4861.30 - Mechanics System 3

Instruction Manual and Experiment Guide

MECHANICS SYSTEM 3

N.B.:

• Pictures, images and descriptions in this manual may not exactly correspond with the actual items supplied.

• It is also important to note that the experiments in this manual are, only, suggestions. They are not meant to indicate the limitation of the equipment, which can be used in wide range of experiments, depending on the educational requirement of the teacher.

4861.30

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GENERAL DESCRIPTION: This kit introduces basic concepts of the mechanics of fluids. By using an air blower it’s possible to explore also experiments conducted with air. It provides a framework for understanding and quantitatively assessing many introductory fluidodynamics questions and problems. LIST OF EXPERIMENTS:

• Sensibility of a manometer • Communicating vessels • Hydrostatic pressure and Pascal’s law • Stevino’s law • Archimede’s law • Bernouilli’s equation • Torricelli’s theorem • Determination of the volume of a solid body • Determination of density and of specific weight of a solid body • Determination of density of immiscible liquids • Capillarity • Boyle’s law • Pumps and siphons • Adhesion e Cohesion • Cartesian diver • Measurement of surface tension • Perfect gas law • Viscosity

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Index of related topics: A

Adhesion and cohesion ..................................................................................................................33 Archimede’s principle......................................................................................................................26

B

Bernouilli’s equation ................................................................................................. 7; 42; 50; 56; 58 Bernoulli’s theorem .....................................................................................................................7; 16 Boyle’s law..................................................................................................................................26; 46 Buoyancy ..........................................................................................................................................26 Buoyant force ...................................................................................................................................61

C

Capillarity ..........................................................................................................................................30 Cartesian diver.................................................................................................................................26 Communicating vessels .....................................................................................................34; 37; 39

D

Density of a solid body ...................................................................................................................55 Density of two immiscible liquids ..................................................................................................44 Drag coefficient ................................................................................................................................61

F

Force pump ......................................................................................................................................53

H

Hagen-Poiseuille law ......................................................................................................................18 Hare’s apparatus .............................................................................................................................44 Hydraulic brake................................................................................................................................48 Hydrostatic pressure .......................................................................................................................39

J

Jurin’s law .........................................................................................................................................30

K

Kutta-Joukowski theorem...............................................................................................................56

L

Lift theorem ......................................................................................................................................56

M

Magnus force ...................................................................................................................................56 Mariotte’s bottle ...............................................................................................................................12

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O

Ostwald viscometer.........................................................................................................................18

P

Pascal’s law............................................................................................................. 12; 15; 39; 48; 50 Perfect gas law ................................................................................................................................59

R

Relative density of two non-mixable fluids ..................................................................................34 Reynold’s number ...........................................................................................................................61

S

Siphon ...............................................................................................................................................50 Stevino’s law ..................................................................................................................18; 26; 34; 37 Stoke’s formula ................................................................................................................................61 Surface tension ..........................................................................................................................30; 33

T

Terminal velocity..............................................................................................................................61 The Archimede’s principle..............................................................................................................23 The Gamow, Oppenheimer, Bloch puzzle ..................................................................................23 Torricelli’s theorem ......................................................................................................................7; 12

U

U-tube manometer ..........................................................................................................................29

V

Venturi’s tube ...................................................................................................................................42

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POTASSIUM PERMANGANATE (CHEMICAL SAFETY DATA): This kit contains potassium permanganate:

Common synonyms Condy's crystals

Formula KMnO4

Physical properties

Form: dark red to purple crystalline powder Stability: Stable, but decomposes if heated above 150 C. Melting point: ca 150 C (decomposes) Water solubility: moderate, produces solutions which are intensely coloured, even when quite dilute Specific gravity: 2.70

Principal hazards

This material is harmful if swallowed or inhaled. It is also harmful if absorbed through the skin. Potassium permanganate is a strong oxidizing agent and may react very exothermically with organic materials.

Safe handling Wear safety glasses and keep the solid or solution from contact with the skin.. Take care not to allow the solid to come into contact with flammable materials.

Emergency

Eye contact: Immediately flush the eye with plenty of water. Continue for at least ten minutes and call for medical help. Skin contact: Wash off with plenty of water. Remove any contaminated clothing. If the skin appears damaged, call for medical aid. If swallowed: Call for immediate medical help.

Disposal

Small amounts of very dilute potassium permanganate solution can be flushed down a sink with a large quantity of water, unless local rules prohibit this. More concentrated solutions and waste solid should be retained for disposal by those in charge of the laboratory.

Protective equipment

Safety glasses. Protective gloves should not normally be necessary. If they are to be used, nitrile will provide some protection, but may degrade upon contact with solid or solution, so should be checked regularly and replaced if damage is apparent.

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Experiment 1. RELATED TOPICS:

• Bernoulli’s theorem • Bernoulli’s equation • Torricelli’s theorem

The purpose of this experiment is to verify the Torricelli’s theorem (or the Bernouilli’s theorem since Torricelli’s theorem is a direct consequence) . ITEMS NEEDED:

• Mariotte’s Bottle • Chronometer (Optional) • Meter stick (Optional) • Calliper (Optional)

THEORY: Let us consider an ideal fluid (incompressible and without viscosity) in a channel at a given time t

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Focus the attention to the part of the fluid that, at a time t, is between section 1 (of area S1) and section 2 (of area S2). We can conjecture that the sections 1 and 2 are sufficiently small so that the velocity v, pressure p, and height z, with respect to an axis system, take the same values. In the time dt the fluid which is contained between 1 and 2 moves to 1’ and 2’. The work done by the gravity dWg is to displace the mass dm from height z1 to height z2: 1 1 2 2dm dV S v dt S v dtρ ρ ρ= = = (1) so 1 2( )gdW z z g dm= − (2) The work done by the pressures acting on the liquid surfaces is

1 1 2 2 1 1 1 2 2 2 1 2( )pdmdW F ds F ds S p v dt S p v dt p pρ

= − = − = − (3)

Analogously the kinetic energy variation for dm between 1 - 1’ and 2 – 2’ is

( )2 22 1

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dT dm v v= − (4)

From the energy conservation principle we can write g pdT dW dW= + (5) we have

2 2

1 1 2 21 22 2

p v p vz zg g g gρ ρ

+ + = + + (6)

that is called Bernouilli’s theorem or Bernouilli’s equation . This equation states that for every section of the channel we have

2

const.2

p vzg gρ

+ + = (7)

Let us apply this equation to the following system

It’s possible to prove that the lines of flux (red lines) start from the surface 1 and end in 2. If the hole in 2 is sufficiently small, the process may be considered as stationary. Let us apply the Bernouilli’s equation (6) between 1 and 2

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2

0 11 2

p vzg gρ

+ +2

02 2

p vzg gρ

= + + (8)

since the velocity v1 is very small if compared to v2 here called v. By posing h = z1-z2 we have

2v gh= (9) and this is called Torricelli’s theorem. Let us apply this theorem to the following figure

The liquid in the container decreases of a quantity S⋅dh in a time dt and, in the same time interval, from the hole flows out a quantity v⋅s⋅dt so, by applying the Torricelli’s theorem

2 2 22dh dhS vs R v r gh rdt dt

π π π= ⇒ = ⋅ = ⋅ (10)

from which we have

2 1 1

2 22

2g rdh h k hdt R

= = ⋅ (11)

where we have posed 2

2

2g rk

R= . The initial condition is h(0)=h0.

By solving the differential equation (11) we have

( )2 20 0

1( ) 4 44

h t h h kt k t= − + (12)

and by posing h(t)=0 we have

2

0 02

2 2h h Rtk g r

= = (13)

this is the time is needed to empty the container. APPARATUS SETTING AND PROCEDURE: Set up the apparatus as shown in the preceding figure. With a meter stick measure h0, with a calliper measure r and R and with a chronometer take the time t the container needs to empty.

