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Steel balls

Topic 3

Taiwan Representative

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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

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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.

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Experimental Setup: vertical collision

1 m

H

Sheet

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Materials 1. Paper (Burning point: 130o~232oC)

Reference:http://wiki.answers.com/Q/What_is_the_Burning_point_of_paper

2 mm 5 mm

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Inflammable materialsPaper

A thin piece of material

left on the ball

Greatly heated

Burned

O2

Greatly heated

Burned

O2

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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 materialsPlastic

Compressed

Into thin layerMelted

Wrinkles

Cooled down

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Materials 3. Aluminum foil (Melting point: 660 C)

5 mm

6 mm

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Non-flammable materials

Aluminum foil

Concentric circles wrinkles

Melted

Compressed

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Materials 4.Copper foil (Melting Point: 1083.0 °C)

Reference:www.chemicalelements.com/elements/cu.html

5 mm2mm

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Non-flammable materialsCopper foil

Concentric circles wrinkles

Depends on condition

Compressed

Melted ??

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H

Plastic Paper

= 7.6 cm= 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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We can melt Al foil . Melting point of Al = 660 .℃

Tmax.= ?

Cooled after melted

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Duration ∆t = ?

Our high speed video: 230 μs/frame

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Material

t=0

Dropping ball

Fixed ball

Edge of pillar

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t=230μs

Material

Dropping ball

Fixed ball

Edge of pillar

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t=230μs x 2

Material

Dropping ball

Fixed ball

Edge of pillar

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t=230μs x 3

Material

Dropping ball

Fixed ball

Edge of pillar

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Duration ∆t = ?

1. Our high speed video: 230 μs/frame

∆t < 230 μs

2. From referenceour mball=287 g

R. Hessel, A. C. Perinotto, R. A. M. Alfaro, and A. A. Freschia, Am. J. Phys. 74 3, 176 (2006).

We can conclude : ∆t < 230 μs

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Collision area A=?

Al foil

Diameter ~ 1.6 mm

Estimate A from the melted Al foil .

Diameter: 1.58± 0.08 mm

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Average pressure P = ?

1F pP

A t A

m=287 g, H=50 cm, g=9.8 m/s2, A< π(0.8mm) 2 , ∆t < 230 μs

2 2 2 2o op m gH p m gH p

)(109.114.30008.0

1

103.2

5.08.92287.02

924 m

NtP

41.9 10P atm

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)(109.3)( 1 JdAPVPW

Work ∆W=?Pressure P > 1.2 x 109 Nt/m2

Collision Area A=πr2 ~ 2x10-6 m2

Thickness of sheet ∆d= 10-4 m

)(108.1~103.21000)102(401 446 JQ

TAt

Q

Conducted heat ∆Q=?∆T < 1000 ℃

A=πr2 ~ 2 x 10-6 m2

σ

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Heat produced

W H m s T

0.39( )W J

Using the lowest production of work,

Increase ofΔT ,

640 25 640 665o oT C T C => 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)

∆W >> ∆Q

Adiabatic process

ThighTlow

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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

O2O2

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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

high is T

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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

high is T