Team of Russia Presented by Nikita Shanin Problem №15 «Slow Descent»: Design and make a device, using one sheet of A4 80 gram per m² paper that will take the longest possible time to fall to the ground through a vertical distance of 2.5m. A small amount of glue may be used. Investigate the influence of the relevant parameters.
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Team of RussiaPresented by Nikita Shanin
Problem №15 «Slow Descent»:Design and make a device, using one sheet of A4 80 gram per m² paper that will take the longest possible time to fall to the ground through a vertical distance of 2.5m. A small amount of glue may be used. Investigate the influence of the relevant parameters.
Plan of solution
Short analysis
Methods of solution definition
Physical model
Creation of devices, that meet demands of model
Perfection of a devices
Comparison of theoretical results
with practicalConclusion
Nikita Shanin IYPT 2011 2
Short analysis
Translational motion:
Ep = Ek trans. + Wdrag
Methods of problem’s solution
decline of velocity’s
vertical component
Nikita Shanin IYPT 2011 3
Short analysis
Translational and rotational motion:
+ Ek rotat.+ Wdrag rotat.
Ep = Ek trans. + Wdrag trans
Methods of problem’s solution
decline of velocity’s
vertical component
Rotation around
an axis
Methods of problem’s solution
decline of velocity’s
vertical component
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0 0.2 0.4 0.6 0.8 1 1.20
5
10
15
20
25
time t, s
ang
ula
r ve
loci
ty ω
, ra
dia
n/s
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Dependence ω(t)
Physical model
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g
?Weaknesses:• instability• fuzzy concept• governed by chance
g
Physical model
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g
Dependence ω on length of stiffener
2 pieces; landscape Length of stiffeners, cm ω, rad/s
0 10
0,5 15,7
1 20,93
1,5 22,43
2 25,12
2,5 22,43
3 19,03
2 pieces; landscape Time using stiffeners Time
6,1 2,7
6,2 2,9
5,9 3,4
5,7 3,2
5,8 3,6
Average time 5,94 ± 0,3 3,16 ± 0,6
0 0.5 1 1.5 2 2.5 30
5
10
15
20
25
30
stiffener’s length, cm
ang
ula
r ve
loci
ty ω
, ra
d/s
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Physical model
g
Fdrag rotat
F’drag rotat
Fdrag trans
Energy balance:Ep = Ek1 + Ek2 + Wdrag1 + Wdrag2
After some transformations:m(rω)2
24mgh = + + + mv2
2 α * v2 * S * l β * (rω)2 * r/4* 2π*N
Slm
mgh
**2/
N*2 *r/4 * )*(r * 2
ν =
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Devices which were used and their size
m
x
r m
r
xx
xformat r, m m, m x cm (ω*r)^2
А4 0,21 0,3 4 3,65
format r m m, m x cm (ω*r)^2
2 p. portrait 0,15 0,42 3 18,07
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m
rx
x
m
rx
x
format r m m, m x сm (ω*r)^2
3 p. portrait 0,1 0,63 2 5,41 4 p. portrait 0,075 0,85 1,5 7,68
Devices which were used and their size
format r m m, m x сm (ω*r)^2
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m
r
x
x
m
rx
x
format r m m, m x сm (ω*r)^2 format r m m, m x сm (ω*r)^2
2 p. landsc. 0,105 0,6 2 6,96 3 p. landsc. 0,07 0,9 1,5 5,96
Devices which were used and their size
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m
rx
x
format r m m, m x сm (ω*r)^2
4 p. landsc. 0,0525 1,2 1 0,07
Devices which were used and their size
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Comparison of theoretical results with practical ones
Slm
mghl
**2/
N*2 *r/4 * )*(r * /
2
t =
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0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.220
2
4
6
8
10
12
theoretical results
Practical results
length of smaller side, d m
Des
cen
t ti
me,
t,
s
ConclusionParameters of the device which take the longest
possible time to fall to the ground through a vertical distance of 2.5m:
• length of sides 15x42 cm.• length of stiffener 3 cm
The relevant parameters:• stiffeners• format of device
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Action of WFdrag
mr2
12I=
r
SWfdrag
trans.
= α * v2 * S * d
Wdrag rotat. = Mf *φ == β * (rω)2 * r/4* 2π*N
mgh = + + + mv2
2m(rω)2
24α * v2 * S * l β * (rω)2 * r/4* 2π*N
Slm
mgh
**2/
N*2 *r/4 * )*(r * 2
ν =
r is a length of a smaller sideω is an angular velocity of a deviceN is a number of complete turns at full time of descent
d is the distance which object has get overv is a linear velocity of a deviceS is a contact area between air and device