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EXPERIMENTAL DATA: By applying equation (13) with h0 = 0.285 m, r=(0.003/2) m, R=(0.039/2) m, g=9.8 m/s2 we have t= 43 seconds. The graph of equation (12)is, for the preceding data, the following

10 20 30 40

0.05

0.1

0.15

0.2

0.25

The experimental value is 40 seconds which agrees with the theoretical prevision. SUGGESTION: Another possibility to verify the Torricelli’s theorem is to observe the parabolic path of the liquid jet from the hole of the container.

For the water jet falling down from the hole we have that:

212

h gt= − (14)

but x v t= ⋅ and so

2

2

12

xh gv

= − (15)

Hence, by measuring a couple of values x1, h1 it’s possible to recover v from equation (15)and to verify if equation (9) holds.

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Interesting is also the important fact that in equation (9)there’s no concern at all about the nature of the fluid. By repeating the experiment, with the same conditions, with another fluid instead of fluid the result must be exactly the same.

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• Mariotte’s bottle • Torricelli’s theorem • Pascal’s law

The purpose of this experiment is to introduce the concept of Mariotte’s bottle as an apparatus for obtaining a constant flux. ITEMS NEEDED:

• Mariotte’s Bottle THEORY: Mariotte siphons or Mariotte bottles are devices that provide a constant pressure that will deliver a constant rate of flow from closed bottles or tanks. The flow rate will depend upon the head as defined in the following figure and not on the height of the free water surface.

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These devices find many uses where a no changing pressure is needed. One application in agriculture is for applying liquid fertilizer to field crops. Because the fertilizer is applied at a constant rate to the field with this device, it is applied more uniformly and there is less of a chance of over fertilizing which can result in contaminating the groundwater. Let us consider the following figure:

The constant flux is given from the Torricelli’s theorem since 2v gh= where h is the difference in level between C (where the atmospheric pressure is established thanks to the Pascal’s law) and A. This difference in level remains constant until the level BB is above C.

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PROCEDURE: Verify the constant flux by observing the shape of the jet from the hole. What happens if C is under A? Here is a picture of this configuration

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• Pascal’s law

The purpose of this experiment is to introduce the concept of Pascal’s law ITEMS NEEDED:

• Pascal’s ball • Syringe • Connecting tube

THEORY:

In the physical sciences, Pascal's law or Pascal's principle gives the fluid pressure at mechanical equilibrium, on which only gravity forces are taken into account. More specifically, this law is actually a set of principles or laws discovered in 1648 by the scientist, Blaise Pascal, summarized as:

In a body of equally dense fluid at rest, the pressure is the same for all points in the fluid so long as those points are at the same depth below the fluid's surface.

In this case we can say:

Any change in pressure applied at any given point on a confined and incompressible fluid is transmitted undiminished throughout the fluid

PROCEDURE:

Fill the syringe with water, push the syringe’s plunger, and watch the water shoot out of every hole equally, even those in the back of the bulb.

This confirms also the independence of the pressure’s direction transmitted by a fluid.

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• Bernoulli’s theorem

The purpose of the experiment is to give another example (counter-intuitive) of the application of Bernouilli’s theorem. ITEMS NEEDED:

• Funnel • Styrofoam ball • Air compressor (Optional)

THEORY: Let us consider the following figure

When a fluid is flowing through a diverging channel as in a funnel the cross-sectional area available to the fluid increases. So it's velocity must decrease to conserve mass (for constant density) 2 2 1 1S v S v= (16)

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Since S2<S1 then v1<v2, otherwise said the air’s velocity must decrease when flowing from 2 to 1. Let us apply the Bernouilli’s equation:

2 2

1 1 2 21 22 2


+ + = + + (17)

by substituting equation (16) in (17) and by posing h = z2-z1 we have

2 2 2

1 2 2 2 22

12 2p S v p vhg gS g gρ ρ+ = + + (18)

from which

2 22 2

1 2 21

12gv Sp p p h gg S

ρ ρ⎛ ⎞

− = ∆ = − +⎜ ⎟⎝ ⎠

(19)

which is a positive quantity since S2<S1. So if we place a sphere at the apex there will be a large pressure drop across the sphere. The net pressure force will balance the weight of the sphere. APPARATUS SETTING: Place the ball under the funnel and blow air inside by observing the ball’s behaviour.

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• Ostwald viscometer • Hagen-Poiseuille law • Stevino’s law

The purpose of the experiment is to explain quantitatively the use of the Ostwald viscometer and to determine the relative viscosity coefficient. ITEMS NEEDED:

• Ostwald viscometer • Water, milk, oil (not included)

THEORY: If we apply a shear stress to an ideal fluid in a direction x the Hooke’s law is no more valid as in solids since the fluid starts to flow with velocity dx/dt in that direction. Anyway real fluids also oppose strains, however it is not the amount of strain that is important but the rate at which the strain is produced. When a real fluid flows, it has an internal resistance to flow, an internal friction, which is called its viscosity. Viscosity is a characteristic property of all real fluids. Let us consider a real fluid between two plates: the bottom plate is held at rest and to the upper is applied a shear stress F/S where S is the plate surface. What is experimentally found is that the liquid flow can be divided in many sheets of fluid that starts to move with a linear decreasing velocity from top as in the following figure:

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The viscosity effect manifest itself with a shear stress drag F/S applied to every single fluid sheet with surface S and velocity v in the opposite direction of motion. The sheets velocity is experimentally found to be v k r= (20) (where r is the normal to the fluid sheet) in some kind of fluids (called Newtonian fluids as air, water)

with1Fk

S η= .

By deriving equation (20) we found that the fluid sheets are subjected to a force drag

F dvkS dr

η η= = (21)

where η is called dynamic viscosity, dxvdt

= is the velocity of the fluid sheet. The forces acting on a fluid

sheet are the shear stress provoked by the hydrostatic pressure and the viscosity friction. Let us study this phenomenon in relation to the Ostwald viscometer

In particular we can focus our attention to the capillary part AC of the viscometer.

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In this case the Bernouilli’s equation doesn’t apply since viscosity effects are relevant. As a matter of fact, we got two different behaviour for a fluid is flowing in tubes with different diameters

Hence we have to be involved in a more subtle analysis because we have to consider the Poiseuille’s region. Let us indicate whit r0 the capillary radius, l its length and 1 2p p p∆ = − the pressure difference that push the liquid forward. We can imagine to divide the tube into many cylindrical coaxial layers. If r is the radius of one of this cylinders, dr the thickness and v(r) the velocity, the carrying capacity of the single cylindrical layer is 2dQ r dr vπ= ⋅ (22) The total carrying capacity of the tube is therefore

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0

0

2r

Q r v drπ= ∫ (23)

Beyond this, in stationary conditions the cylindrical column of liquid (the blue column in the picture) moves without acceleration so the force exerted by the pressure gradient must be equal to the drag force (see equation (21)) exerted by the liquid sheath that surround the column

( ) 21 2 2dv dvp p r S rl

dr drπ η η π− = − = − (24)

or

( )1 2 2dv rp pdr lη

= − − (25)

By integrating this differential equation with v=0 for r=r0, we get

( ) ( ) ( ) ( )0 2 20

1 2 1 202 4

r

r

r rrv r p p dr p pl lη η

−− = − − = − −∫ (26)

from which

( ) ( ) ( )2 20

1 2 4r r

v r p plη

−= − (27)

It’s clear how the velocity distribution is parabolic with a maximum on the capillary’s axis

( )2

0max 1 2 4

rv p plη

= − . If now we substitute (27) in (23)we get

( ) ( ) ( )0 2 2 40 1 2 0

1 20

24 8

r r r p p rQ r p p drl l

ππ

η η

− −= − =∫ (28)

Hence, in the time t the liquid volume streamed out is

( )1 2 4

08p p

V Qt r tl

πη−

= = (29)

which is the Hagen-Poiseuille law. If we use a liquid with a known viscosity η1 (e.g. water at a certain temperature) we have from (29)

411 0 18

p r tlV

η π∆= (30)

for a liquid with an unknown viscosity we have

422 0 28

p r tlV

η π∆= (31)

By dividing both members we get

1 1 1

2 2 2

p tp t

ηη

∆=∆

(32)

but for the Stevino’s law we have i ip ghρ∆ = and hence

2 2 2

1 1 1

tt

η ρηη ρ

= = (33)

which express the relative viscosity coefficient. APPARATUS SETTING: Fill the viscometer with water until a level F is reached. Wait few minutes and measure the temperature of the water inside the apparatus.

