Steel balls
Topic 3
Taiwan Representative
1
Outline
Question
 Experimental set up:

Vertical collision

Materials:
Paper, Plastic, and Metal sheets

Experimental results:
Hole size v.s materials, height…
What we learned & discussion
 Conclusion

2
Question

Colliding two large steel balls with
a thin sheet of material (e.g. paper)
in between may "burn" a hole in the
sheet. Investigate this effect for
various materials.
3
Experimental Setup: vertical collision
H
Sheet
1m
4
Materials

o
o
1. Paper (Burning point: 130 ~232 C)
2 mm
5 mm
Reference:http://wiki.answers.com/Q/What
_is_the_Burning_point_of_paper
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Inflammable materials
Paper
Burned
O2
Greatly heated
A thin piece of material
left on the ball
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Materials

2. Plastic (Melting Point: 106~114℃)
(Soften temperature: 80~98℃)
3 mm
4 mm
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Non-flammable materials
Plastic
Cooled down
Wrinkles
Melted
Compressed
Into thin layer
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Materials

3. Aluminum foil (Melting point: 660 C)
6 mm
5 mm
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Non-flammable materials
Aluminum foil
Melted
Compressed
Concentric circles
wrinkles
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Materials

4.Copper foil (Melting Point: 1083.0 °C)
2mm
5 mm
Reference:www.chemicalelements.com/elements/cu.html
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Non-flammable materials
Copper foil
Melted ??
Depends on condition
Compressed
Concentric circles
wrinkles
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Plastic
Paper
= 7.6 cm
H
= 7.6 cm
= 7.6 cm
= 7.6 cm
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Height v.s. Materials
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Height v.s. Materials
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Height v.s. Materials
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Height v.s. Materials
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Estimation of physical parameters
during collision
The highest temperature Tmax.= ?
Duration ∆t = ?
Collision area A=?
Average pressure P = ?
Conducted heat ∆Q = ?
Adiabatic process? i.e. ∆W >> ∆Q
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Tmax.= ?
 We can melt Al foil .
Melting point of Al = 660 ℃.
Cooled after melted
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Duration ∆t = ?
Our high speed video: 230 μs/frame
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t=0
Dropping ball
Edge of pillar
Material
Fixed ball
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t=230μs
Dropping ball
Edge of pillar
Material
Fixed ball
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t=230μs x 2
Dropping ball
Edge of pillar
Material
Fixed ball
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t=230μs x 3
Dropping ball
Edge of pillar
Material
Fixed ball
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Duration ∆t = ?
1. Our high speed video: 230 μs/frame
 ∆t < 230 μs
our mball=287 g
2. From reference
We can conclude :
∆t < 230 μs
R. Hessel, A. C. Perinotto, R. A. M. Alfaro,
and A. A. Freschia, Am. J. Phys. 74 3, 176 (2006).
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Collision area A=?
Estimate A from the melted Al foil .
Diameter ~ 1.6 mm
Diameter: 1.58± 0.08 mm
Al foil
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Average pressure P = ?
F p 1
P 

A t A
2 po  2m 2gH  p  m 2gH  po
m=287 g, H=50 cm, g=9.8 m/s2, A< π(0.8mm) 2 , ∆t < 230 μs
0.287 2  9.8  0.5
1
4
9 Nt
P

 1.9 10 ( 2 )
4
2
2.3 10
0.0008  3.14
m
P  1.9 10 atm
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Work ∆W=?
Pressure P > 1.2 x 109 Nt/m2
Collision Area A=πr2 ~ 2x10-6 m2
Thickness of sheet ∆d= 10-4 m
W  P  V  P  ( A  d )  3.9 101 ( J )
Conducted heat ∆Q=?
Q
   A  T
t
∆T < 1000 ℃
A=πr2 ~ 2 x 10-6 m2
σ
Air
0.026
Al
237
Cu
401
Fe
80.4
Steel
46
Water 0.61
Wood 0.11
Q  401 (2 106 ) 1000 2.3104 ~ 1.8 104 ( J )
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Heat produced
W  H  m  s  T
Using the lowest production of work,
W  0.39( J )
Increase ofΔT ,
T  640 C  T  25  640  665 C
o
o
=> Higher than melting point of Al: 660
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Adiabatic process??
Work done during collision ∆W > 3.9 ×10-1 (J)
 Heat conducted during collision :
∆Q < 2.9 ×10-4 (J)

Tlow
Thigh
∆W >> ∆Q
 Adiabatic process
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Conclusion


Height: Diameter of holes increases as
height increases
Collision Area:
 Burning
 Melting
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Colliding process
P
T
P
T
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Burning
O2
O2
33
Melting
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Conclusion


Height: Diameter of holes increases as
height increases
Collision Area:
 Melting
 Burning

Adiabatic Process is the main reason holes
are burned because:
 Δt
is small
 Area is small
 Pressure is high
 T is high
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Conclusion


Height: Diameter of holes increases as
height increases
Collision Area:
 Melting
 Burning

Adiabatic Process is the main reason holes
are burned because:
 Δt
is small
 Area is small
 Pressure is high
 T is high
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