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Bill

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

back to its initial height after being dropped. To list a

few: the deformation of the ball and internal friction of

the ball during deformation (Cross, Rod 1999). The

degree of deformation that the ball experience is related

to the elasticity of the ball.

Hertz Theory of Elasticity of Steel Balls

By Free-Fall and Applied Pressure

Figure 1

Transverse

Force

Longitudinal

Force

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

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

Figure 1

Figure 2

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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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Figure 5: Example Oscilloscope reading

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.

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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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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

x

x’

Figure 6:

Uncompressed

Figure 7:

Compressed

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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.

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

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

Table 1: Contact Times at known heights

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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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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: Compression Size to Initial Size Ratio Percents

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

y = -0.0001x + 0.0095 R² = 0.9939

0.0088

0.0089

0.009

0.0091

0.0092

0.0093

0.0094

0.0095

0 1 2 3 4 5

Co

mp

ress

ion

Le

ngt

h (

m)

Metric Tons

19.1mm Ball Compression vs. Force

y = -7E-05x + 0.011 R² = 1

0.01082

0.01084

0.01086

0.01088

0.0109

0.01092

0 0.5 1 1.5 2 2.5

Co

mp

ress

ion

Le

ngt

h (

m)

Metric Tons

22.1mm Ball Compression vs. Force

y = -1E-04x + 0.0127 R² = 0.828

0.01235

0.0124

0.01245

0.0125

0.01255

0.0126

0.01265

0 1 2 3 4

Co

mp

ress

ion

Le

ngt

h (

m)

Metric Tons

25.4mm Ball Compression vs. Force

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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.

y = -3E-05x + 0.0254 R² = 0.9932

0.025260.02528

0.02530.025320.025340.02536

0 1 2 3 4Co

mp

ress

ion

Le

ngt

h (

m)

Metric Tons

50.8mm Ball Compression vs. Force

y = -2E-05x + 0.0508 R² = 0.9763

0.05065

0.0507

0.05075

0.0508

0 1 2 3 4 5 6

Co

mp

ress

ion

Le

ngt

h (

m)

Metric Tons

101.6mm Ball Compression vs. Force

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Height

(cm)

19.1mm 22.1mm 25.4mm 50.8mm 101.6mm

50 .038 cm2

.011 cm2 .031 cm

2 .096 cm

2 .189 cm

2

70 .035 cm2

.031 cm2

.031 cm2

.12 cm2 .292 cm

2

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 cm

2 .031 cm

2 .139 cm

2 .478 cm

2

Height (cm) Velocity (m/s)

50 3.13

70 3.71

80 3.96

100 4.43

120 4.85

Table 3 displays

the contact area, A

of each ball at each

height.

Table 3: Contact Area of each ball

Table 4: Velocity of each ball at each Height

Table 4 displays the

velocity of each ball

at each height prior

to their collision

with the steel plate.

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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.

Table 5: Calculated Radius of each ball

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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>.