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PROCEDURE: By sucking on the branch with the capillary tube bring the water above the mark B. Let the water flowing down and measure the transit time between B and A. Repeat the same procedure for the viscosity unknown fluid (e.g. milk, oil) and measure again the transit time. After that, measure the density of the viscosity unknown fluid, and by knowing the density of water at the given temperature Its possible to apply equation (33). EXPERIMENTAL DATA:

• For water at 18°C we obtain a mean time of 77 seconds. • For milk (semi skimmed) at 18°C we obtain a mean time of 200 seconds. • Density of water at 18°C is 1008 Kg/m3 • Density of milk (semi skimmed) at 18°C is 1026 Kg/m3 (measured with a balance and a graduated

cylinder) • Hence by applying equation (33)we get, for semi skimmed milk η=2.64

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• The Gamow, Oppenheimer, Bloch puzzle • The Archimede’s principle

This is a problem that embarrassed physics greats George Gamow, Robert Oppenheimer and Felix Bloch. ITEMS NEEDED:

• Two transparent vessels one containing the other (not included) • Object more dense than water (not included) • Oil (not included)

THEORY: Imagine you are in a small boat with a large stone sitting in the bottom of it. The boat is floating in a swimming pool. What happens if you throw the stone overboard? Does the level of the water in the pool go up, down, or stay the same? As a warm up, what happens to the level of the boat on the water when the stone is thrown overboard? Well for the sake of simplicity we can start to think the boat and the pool having cubical shape

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If the boat (mass Mboat) is charged with a stone (mass Mstone) the gravity force acting on the boat is ( )g stone boatF M M g= + (34) the buoyancy force is given by the Archimede’s principle and is equal to A liquid liquidF V gρ= (35)

where ρliquid and Vliquid = l2⋅h3 are the liquid density and the displaced liquid volume. In equilibrium conditions we can write that gravity force must balance the buoyancy force or, by reasoning in terms of masses ( ) 2

3stone boat liquidM M l hρ+ = (36) If the boat is empty and the stone thrown overboard inside the pool

In this case the gravity force is g boatF M g= (37) and the buoyancy force 2

4A liquid liquid liquidF V g l hρ ρ= = (38) with the same reasoning of the preceding situation we have 2

4boat liquidM l hρ= (39)

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By substituting (39)in (36) we obtain ( )2

3 4stone liquidM l h hρ= − (40) and since Mstone is always positive, this means 3 4h h> (41) Lets take in consider the volume of displaced water. In the first case (stone onboard) we have that the volume of displaced water is 2 2

1 3h L h l− (42) In the second case (stone overboard) we have that the volume of displaced water is 2 2

2 4 stoneh L h l V− − (43) where Vstone is the stone’s volume.

But from equation (40) we have that 4 3 2stone

liquid

Mh hlρ

= − and inserting it into equation (43) we get

2 22 3

stonestone

liquid

Mh L h l Vρ

− + − (44)

By equating equation (44) with equation (42) we have

2 21 3h L h l− 2 2

2 3h L h l= − stonestone

liquid

M Vρ

+ − (45)

or

1 2 2 2 2

1 1stone stone stone

liquid stone liquid stone

M M Mh hL L Lρ ρ ρ ρ

⎛ ⎞− = − = −⎜ ⎟⎜ ⎟

⎝ ⎠ (46)

and since ρliquid < ρstone this means 1 2h h> (47) Now we can abandon the hypothesis on the cubical shape. As a matter of fact, exactly the same arguments remains if we substitute hi with f(hi) or g(hi) where f and g are monotonically growing functions expressing respectively the volume of the boat’s displaced water and the volume of the pool. Both f and g are depending on the shapes of the pool and boat. In this way we have (by calling object what we called earlier stone)

3 4( ) ( ) object

liquid

Mf h f h

ρ− = (48)

and

1 21 1( ) ( ) stone

liquid object

g h g h Mρ ρ

⎛ ⎞− = −⎜ ⎟⎜ ⎟

⎝ ⎠ (49)

this means that it’s always h3>h4 but h1>h2 only if the object is less dense than water. Otherwise it could be h1=h2 or h1<h2. PROCEDURE: Try first with an object more dense than water and observe the level difference on the pool and on the boat. Then try with water and finally with oil.

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• Cartesian diver • Archimede’s principle • Boyle’s law • Stevino’s law • Buoyancy

This is an experiment due to develop intuition for how fluids at rest behave under gravity and pressure forces. ITEMS NEEDED:

• Plexiglas tube with rubber stop without hole • Test tube with rubber stop with hole • Salt (not included)

THEORY: Changes in pressure change the volume of the air/test tube unit so that it alternately becomes more or less dense than the surrounding water. This experiment demonstrates the property of buoyancy. An object is buoyant in water due to the amount of water it displaces or ‘pushes aside‘. If the weight of water that is displaced by an object in water exceeds the weight of the object then the object will float. As you apply pressure to the long tube, you apply pressure to the air bubble in the test tube effectively reducing its size. As the bubble’s size reduces, the test tube becomes less buoyant and begins to sink.

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In general, thanks to the Archimede’s principle, the buoyant force B(z) on the diver at a depth z is ( ) ( )B z gV zρ= (50) where ρ is the density of water and V(z) volume of the air in the diver at depth z. (Note that, as a simplifying assumption, we’re ignoring the volume of the glass in the test tube, which we’ll assume is small relative to the volume of the air it holds.) By Boyle’s law we have

0 00 0 ( ) ( ) ( )

( )PVPV P z V z V zP z

= ⇒ = (51)

The pressure P(z) at a depth z is just the surface pressure plus the weight per unit area of the water above it, so (by the Stevino’s law) we have 0( )P z P gzρ= + (52) hence from (51)

0 0

0

( ) PVV zP gzρ

=+

(53)

and finally from (50)

0 0

0

( ) PVB z gP gz

ρρ

=+

(54)

The critical depth is the depth beyond which the weight of the diver exceeds the buoyant force, or

0 0

0diver

PVm g gP gz

ρρ

>+

(55)

solving for z, the critical depth is

0 0 0critical

diver

PV Pzm g gρ

= − (56)

From equation (7) it results that when 0divermVρ

≥ the Cartesian diver can’t sink. The unique way to observe

this is to increase the liquid density (e.g. by adding salt).

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PROCEDURE: Fill the long tube with water. Fill the test tube partially with water, then turn it upside down without spilling the water and put it in the long tube so it just floats at the top. Experiment a few times to get the right amount of air into the test tube. Eventually put a rubber stopper on the top of the long tube. Press down with hand or on the rubber stopper to see how the diver move up and down. After that, add salt in the tube and observe how is more and more difficult to sink the diver. EXPERIMENTAL DATA: In a test case V0 = 8.5 cm3, mdiver=17g, g=9.81 m/s2, ρ=1000 Kg/m3, P0=101300 Pa (N/m2), hence by applying equation (56) we get zcritical ≅ 5 m. Since the tube is lower than 5m there’s no way to loose the Cartesian diver. Conversely, by adding salt, it’s possible to make impossible the Cartesian diver sink.

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• U-tube manometer

The purpose of this experiment is to explain the functioning of the U-tube manometer. ITEMS NEEDED:

• U-tube manometer • Potassium permanganate

THEORY: The instrument is made up by a U-shaped glass tube fixed on a vertical panel as air gauge and differential air gauge. Fill the tube to a half with mercury or a liquid of smaller specific weight, depending on the sizes of the pressures to be measured. With an adequate tube connect one of the two limbs of the gauge to the vessel containing the fluid the pressure of which must be measured; the difference between that pressure and the atmospheric is determined (expressed in centimetres or millimetres of liquid) by reading the difference in level of the liquid between the two limbs:

If also the other limb is connected to a vessel the difference in level will show the difference in pressure between the fluids contained in the two vessel (differential gauge)

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• Capillarity • Jurin’s law • Surface tension

The purpose of this experiment is to introduce the concepts of capillarity. ITEMS NEEDED:

• Capillary tubes THEORY: The general laws of the heavy fluids static must be modified when the existence of phenomena linked to the cohesion force of the molecules that compose them is considered. In particular, the surface tension along the dividing line between an aeriform fluid A, a liquid B and the wall C of the container causes the bending of the surface at rest of B (fig 1).

