Small effects in the class-room experiments. Ivan Lomachenkov. Some physical projects have been realized at University centre of JINR. . Introduction 1. The main idea: let’s contrast the ’serious’ physics (the physics of microcosm) and the ordinary physics (”the physics at the kitchen”).
Small effects in the class-room experiments.Ivan Lomachenkov
Some physical projects have been realized at University centre of JINR.
The answer is YES.
Some criterions: a) the available and simple equipment; b) not complicated physical model of phenomenon; c) the opportunity to repeat phenomenon many times; d) class-room experiments in addition to basic course of physics; e) not only computer
animation of the phenomenon.
Physics at the kitchen
Part 1. Surface tension: the intensification of the molecular forces.
The surface tension forces support an elementary cell: F0= 4s× l. Summary surface tension forces support a sieve: F=F0×N, N=S/s0, N – the number of the elementary cells;
s0=1 mm2, S=pD2/4,
N»16000 ! – the intensification factor of the surface tension forces.
Equilibrium condition of the sieve:
4sl×N = G, G = (M+4m+3m0)g – the weight of all bodies, g – acceleration of gravity.G » 3.4 N.
We can extract the estimation for s from this
experiment: sext »53 mN/m.
The precise value is s=73 mN/m.
The reason of discrepancy: there is the partial
wetting between water and wire netting.
Strong pressure between
plates is induced by
pressure fall under
curved surface of water.
There is almost absolute
wetting between water
on a large scale
P- pressure inside of water;
P =P0 - 4s/d
d – thickness of water
P0=105 Pa, P=P0 – DP,
d»0.02 – 0.08 mm;
F=DP×S, S=0.13×0.18 m2;
Fmax »336 N!
34 kg !
We can hang up!
There’re two questions: a) can we increase the life-
time of the soap-bubble?; b) what’s the main reason which
restricts this time?
j0=70%, t0»1 min
j=85%, t»2 min
j=90%, t»2.5 min
j=95%, t»3.5 min
There’re in the class-room: j0»70%, t0» 1min.
In the frame of the simple model we can obtain the formula:
t= t0×(1 – j0)/(1 – j), j – the humidity of air alongthe pipe.
Let’s suppose: we’ve created the ideal
conditions for the soap-bubble (there
aren’t air flows and speck of dusts,
j=100% at al.). Can the soap-bubble
”lives” for ever?
The answer is NO.
process of diffusion
P=P0 + 4s/r, r – radius of the bubble
P0 - atmospheric pressure
According to observations the “life-time” of soap-bubble in closed
vessel may be more than 10 hours!This time drastically depends on
There’re two main reasons why the soap-bubble
can’t “live” for ever: a) the molecules of water slide
down on the surface of soap-bubble and the
thickness of the wall of bubble is decreasing drastically;
b) the pressure inside of the soap-bubble is greater
then atmospheric pressure by a factor 4s/r ( r-radius
of the bubble). Therefore there’s the process of diffusion
molecules of air outside of the soap-bubble
(“diffusion wasting away process”).
The objects of investigations are the air and
water streams. There are some opportunities
to intensify the oscillations of air stream
inside glass tube (Rieke’s effect) and to display
the structure of water stream. In addition to we
can discuss the influence of sound field on the
air flow (draught)
( L» 80 cm, Æ»35 mm)
There’s air flow through the tube forming of the standing wave
inside the tube. The heater provides the positive feed-back.
Dp=0 (node of pressure)
Dx=0 (displacement of air)
stage of pressure
stage of rarefaction
The positive feed-back extremaly depends on location
of the heater. There’s an effect (sound) only in case when the
heater is located in lower part of the tube.
In accordance with the experiments h=L/4.
l=2L – the wave-length of standing wave;
c – the velocity of sound in the air;
f0 = c/l = c/2L – the frequency of main
displacement of air
The directions are opposite: there’s the negative feed-back the oscillations of air will be suppressed.
The directions are the same: there’s the positive feed-back the oscillations of air won’t be suppressed.
displacement of air
stage of pressure
In this case the effect of the sounding
tube can’t be found. This experiment
demonstrates that really there’s
the pressure antinode in the centre of
the tube.The positive feed-back
There are some questions: a) can we observe the
process of disintegration of water stream?
b) can we influence on this process? C) can we
extract some physical quantities from these
There’s the capillary wave on the surface of water stream. The
direction of motion of the capillary wave is opposite the water stream
one. But the velocity of the capillary wave always equals the water
stream one: c = v. So we can observe the capillary wave like
the standing wave. The reason of the existence of the capillary waves
is the surface tension.
droppings structure of stream
There’s the simple estimation for l: l>9/2×r, r »0.5 mm –
radius of the nozzle. Hence l>2.25 mm.
It’s easy to determine the velocity of the stream: v » 2 m/s,
therefore c » 2 m/s.
According to the observations the resonance frequency of the
dropping process is about 300 Hz: fres » 300 Hz. Therefore we
can calculate the wave-length of the capillary wave: l = c/fres,
lobs » 6.6 mm.