
area of contact that the ball makes with the carbon paper on the steel surface. The smaller balls
permanently deformed, but the larger balls showed no obvious signs of deformation.
Background Information
In elementary classical mechanics, collisions are studied using Newtonian methods and
laws (Gugan, D. 2000). In a classical free-fall problem, an object can have two kinds of energy;
kinetic and potential. When energy is placed into an object, it is conserved until some outside
force acts upon it. For example if a ball is held still at a height h above the ground, it has 100%
gravitational potential energy. Once the ball is dropped and during its path toward the ground, it
loses gravitational potential energy and gains kinetic energy. Kinetic energy becomes 100% just
before the all collides with the ground (or when h=0). In addition, Heinrich Hertz introduced an
elastic theory to describe how they tended to behave during collision. Many dynamics of a
collision between a ball and a surface can be determined by knowing the initial conditions and
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the functional form of the force that acts on the ball (Cross, 1999). Measuring this force not only
provides information concerning the behavior of the ball, but it also sheds light on the duration
of the collision and its elastic properties (Cross, 1999). Since the collision is elastic, the force on
the ball obeys Hooke’s Law, F=kx (Cross, 1999). However, the spherical geometry of the balls
become a factor; the displacement “x” gets a root of (3/2), then the force equation to describe the
elasticity of the ball becomes(Cross, 1999):
………………..(1)
Where “F” is the force and “k” is the spring or elasticity constant.
When a ball undergoes collision, it loses energy. In general, if a ball is dropped onto a
surface from a known initial height (h0) and it rebounds to a height (h), the loss of energy is
equal to mg(h0-h), which can be described by the restitution coefficient, “e”(Cross,1999). The
coefficient of restitution (COR) is the ratio of the relative velocity after impact to the relative
velocity before the impact of two colliding bodies, equal to 1 for an elastic collision and 0 for an
inelastic collision (dictionary.reference.com). So the expression for the COR is:
…………….(2)
Where V2 is the rebound velocity and V1 is the velocity prior to impact.
Cross equates the COR to the square root of the quotient of the final height and the initial
height. This is plausible because the gravitational and potential energy of the ball is related to
the velocity and height of the ball. So the COR becomes:
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√ ………….. (3)
where H2 is final height of the ball after collision and H1 is initial height of the ball prior to
collision.
As mentioned earlier, the most efficient way to test Hertz theory is to measure the ratio of the
circle of contact, to the product of the impact speed and the time of contact and equate this value
to the product of a constant and the radius of the ball (Gugan, 2000). Thus we can test Hertz
theory by using the equation:
………………… (4)
where A is the area of contact that the ball made with the steel plate, T is the amount of time that
the ball spent on the steel plate (contact time), U is the final velocity prior to collision of the ball
and the plate, and R is the radius of the ball (Gugan, 2000). I am not positive as to how the
constant, 1.068 arises. I have already calculated the radius of each ball, R, and I have the contact
time, T, of each ball from each height from the oscilloscope readings. In order to find the
remaining variables, I had to perform some separate calculations.
I plan to find the contact area, A by measuring the radius r, that each ball creates on the
carbon paper and using the area of a circle:
A = πr2……………… (5)
To find the final velocities prior to collision, U, of each ball, I will use the conservation
of energy equation:
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Solving for v,
√
………………(6)
where m is the mass of the ball, g is the acceleration due to gravity, h is the ball’s initial height
before free-fall and v is the velocity.
Experimental Procedure
For the experiment, I decided to use an inconsistent, but yet wide range of various sized
steel balls. Readings from a vernier caliper measures the diameter of the five experiment balls to
be 19.1mm, 22.1mm, 25.4mm, 50.8mm (Figure 3), and 101.6mm (Figure 4). The mass of the
balls were .028 kg, .045 kg, .066 kg, .527 kg, and 1.79kg respectively. Each ball was dropped
three times at five different heights: 50 cm, 70 cm, 80 cm, 100 cm, and 120 cm.
Figure 3: 50.8 mm (2 in.) Ball
Figure 4: 101.6 mm (3 in.) Ball
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The Circuit
I managed to design a simple circuit that will allow me to calculate the contact time of
the each ball during its collision. Using a voltage generator supplied by the Physics department,
I was able to attach one probe to a steel conducting plate that was 3mm thick, while the other
probe was channeled to a digital oscilloscope. A conducting wire was then used to connect the
oscilloscope to the steel ball. Therefore, when the steel ball is dropped and comes into contact
with the steel plate, the circuit is complete, and the oscilloscope reads a voltage of magnitude 10.
The duration of time that the circuit was closed and read 10V is displayed on the oscilloscope by
a function (Δx), which effectively measures the steel ball’s contact time. (Figure 5) displays a
sample ball drop reading extracted directly from the oscilloscope and the Δx (contact time) value
of the ball with the steel plate. The scale that was being used was 2 volts per square and the
contact time reads 960 microseconds.