It is said that the liquid “wets” (for instance water, fig 1a) or does not “wet” (for instance mercury, fig. 1b) the wall, depending if it results α<90° or α>90°, where α is the connecting angle. The surface tension is the cause of the existence of the phenomena of capillarity. For example, in a U-tube, where one of the two branches has the radius of some tenths of mm, the surface of the two branches will not be at the same level; the shape of the liquid surface in the capillarity will be concave or convex for α<90° and α>90° respectively (fig 2a and 2b).

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The surface tension τ of the liquid intervenes to determine the equilibrium of the column by exerting a pressure upward in the first case, downward in the second: the level in the capillary is therefore higher or lower than the level in the wider branch, depending if liquid wets or does not wet the wall

By referring us to the preceding figure we can indicate with τ the superficial tension acting on the perimeter’s unit length of the circle TT. The vertical component of the superficial tension is cosτ α . Hence, along all the liquid meniscus we have the force 2 cosrπ τ α⋅ , this force is directed upwards and raises the liquid column of height h. The volume of the raised liquid is evidently composed of two parts: one cylindrical part of volume 2r hπ and one crown shaped part (generally neglectable) around the meniscus of volume between MM and TT

32 31 4

2 3 3rr r r ππ π⋅ − = .

If ρ is the density of the liquid at the equilibrium we must have

( )3

22 cos3rr r h r gππ τ α π ρ

⎛ ⎞⋅ = − +⎜ ⎟

⎝ ⎠ (57)

and if α<<1 (hence cos α ≅1) we can write

23 2 2

gr grhh r ρ ρτ ⎛ ⎞= −⎜ ⎟⎝ ⎠

(58)

by supposing r<<h and hence by neglecting the crown shaped volume. This is the Jurin’s law. With this formula it’s possible to calculate the surface tension by measuring h, r and ρ. APPARATUS SETTINGS: Observe that the liquid reaches different levels in the different capillary tubes and that the differences in heights increase as the radius decreases, according to the Jurin’s law:

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EXPERIMENTAL DATA: With a tube of 1 mm. diameter it’s possible to observe an height of 15 mm. By applying equation (58) we get a surface tension τ of nearly 40 N/m (at 16 °C). The order of magnitude agrees well with the tabulated values.

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• Adhesion and cohesion • Surface tension

The purpose of this experiment is to illustrate the ITEMS NEEDED:

• Mercury (not included) • Distilled water (not included) • Potassium permanganate

THEORY: Surface tension is a phenomenon observed at a liquid-gas interface. It is caused by the difference in forces of adhesion and cohesion. The amount of surface tension is caused by the magnitude of the difference in these forces. The mercury shown in the picture at left has a high surface tension, resulting in small drops that are almost spherical. The larger drops are flattened because mercury has a high specific gravity (density), and gravity is acting on a larger volume of mercury. The image at right has two drops of distilled water, four drops of distilled water with green colouring added in a line through the centre, and an outer ring of distilled water with green colouring and dishwashing detergent added. The shape of the drops is an indication of the amount of surface tension. The drops with the highest domes have the greatest surface tension. So it appears the colouring enhanced the surface tension of the distilled water by increasing cohesion within the drop. Adding detergent relaxes the surface tension by decreasing the cohesion of water molecules for themselves. Surfactants are materials that decrease the surface tension of water. Surfactants are regularly used when mixing dissimilar liquids or liquids with dry powders. The decrease in cohesion allows easier and more thorough mixing

34


• Stevino’s law • Communicating vessels • Relative density of two non-mixable fluids

The purpose of this experiment is to study the Stevino’s law that allows us to get in touch with the communicating vessels principle. ITEMS NEEDED:

• U-tube THEORY: The pressure exerted within a liquid depends only from the free surface and from the liquid density. Let us consider the following figure:

35

We want to determine the hydrostatic pressure in the point p at a deep h with respect to the free surface of the liquid. If p0 is the pressure exerted from the environment to the free surface (atmospheric pressure) we can isolate a fluid cylinder of volume V and apply to it the first law of dynamics 0i

iF =∑ (59)

In the y-direction we have 1 2 3 0F F F+ − = (60) where

1 0F p A= , 2F mg Vg Ahgρ ρ= = = , 3F pA= . By substituting these values in equation (60)we have

0 0p A Ahg pAρ+ − = (61) or 0p p hgρ= + (62) which is called the Stevino’s law. By analyzing that law we can conclude that:

• The pressure p is the same in all the points at the same deep h • The pressure p is linearly growing with deep h • Whatever will be the vessel shape we have

or (another form of the Stevino’s law)

2 1p p g hρ− = ∆ (63)

PROCEDURE: Pour some liquid inside the U-tube (water with potassium permanganate is more clearly visible). If at the equilibrium there was a level difference between the two arms:

36

by applying equation (63)we would get a net force on the free surface and the equilibrium could not be reached. So the only solution for the equilibrium is to have the liquid at the same level in both the arms. Now pour two non-mixable fluids inside the U-tube (e.g. water and oil):

By applying equation (62) we get ' ' '' ''atm atmp gh p ghρ ρ+ = + (64) or

' '''' '

hh

ρρ

= (65)

hence by measuring h and h’ we can find the relative density of two non-mixable fluids.

37


• Stevino’s law • Communicating vessels

The purpose of this experiment is to get more involved with the communicating vessels principle. ITEMS NEEDED:

• Communicating vessels • Water (not included) • Oil (not included)

THEORY: Observing the level reached by the liquid in the four glass tubes of the apparatus, one can check the so-called communicating vessels principle. According to it, the level of the liquid surface in a set of tubes communicating with one another is the same in all tubes if the liquid contained in each tube is the same. That is a direct consequence of Stevino's law (see preceding experiments), according to which the pressure inside a liquid of density ρ at a point at depth h from the liquid surface is 0p gh pρ= + (66) where p0 is the atmospheric pressure on the liquid surface and g the local acceleration of gravity. In fact, since at equilibrium pressure must be the same at all points of the liquid at the same depth, the height of the liquid columns in the various tubes with respect to some horizontal plane must be the same. If the tubes about contain the same liquid, the heights of the surfaces are the same for the tubes containing the same liquid. APPARATUS SETTING: Lay the apparatus on a horizontal plane and fill it with water up to 4-5 cm from the edge of the tubes. PROCEDURE: It can be immediately verified that the level of the water surface is the same in all tubes and it remains so even if the apparatus is inclined (if the water level is such that in the second tube the surface is in the curved

38

section, the separating surface with air results slightly inclined; this depends upon the presence of a surface tension determining the patched angle between the glass surface and the water). By partially empty in the apparatus from the water and by substituting in a tube with a liquid non-miscible with water (for instance an oil) it can be seen that, while the level in the tubes containing water coincide, the level in the tube containing the other liquid is different:

This is exactly the same experiment we did before with the U-tube.

39


• Communicating vessels • Hydrostatic pressure • Pascal’s law

The purpose of this experiment is to show how the hydrostatic pressure at a given depth does not depend upon the shape of the vessel containing the liquid or the amount of liquid in the vessel.. ITEMS NEEDED:

• Communicating vessels THEORY: If the height of the fluid's surface above the bottom of the five vessels is the same, in which vessel is the pressure of the fluid on the bottom of the vessel the greatest ? The amount of liquid in each vessel is not necessarily the same. Let us imagine to have the following vessels:

The pressure P is the same on the bottom of each vessel. Why the pressure does not depend upon the shape of the vessel or the amount of fluid in the vessel rests upon three things:

• Pressure is force per unit area and this is not same as the total weight of the liquid in a vessel. • A fluid can not support its self without a container. Thus the walls of the container exert a pressure

on the fluid equal to the pressure of the fluid at that depth. • The pressure at given level is transmitted equally throughout the fluid to be the same value at that

level.

40

Vessel A No matter how wide the vessel, the pressure is just the weight of the fluid above unit area on the bottom. Even if you take the whole weight of the fluid in the container mg and divide by the area of the bottom A, you still get the same results since the vessel is equivalent to a column of water:

In formulas we have:

mg Vg AhgP ghA A A

ρ ρ ρ= = = = (67)

Vessel B Vessel B could be divided into three parts.

The fluid in parts 1 and 3 is supported by upward force of the vessel on the fluid. Part 2 could be though of a vertical column of liquid similar to vessel A. One could ask why doesn't the fluid in parts 1 and 3 (which are much bigger) not squeeze column 2 where they meet (along the dashed line) and there by increase the pressure on the bottom of the column? The answer is that the fluid in column 2 exerts and equal but opposite pressure outwards on the two other liquids to support itself. From the point of view of column 2, the water outside the dashed lines in sections 1 & 3 could be replace by solid vertical walls (along the dashed lines) and column 2 would still be in equilibrium. Vessel C Again we could divide the water into three sections.

41

The middle section is similar to that of vessel A or B. Since the height of the fluid in section 1 or 3 is not high enough to produce the same pressure as the height of the fluid in 2, how does the pressure on the bottoms of section 1 and 3 get to be the same as that of 2 ? The answer is that top of the container's walls in sections 1 and 3 produce a downward pressure that is equal to the fluid pressure in the middle section at the same level. If you poked a hole in the top of the container in sections 1 or 3, water would fountain upwards from the hole under pressure (see preceding experiment). From Pascal's principle, this pressure has to be that of the fluid in the middle section at the same level. Vessel D Again the center column is similar to vessel A.

The pressure of the vessel's wall creates a pressure that vertically supports the fluid in sections 1 and 3. At the same time the pressure of the walls create a horizontal component of pressure that sustains the fluid in the center column. Vessel E While one can not offer simple arguments like those for the other vessels, the pressure on the bottom is still the same basically because of Pascal's principle:

You go down from the surface to some depth, then move sideways until you can go down again. Repeat the process until you reach the bottom. Since the pressure at the same depth is the same, moving sideways does not change the pressure. Only downwards motion increases the pressure.

42


• Venturi’s tube • Bernouilli’s equation

The purpose of this experiment is to explain the venturi effect. ITEMS NEEDED:

• Venturi’s tube • Air blower

THEORY AND PROCEDURE: A fluid passing through smoothly varying constrictions experience changes in velocity and pressure. These changes can be used to measure the flowrate of the fluid. To calculate the flowrate of a fluid passing through a venturi, let us consider the following image:

From the Bernouilli’s equation we have

2 2

1 1 2 21 22 2


+ + = + + (68)

43

and since z1 = z2 we get

2 21 12 2a b b ap p p v vρ ρ− = ∆ = − (69)

From the continuity equation ( a a b bA v A v= ), the throat velocity vb can be substituted out of the above equation to give

2

22

1 12

aa

b

Ap vA

ρ⎛ ⎞

∆ = −⎜ ⎟⎝ ⎠

(70)

From equation (70) we see that in A the pressure is greater than in B:

Solving for the upstream velocity va and multiplying by the cross-sectional area Aa gives the volumetric flowrate Q,

2

2

2

1

a

a

b

ApQAA

ρ∆

=⎛ ⎞

−⎜ ⎟⎝ ⎠

(71)

Ideal, inviscid fluids would obey the above equation. The small amounts of energy converted into heat within viscous boundary layers tend to lower the actual velocity of real fluids somewhat. A discharge coefficient C is typically introduced to account for the viscosity of fluids

2

2

2

1

a

a

b

ApQ CAA

ρ∆

=⎛ ⎞

−⎜ ⎟⎝ ⎠

(72)

C is found to depend on the Reynolds number of the flow, and usually lies between 0.90 and 0.98 for smoothly tapering venturis. The mass flowrate can be found by multiplying Q with the fluid density.

44


• Hare’s apparatus • Density of two immiscible liquids

The purpose of this experiment is to determine the density of a liquid using the Hare’s apparatus. ITEMS NEEDED:

• Hare’s apparatus • 2 beakers • some liquid with unknown density

THEORY: The Hare’s apparatus consists of an upside-down U-tube the free ends of which can draw on two containers filled one with water and the other with the liquid whose density must be determined; on the bend of the U-tube there is a third shorter limb through which is possible to obtain a depression in the tube sucking off the air. That depression makes the liquids rise in their respective limbs so to balance the difference between pressure inside the tube and pressure outside. Now, since such difference has the same value for both liquids, being equal for both the value of the internal pressure (the produced depression) and that of the external pressure (atmospheric), from the comparison between the heights of the columns in the two limbs of the U-tube it is possible to derive the relative density respect with water of the liquid under exam. APPARATUS SETTING: Put the two glasses under the free ends of the U-tube as shown in the preceding picture; if necessary make the U-tube slide along its support. Pour water in one beaker and the liquid whose density we are looking for in the other, in such a quantity to wet the two ends for some centimeters. Suck from the rubber tube starting from the third limb of the U-tube.

45

PROCEDURE: Suck off air from the U-tube until an appreciable raise in the height of the liquids in the two limbs is obtained; using two fingers stop the rubber tube used to suck. By using a stick meter measure the heights of the two columns respect to the surfaces of the liquid in the two glasses: the density of the liquid under exam is given by the ratio between the measurements of the two heights, with that of water at the numerator. In fact the pressure, expressed in dine/cm2, exerted by a column of liquid with an height of h cm and a density of ρ grams/cm3 is equal to hρg where g is the gravity acceleration. Therefore if two columns of different liquids exert equal pressure, the following relationship will hold: 1 1 2 2h g h gρ ρ= (73) or

1 2

2 1

hh

ρρ

= (74)

otherwise said the inverse of ratio between two different liquids heights gives us the relative density.

46


• Boyle’s law

The purpose of this experiment is to determine the relationship between the volume and pressure for a fixed mass of gas at constant temperature (Boyle’s law). ITEMS NEEDED:

• U-tube manometer • Syringe • Potassium permanganate

THEORY: Boyle's law is a basic law in chemistry describing the behaviour of an ideal gas under a constant temperature. An ideal gas is a perfectly pure gas undergoing perfect elastic collisions with its container. No such gas actually exists, but real gases behave closely enough to ideal gases that we can use theories applying to the latter to describe the former. Boyle's law states that at a fixed temperature the volume of a gas is inversely proportional to the pressure applied to it. In other words, the pressure of a gas times its volume is a constant PV const= (75) It is not necessary to know the exact value of this constant to meaningfully understand the behaviour of gases. The law was discovered by Robert Boyle in 1662. The standard measurements for volume and pressure are cubic meters and pascals (or atmospheres). One obvious example of Boyle's law in action is in a syringe. In a syringe, the volume of a fixed amount of gas is increased by drawing the handle back, thereby lessening the pressure. The blood in a vein has higher pressure than the gas in the syringe, so it flows into the syringe, equalizing the pressure differential. Boyle's law is one of three gas laws which thoroughly describe the behaviour of gases under varying temperatures, pressures and volumes. Another way of describing Boyle's law is that when you push a gas, it tends to push back. Without the massive amount of gravity holding them together, the solar system's gas planets would rapidly diffuse in all directions, quickly depressurising. There is a limit to the amount we can compress any given gas sample, because eventually the pressure becomes so great that it bursts out of any container we can create for it.

47

PROCEDURE: Connect the syringe to the U-tube manometer previously filled with coloured water (use the potassium permanganate)

Take care that the syringe has no air losses in the connection point. While pushing on the syringe plunger, record simultaneously the volume of the syringe expressed in ml and the pressure on the column of coloured water expressed in mm. You should obtain a graph like this:

10 20 30 40Volume

0.2

0.4

0.6

0.8

1

Pressure

48


• Hydraulic brake • Pascal’s law

The purpose of this experiment is to explain the principle of the hydraulic brake. ITEMS NEEDED:

• 2 syringes • connecting tube for syringes • 2 extension clamp • universal base • base clamp • weights (not included)

THEORY: One of the technological application of the hydraulic press, and therefore of the Pascal’s law, is the hydraulic transmission brake that is used in many machines. In this kind of brake a force f, of very small intensity is exerted on a plunger.