Figure 5: Example Oscilloscope reading
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Control Factors
There were a few factors that I had to control for the experiment that I noticed had the
potential to affect my data. One factor was to control the consistency and effectiveness of
dropping the balls. I controlled this by using a rod that was stabilized by the table and a clamp
that pinched the ball snuggly. The screw on the clamp is then unscrewed and the ball is dropped
from the intended height. Also, when the balls were dropped, I had to ensure that they did not
land on the side where the conducting wire was located. Another major factor that I discovered
was that the plate would rebound and absorb a lot of the energy of the ball (the ball had little to
no rebound height). To fix this, I used two C-clamps to press the plate against the table so that
the ball can conserve a fair amount of its energy.
Hydraulic Press
Another way I chose to explore the elasticity of the steel balls was to place them under a
force (pressure). This idea is essentially the same as the force applied to the balls dropping onto
a steel plate, except the force being applied here is not gravity, but instead a known pressure.
Using carbon paper, I applied pressure on each ball so that it will leave a small circle on the
carbon paper. Measuring the area of this circle and using simple geometry will help me
determine how much of the ball came into contact with the surface and will effectively allow me
to measure how much compression the ball witnessed at the specific pressure magnitude.
Figures 6 and 7 below show a schematic of how the balls are compressed in the hydraulic system
from its original diameter, x to a new compression factor, x’.
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Figure 6:
Uncompressed
Figure 7:
Compressed
x
x’
Data Analysis
Ball Drop
I chose to drop each of the five balls from a total of five different heights 50, 70, 80, 100,
and 120 centimeters. The readings for the time of contact of each ball began to become slightly
skewed when the height reached 100 and 120 centimeters. The data that the oscilloscope offered
for the 19.1mm ball seemed defective so I omitted that portion of the data. Initially, I
hypothesized that the contact time for the balls would increase as the height of release is
increased. Instead, I found that the contact time decreased with increased height. I believe this
is true because there is more force applied to the ball which increases the compression, thus
increasing its speed that it compresses and decompresses. Due to this, the ball spends less time
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in contact with the steel plate. (Table 1) lists the contact time of each ball at each height. Note:
There are some data that are missing because the data was never collected due to its defects.
After analyzing the data, the smaller steel balls showed a substantial tendency of
spending less time in contact with the plate in comparison to the larger balls. The balls’ contact
time ranged from 120 microseconds to about 1.08 milliseconds. The contact times increased
with increase in mass. The larger balls had less rebound height than the smaller balls. I believe
this is due to the massiveness of the ball and the fact that the rebound force was insufficient
enough to cause the ball to bounce up to a relatively large height.
Table 1: Contact Times at known heights
Height (cm)
19.1mm
22.1mm
25.4mm
50.8mm
101.6mm
50
220µs
240µs
360µs
660µs
940µs
70
140µs
220µs
360µs
600µs
920µs
80
180µs
200µs
220µs
520µs
920µs
100
-----
180µs
200µs
440µs
800µs
120
-----
240µs
-----
560µs
1080µs
Hydraulic System
My plan with the hydraulic system was merely to explore a different way of finding the
elasticity of the balls. With the use of carbon paper and the hydraulic press, I was able to apply a
pressure to each ball. The carbon paper then obtains a circle from the balls which signifies how
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much of the ball was in contact with the surface. Using
Pythagorean’s Theorem, I am able to find how much the ball
compressed. There are pictures of the hydraulic press and dial
below. The 19.1, 22.1 mm balls permanently deformed a bit
when put under a small amount of pressure. However, the 1,
2, and 3 inch balls showed no signs of permanent deformation
when put under a great amount of pressure. The range of
pounds of pressure that I used was from a quarter ton to five
tons.
When I plotted all of the balls’ compression versus the
amount of force (Figure 8), I noticed that the data points
displayed some linear behavior. The ball’s height of each ball
got smaller and smaller as the amount of force was increased.
Also, the slope of each graph changed as the size of the ball
changed. According to the data, as the size of the balls increased, the slope of the graphs seemed
to decrease. In other words, the bigger the ball, the more difficult it will be to compress it.
Below (Table 2) displays the percent of the size of each ball after compression in relation to each
of their sizes prior to compression. Some data is not present, especially for the smaller balls
because too much applied force caused them to exceed their elasticity limit and deform
permanently.
Ratio = Z2/Z1 x 100%………………….(5)
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Table 2: Compression Size to Initial Size Ratio Percents
Metric Tons
19.1mm
22.1mm
25.4mm
50.8mm
101.6mm
.25
99.4%
------
------
------
------
1
98%
98.6%
99.13%
99.7%
99.92%
2
96.52%
98%
98.9%
99.6%
99.9%
3
94%
------
97.6%
99.4%
99.84%
4
------
------
------
------
99.81%
5
------
------
------
------
99.7%
Table 2: As the size of the balls increased, so did its compression
size versus initial size ratio. As the amount of force applied to each
ball increased, its compression size versus its initial size decreased.