The associated pressure P propagates along an hydraulic circuit until it reaches a second piston, normally of much larger section.

49

The ratio between the intensity of the thrust F of this second piston and that of f is equal to the ratio A/a between the surfaces of the plungers, that is between the squares of the radii in the case of cylindrical plungers. PROCEDURE: The two syringes (and the connecting tube) must be filled with water eliminating completely the air. After that, by using suitable weights, the preceding equation can be verified:

F A Fg A M Af a fg a m a= ⇒ = ⇒ = (76)

where M is the mass acting on A and m is the mass acting on a.

50


• Siphon • Bernouilli’s equation • Pascal’s law

The purpose of this experiment is to explain the principle of the siphon. ITEMS NEEDED:

• 2 beakers • silicon tubing

THEORY: A siphon (also spelled syphon) is a continuous tube that allows liquid to drain from a reservoir through an intermediate point that is higher than the reservoir, the up-slope flow being driven only by hydrostatic pressure without any need for pumping. It is necessary that the final end of the tube be lower than the liquid surface in the reservoir. Once started, a siphon requires no additional energy to keep the liquid flowing up and out of the reservoir. The siphon works because the ultimate drain point is lower than the reservoir and the flow of liquid out the drain point creates a vacuum in the tube such that liquid is drawn up out of the reservoir. The maximum height of the intermediate point (the crest) is limited by atmospheric pressure and the density of the liquid. At the high point of the siphon, gravity tends to draw the liquid down in both directions creating a vacuum. Atmospheric pressure on the top surface of the higher reservoir is transmitted through the liquid in the reservoir (Pascal’s law) and up the siphon tube and prevents a vacuum from forming. When the pressure exerted by the weight of the height of the column of liquid equals that of atmospheric pressure, a vacuum will form at the high point and the siphon effect ended. For water at standard pressure, the maximum height is approximately 10 m; for mercury it is 76 cm. An analogy to understand siphons is to imagine a long, frictionless train extending from a plain, up a hill and then down the hill into a valley below the plain. So long as part of the train extends into the valley below the plain, it is "intuitively obvious" that the portion of the train sliding into the valley can pull the rest of the train up the hill and into the valley. What is not obvious is what holds the train together when the train is a liquid in a tube. In this analogy, atmospheric pressure holds the train together. Once the force of gravity on the couplings between the cars of the train going up the hill exceeds that of atmospheric pressure, the coupling breaks and the train falls apart. The train analogy is demonstrated in a "siphon-chain model" where a long chain on a pulley flows between two beakers. A plain tube can be used as a siphon. An external pump has to be applied to start the liquid flowing and prime the siphon. This can be a human mouth and lungs. This is sometimes done with any leak-free hose to

51

siphon gasoline from a motor vehicle's gasoline tank to an external tank. If the tube is flooded with liquid before part of the tube is raised over the intermediate high point and care is taken to keep the tube flooded while it is being raised, no pump is required. Devices sold as siphons come with a siphon pump to start the siphon process. Among some physicists there is some dispute as to what causes the siphon to lift liquid from the upper reservoir to the crest of the siphon. They argue that theoretically, internal molecular cohesion is sufficient to pull the liquid up the intake leg of the siphon to the crest. Furthermore, some argue that theoretically a siphon will operate in a vacuum. In practice atmospheric pressure is required, to prevent the liquid from boiling. Bernoulli's equation may be applied to a siphon to derive the flow rate and maximum height of the siphon.

Let the surface of the upper reservoir be the reference elevation. Let point A be the start point of siphon, immersed within the higher reservoir and at a depth −d below the surface of the upper reservoir. Let point B be the intermediate high point on the siphon tube at height +hB above the surface of the upper reservoir. Let point C be the drain point of the siphon at height −hC below the surface of the upper reservoir. The Bernouilli’s equation is

2

const.2

p vzg gρ

+ + =

Applying Bernoulli's equation to the surface of the upper reservoir we get

200 const.

2atmpg gρ

+ + = (77)

52

since we said that the surface of the upper reservoir is the reference elevation (z=0) and v that is the velocity of the upper reservoir’s surface may be set to zero Apply Bernoulli's equation to point A at the start of the siphon tube in the upper reservoir where p = pA, v = vA and z = −d

2

const.2

A Ap vdg gρ

− + + = (78)

Apply Bernoulli's equation to point B at the intermediate high point of the siphon tube where p = pB, v = vB and z = hB

2

const.2

B BB

p vhg gρ

+ + = (79)

Apply Bernoulli's equation to point C where the siphon empties. Where p = pC=patm, v = vC and z = - hC

2

const.2

atm CC

p vhg gρ

− + + = (80)

As the siphon is a single system, the constant in all four equations are the same. Setting equations (77) and (80) equal to each other gives

2

2atm atm C

Cp p vh

g g gρ ρ= − + + (81)

or

2C Cv gh= (82) The velocity of the siphon is thus driven solely by the height difference between the surface of the upper reservoir and the drain point. The height of the intermediate high point, hB, does not affect the velocity of the siphon. However, as the siphon is a single system, vB = vC and the intermediate high point does limit the maximum velocity. The drain point cannot be lowered indefinitely to increase the velocity. Equation (79) will limit the velocity to a positive pressure at the intermediate high point to prevent cavitations (the phenomenon where small and largely empty cavities are generated in a fluid, which expand to large size and then rapidly collapse, producing a sharp sound). The maximum velocity vC=vB=vmax may be calculated by combining equations (77) and (79) with pB=0:

2

max02

atmB

p vhg g gρ ρ= + + (83)

or

max 2 atmB

pv g hgρ

⎛ ⎞= −⎜ ⎟

⎝ ⎠ (84)

and since this quantity must at least be 0 we have that

0 atmB

phgρ

≤ ≤ (85)

This gives us the maximum height that a siphon will work. It is simply when the weight of the column of liquid to the intermediate high point equates to atmospheric pressure. Substituting values for water will give 10 metres for water and 0.76 metres for mercury. PROCEDURE: Try to realize a siphon with the items supplied.

53


• Force pump

The purpose of this experiment is to explain the principle of the force pump. ITEMS NEEDED:

• Force pump THEORY: Pumps are machines that through work provide potential and kinetic energy to a liquid

54

A and B are two valves, that is devices that permit the liquid flow in one direction only. Raising the piston, a depression is created in the chamber C1 and the water, pushed by the atmospheric pressure, opens the valve A and it enters inside the chamber. The next compression of the piston first causes the closing of valve A, then the opening of valve B and the consequent passage of the liquid from chamber C1 to C2. Repeating the cycle until chamber C2 is entirely filled, in the compression phase of the piston the liquid comes out. APPARATUS SETTING: Fill a small tank of adequate dimensions with water and immerge the suction tube in the liquid. PROCEDURE: Try to verify the functioning principle of the force pump.

55


• Density of a solid body The purpose of this experiment is to explain the difficulties underlying this apparently simple experiment. ITEMS NEEDED:

• Electronic balance (not included) • Meter stick (not included) • Vernier calliper (not included)

THEORY: See “Basics of the experimental error theory” in the appendix.

56


• Bernouilli’s equation • Magnus force • Kutta-Joukowski theorem • Lift theorem

The purpose of this experiment is to explain a dramatically clear effect due to an underlying complex theory starting from the Bernouilli’s equation. ITEMS NEEDED:

• Air blower • Voltage regulator • Styrofoam ball • Ping pong ball

THEORY: The details of how a spinning ball creates lift are fairly complex. Next to any surface, the molecules of the air will stick to the surface. This thin layer of molecules will pull the surrounding air for a spinning ball in the direction of the spin. If the ball were not moving, we would have a spinning, vortex-like flow set up around the spinning ball. If the ball is moving through the air at some velocity, on one side of the ball the air will oppose the flow. On the other side of the ball, the air will flow in the same direction. The spinning ball will then turn the flow, and a force will be generated. Because of the change in the air velocity, the air pressure will also be altered around the ball. The magnitude of the force can be computed by multiplying the surface pressure times the area around the ball. The direction of the force is perpendicular to the flow direction. This is due to the following qualitative reasoning. Assume you have a sphere around which a flow streams. The flow velocity is denoted by U. The pressure on the upper part of the sphere is denoted by pt and at the bottom part pb. According to Bernoulli’s equation the pressure difference is given by

( ) ( )( ) ( )2 2

2 2b t t b t b t b t bp p v v v v v v U v vρ ρ ρ− = − = + − = − (86)

since tv U Rω= + and bv U Rω= − where R is the sphere radius and ω the angular velocity. The force acting on the sphere (or Magnus force), L, called also lift if the force acts upwards, is obtained by integrating along the upper part of the sphere. The integration here should be intended just as symbolic since we should perform some coordinates system change:

( )2

0t bL U v v

π

ρ= −∫ (87)

A close inspection of the integral reveals that it corresponds to the negative circulation around the sphere. The circulation is given by (always symbolic):

57

( )2 0 2 2 2

0 2 0 0 0

= b t b t t bv v v v v vπ π π π

π

Γ + = − = − −∫ ∫ ∫ ∫ ∫ (88)

Hence, we can establish a relation between the lift and the circulation: L Uρ= − Γ (89) this is called lift theorem or Kutta – Joukowski theorem. This implies that if there is lift, the circulation is non-vanishing and negative, and the flow velocity above the sphere is larger than beneath the sphere. PROCEDURE: Incline of an adequate angle the air blower and turn it on at a fixed air speed. This is important to keep more evident the fact that there’s a lift acting on the ball.

Try to release from your hands the styrofoam ball in order to keep it self stable in the air flux produced by the air blower. If the air flux is too strong or too weak please act on the voltage regulator. Observe how the ball is spinning. The best think to visualize it is to mark a point on the ball with a pencil. Observe how the ball is in a potential well due to the combinations of all the forces acting on it since if we gently touch the ball, it recovers its original position. Observe the differences by using the ping pong ball instead of the styrofoam one

58


• Bernouilli’s equation

The purpose of this experiment is to show another effect that implies the Bernouilli’s equation. ITEMS NEEDED:

• Air blower • Voltage regulator • Toilet paper (not included)

THEORY: The pressure difference obtained by the Bernouilli’s equation can be visualized also by using the air blower with a roll of toilet paper.

59


• Perfect gas law

The purpose of this experiment is to verify the perfect gas law validity by measuring the force needed to pull the piston at a certain distance ITEMS NEEDED:

• Syringe • Data-acquisition system (not )

THEORY: Let us consider the following picture, where the syringe is kept in a vertical position and the temperature is assumed constant

If we start from the perfect gas law PV NkT= (90) and we apply it to the air contained in the syringe we have

60

inside airNkT NkTPV xS

= = (91)

from which

inside airNkTF

x= (92)

By applying Newton’s second law it results 0inside air atmF F P S+ − = (93) or, by substituting equation (92) in equation (93)

0

atm

atm

NkTF P S NkTx xP S

= −− +

(94)

where we have rescaled the force since we want that when x = x0 , F= 0 APPARATUS SETTING: With the syringe’s cap open place the piston at a fixed distance x0. Hold the syringe with one hand while keeping the cap closed with a finger. With the other hand pull the force sensor until the desired position on the scale is reached. At this point keep the experimental data of the force. EXPERIMENTAL DATA: Here’s a graph of the data collected with a force sensor

from which it’s clear the hyperbolic behaviour predicted by equation (94).

61


• Reynold’s number • Drag coefficient • Stoke’s formula • Buoyant force • Terminal velocity

The purpose of this experiment is to show a uniform rectilinear motion by using a tube and an air bubble. ITEMS NEEDED:

• Perspex tube (100 cm) • Water (not included)

THEORY: One of the classic problems of physics concerns a spherical object moving vertically through a resistive medium. One may think for instance of a cannon ball shot up into the air, or of a grain of sand sinking slowly to the bottom of a lake. If the resistive force F(v) is either linear or quadratic in the velocity v, the problem admits an analytical solution. First of all, however, we discuss when these two particular forms of resistive force actually occur. Both correspond to a specific interval of the Reynolds number and can be derived from the following general formula:

( ) 212 dF v C Avρ= (95)

Here Cd is the so-called drag coefficient, ρ the density of the medium, A the object’s cross-sectional area (in the case of a sphere 2Rπ ), and v its velocity. It should be noted that equation (95) is actually only valid in the case of constant v, the so-called stationary situation, and that in general (if v varies) one should also take into account the Boussinesq–Basset viscous memory force and the ‘‘added mass’’ term. These terms are especially important if the density and the viscosity of the medium are large. On the other hand, they complicate the calculations to such a high degree that we choose to ignore them. The most intriguing element in equation (95) is the drag coefficient; it depends in a complicated way on the Reynolds number (Re), as depicted in the following figure

62

This figure can be found in almost any textbook on hydrodynamics. So in order to use the above equation one first has to know the value of Re. This dimensionless number is defined as follows:

Re lvρη

= (96)

where l represents the characteristic length scale of the object in the cross-sectional plane (in the case of a sphere this is just the diameter 2R) and η the dynamic viscosity of the medium. In the hydrodynamic literature one finds several formulas to describe the curve of the preceding figure, most of which are restricted to a relatively small range of Re. An exception to this rule is the following curve-fit formula which holds for all 0<Re<2x105:

24 6(Re) 0.4Re 1 RedC ≈ + +

+ (97)

In the limit for very small Reynolds number, say Re<1, the first term dominates and we may ignore the second and third terms.

With 24 24 12(Re)RedC lv Rv

ηρ ρη

= = = and 2A Rπ= the general formula in equation (95) then reduces to

2 21 12( ) 62

F v R v RvRvη ρπ πη

ρ= = (98)

This is known as Stokes’ formula. It tells us that the drag on a (very) slowly moving sphere is linearly proportional to its velocity. The formula applies to the sinking grain of sand mentioned above, to micro-organisms in water, to tiny dust particles floating in air and we try to use it to an air bubble rising in a tube filled with water. Generally speaking, though, the condition Re<1 is rather restrictive and rarely met in practice. This has not prevented the linear drag force or ‘‘viscous friction,’’ with its appealing simplicity and nice theoretical properties (e.g., it can be incorporated in the Lagrangian formalism), from becoming the favorite type of damping in the physics literature. If we imagine an air bubble like a sphere rising in a tube filled with water along a vertical path, the equation of motion reads

( ) 6air air waterdvm m m g Rvdt

πη= − − − (99)

Here mwaterg, with mwater the mass of the displaced water is the buoyant force. The above equation may also be written as

61 water

air air

mdv Rg vdt m m

πη⎛ ⎞= − − −⎜ ⎟

⎝ ⎠ (100)

or equivalently

dv g vdt

α= − −% (101)

where g% (the effective gravitational acceleration) is given by

1 1water water

air air

mg g gm

ρρ

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

% (102)

which is clearly a negative quantity since water airρ ρ>> . Moreover we have, with 343air airm Rπ ρ=

2

6 92air air

Rm Rπη ηα

ρ= = (103)

Equation (101) is readily integrated [with initial condition v(0)=0] to give the velocity

( ) tg gv t e α

α α−= −

% % (104)

63

This looks quite different from the solution without drag but if we take the Taylor expansion we have 2 3( ) ( )v t gt gt o tα= − + +% % (105) and for 0, g gα → →% indeed reduces to the frictionless velocity v gt= − .