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Figure 8
0.0095
19.1mm Ball Compression
vs. Force
Compression Length (m)
0.0094
0.0093
y = -0.0001x + 0.0095
R² = 0.9939
0.0092
0.0091
0.009
0.0089
0.0088
Compression Length (m)
0
1
2
3
Metric Tons
4
5
22.1mm Ball Compression
vs. Force
y = -7E-05x + 0.011
R² = 1
0.01092
0.0109
0.01088
0.01086
0.01084
0.01082
0
0.5
1
1.5
2
2.5
Metric Tons
25.4mm Ball Compression
vs. Force
0.0126
Compression Length (m)
0.01265
0.01255
0.0125
y = -1E-04x + 0.0127
R² = 0.828
0.01245
0.0124
0.01235
0
1
2
Metric Tons
3
4
Compression Length (m)
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50.8mm Ball Compression vs.
Force y = -3E-05x + 0.0254
0.02536
0.02534
0.02532
0.0253
0.02528
0.02526
R² = 0.9932
0
1
2
3
4
Compression Length (m)
Metric Tons
101.6mm Ball Compression
vs. Force
0.0508
y = -2E-05x + 0.0508
R² = 0.9763
0.05075
0.0507
0.05065
0
1
2
3
4
5
6
Metric Tons
Testing Hertz Theory Using the Data Collected
Earlier I made mention of using equation (4) to test Hertz theory. Here I will use this
equation and apply this equation to the data collected. What is expected is that if I take the ratio
of the area of contact versus the product of the velocity and the contact time, then it should
generate the radius of the ball multiplied by the constant 1.068. I decided to use the contact area,
the contact time, and the velocity for each ball at each height that I personally computed and
measured to obtain a theoretical value for the ball’s radius. I then compared the theoretical value
of the ball’s radius to the actual value of the ball’s radius.
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Table 3: Contact Area of each ball
Height
(cm)
19.1mm
22.1mm
25.4mm
50.8mm
101.6mm
50
.038 cm2
.011 cm2
.031 cm2
.096 cm2
.189 cm2
70
.035 cm2
.031 cm2
.031 cm2
.12 cm2
.292 cm2
80
.028 cm2
.028 cm2
.035 cm2
.04 cm2
.283 cm2
100
.023 cm2
.028 cm2
.031 cm2
.196 cm2
.37 cm2
120
.025 cm2
.031 cm2
.031 cm2
.139 cm2
.478 cm2
Table 4: Velocity of each ball at each Height
Height (cm)
Velocity (m/s)
50
3.13
70
3.71
80
3.96
100
4.43
120
4.85
Table 4 displays the
velocity of each ball
at each height prior
to their collision
with the steel plate.
Table 3 displays
the contact area, A
of each ball at each
height.
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Table 5: Calculated Radius of each ball
Height (cm)
19.1mm
22.1mm
25.4mm
50.8mm
101.6mm
50
.0052 m
.0012 m
.0026 m
.0043 m
.006 m
70
.0063 m
.0036 m
.0022 m
.005 m
.008 m
80
.0037 m
.0033 m
.0038 m
.0018 m
.0073 m
100
--------
.0033 m
.0033 m
.0094 m
.0098 m
120
--------
.0025 m
----------
.0048 m
.0085 m
The actual radii of the balls are .0096m (19.1mm ball), .0111m (22.1mm ball), .0127m
(25.4mm ball), .0254m (50.8mm ball), .0508 (101.6mm ball). The percent differences average
around 80 and 90 percent. I believe the percent differences are extremely high because of errors
such as retrieving a flawed contact time. Error from measuring the area of contact could also be
a major factor because some of the balls did not make a clear and uniform circle on the carbon
paper so it was slightly difficult to measure the radius of the circle.
Conclusion
After analyzing the calculated radii of each ball versus the actual radii and calculating a
percent difference for each ball at each height, each percent difference was rather large; they
averaged around 80 and 90 percent. However, due to the possibility of error in the experiment
and the equipment such as the oscilloscope providing faulty data, the measured value of the
balls’ radii were extremely inaccurate.
Nevertheless, I feel as though if the experiment was controlled in excess, such as a better
way to place the balls in free-fall, using a thicker more durable plate of steel since this plate bent
and dented under the pressure of the C-clamps and the balls constantly colliding into it, then the
data could come back fairly precise.
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References
Cross, Rod (1999) The bounce of a ball. University of Sydney, Australia: American Journal of
Physics, Vol. 67, Issue 3, pp. 222 August 1998.
Gugan, D (2000) Inelastic collision and Hertz Theory of Impact. H.H. Willis Physics
Laboratory, Royal Fort, Bristol BS8 1TL, United Kingdom: American Journal of Physics, Vol.
68, Issue 10, pp. 920-924 October 2000. http://oxfordcroquet.com/tech/gugan/
“coefficient of restitution." Dictionary.com Unabridged. Random House, Inc. 20 Apr. 2013.
<Dictionary.com http://dictionary.reference.com/browse/coefficient of restitution>.