Equation (104) shows that the velocity during the upward journey cannot grow beyond gα

−%

, the so-called

terminal velocity

( ) 22

9water air

term

gRgvρ ρ

α η−

= − =%

(106)

this also follows directly from equation (101), when dv/dt is set equal to zero. Let us consider a water density of ρ = 103 Kg/m3 , a viscosity of η = 10-3 Kg/ms and an air density of 1.29 Kg/m3. If we apply equation (106) we obtain a terminal velocity 6 22.18 10termv R= × (107) With this velocity the Reynolds number becomes

12 3Re 4.35 10lv Rρη

= = × (108)

which means that only sufficiently small air bubbles with radius R < 0.6 x 10-4 m will raise according to the Stoke’s law. If, as in our case, this condition is not verified we have to use another expression for the drag coefficient. The second interval we focus upon is 103<Re<2x105 where the drag coefficient is seen from equation (97) to be approximately constant: Cd~0.4. In that case equation (95) reduces to ( ) 2 20.2F v R vρπ= (109) So in this regime the drag on the sphere is quadratic in the velocity. It applies for instance to a pebble dropped from the Leaning Tower of Pisa (except for a fraction of a second at the very start) or a sky diver and is commonly known as ‘‘air drag.’’. Now the equation of motion reads

( ) 2 20.2air air water waterdvm m m g R vdt

ρ π= − − − (110)

this equation can be written as follows

2 2

3

0.2 314

water water

air air

R vdv gdt R

ρ ρ πρ ρ π

⎛ ⎞ ⋅= − − −⎜ ⎟

⎝ ⎠ (111)

or

( )2 21dv g vdt

γ= − +% (112)

where

2 0.2 3

4 1air

water

R gγ

ρρ

⋅=

⎛ ⎞−⎜ ⎟

⎝ ⎠

(113)

By posing 0dvdt

= in equation (112) we have that the terminal velocity is

22

4 11

0.2 3

air

waterterm

R gv

ρρ

γ

⎛ ⎞−⎜ ⎟

⎝ ⎠= − =⋅

(114)

or

64

6.67 1 airterm

water

v gRρρ

⎛ ⎞= −⎜ ⎟

⎝ ⎠ (115)

Let us consider the same values as before: water density of ρ = 103 Kg/m3 ,and an air density of 1.29 Kg/m3. If we apply equation (106) we obtain a terminal velocity

8.08termv R= ⋅ (116) With this velocity the Reynolds number becomes

3

7 2Re 1.62 10lv Rρη

= = × (117)

which means that only air bubbles with radius 1.5 mm < R < 53 mm will raise according to the quadratic drag. This is our case. For example if we consider an air bubble of 2 mm we have from equation (115) that vterm is nearly 36 cm/s. PROCEDURE: Try to verify the preceding equations with the perspex tube filled with water and an air bubble.

65

APPENDIX Basics of experimental error theory

We can say that everything we known about the physical world has an inherent uncertainty. In particular, when we experimentally investigate something there is always an “experimental error” and an “experimental precision”. Since one of the main features of experiments is their reproducibility, it is very important to deal with this subject in order to be able to explain how good our results are. This is possible with “experimental error theory”, a scientific approach to this problem.

Let us consider the following example: find the density of a solid rubber cube.

o First trial (with very raw instrument). We can estimate that the mass of the cube is nearly 50 g and the length of a side is nearly 6 cm. So the density would be:

3 0,23148...M MV L

ρ = = = . There are many questions: “Where can I stop with

decimal digits to communicate my result?” “Is it better to have precision on the mass measurement or on length measurement?” “How do we combine our experimental error on the mass measurement with the experimental error on length measurement?”

o Second trial (with more accurate instruments). By using an electronic balance and a meter stick I find a mass of 60g and a side length of 5,4 cm. So the density would be:

3 0,381039475...M MV L

ρ = = = . We still need to answer the questions posed above

but we also have to answer a new question: “What makes this trial better than the first one?”

o Third trial (with much more accurate instruments to measure the side length). If we improve the accuracy of the length measurement, for example by using a vernier caliper, the problem becomes more involved. This is due to the fact that we do not get the same result if we make more than one measurement. Instead we have a set of different measurements like (5,455 cm; 5,425 cm; 5,465 cm; …). So we are again faced with the question: “Which one of the measurements (5,455 cm; 5,425 cm; 5,465 cm; …) do should I take?”

Therefore, the more we analyse the problem the more it gets involved. To search for a possible solution we can start from the third trial and observe that, generally speaking, when we improve the accuracy of an instrument we reach a point at which the experimental results are not unique but are scattered around some value as illustrated in this graph:

66

If the number of measurements N is greater then about 30, the distribution of the experimental data is bell-shaped and has a value X for which there is a maximum and around which the data are scattered in a nearly symmetrical way. It is also possible to distinguish a value that determines an interval around X into which a significant percentage of the measurements are placed. We need to answer the questions: “Is X the best estimate of our measurement?”, “How much can we rely on this value?” and “What percentage of the measurements are in the interval X-σ and X+σ ?”

To express these questions mathematically, we could try a prototype function that fits our data and that expresses the probability to get a particular measurement value:

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1

this is the graphical representation of the function

2

( ) xf x e−= If we want to centre the function around the value X we use the expression x-X in place of x, and if we want to control the scattering of the measurements around X it’s possible to divide (x-X)2 by 2σ2. The following figure shows f(x) with X=2 and σ=1;1.5;2

67

-2 2 4 6 8 10

0.2

0.4

0.6

0.8

1

-2 2 4 6 8 10

0.2

0.4

0.6

0.8

1

-2 2 4 6 8 10

0.2

0.4

0.6

0.8

1

Finally, if we want to control the area under the curve we have to multiply it by a normalization factor Aσ that would depend on σ . Therefore our prototype function is:

( )2

22( )x X

f x A e σσ

−−

⋅= ⋅ (118)

where X is the value for which we have the maximum and σ determines how the measurements are scattered around X. This is called a “Gaussian function” or a “Normal function”, but the underlying data represent a distribution (still called Gaussian) and not a function. It can be proved that the Gaussian distribution is derivable from the binomial distribution assuming that the number of measurements N ∞ and σ remains constant. The physical meaning of all this is that we do not describe a measurement with a single number but rather with a set of values each one with its own probability to appear as an experimental datum. This probability is governed by the “Gaussian distribution”. There is an analogy with quantum mechanics (for example with the wave packet of a particle) where the interpretation is that if we make a measure of the position of the particle then the probability to obtain a particular value is governed by the Gaussian function and is never a well defined fixed value. Let us determine the value of Aσ in (118). We must have a probability of 1 to get a measurement in the range from -∞ to +∞ (that is, if we perform a measurement we are certain to get some kind of result no matter how large or how small that result is):

1( ) 12

f x dx Aσ π σ

+∞

−∞= ⇒ =

⋅∫

To give an interpretation of σ we can ask what happens if we are only interested in the probability of finding measurements in the range from X-σ to X+σ instead of the range from -∞ to +∞:

21

21

1( ) 0.682

tX

Xf x dx e dt

σ

σ π+ + −

− −= ⋅∫ ∫

so σ, also called then “standard deviation” (σ2 is called “variance”), is the amount of uncertainty we have to allow for, in the most probable value X, if we want to claim a roughly 68% chance of correctly predicting the result of any single measurement. To determine X, also called the “mean value”, we consider a set of N measurements x1, x2, …, xN. The probability to get a single result between xi and xi+dx is:

( )2221

2

ix x

iP e σ

σ π

−−

⋅= ⋅

68

so the probability to get all the results (viewed as independent events) is:

( )21

221 2

1...

N

ii

x x

N NP P P P e σασ

=

−

−⋅

∑

= × × × ⋅

Since we are speaking about the probability P to get all the results and we can suppose to have already done our experiment with a set of real results what should be the value of P? If we accept the maximum likelihood principle we can make an analogy with entropy and say that P is proportional to the entropy obtained from our experiment. The value X must be a point of maximum entropy. By the second principle of thermodynamics we have to maximize P, otherwise

said X is the value of x that minimize the exponent: ( )2

1

0N

ii

d x xdx =

⎡ ⎤− =⎢ ⎥⎣ ⎦∑ from which it results:

1

1 N

ii

X xN =

= ∑ (119)

that is, the mean value X is the arithmetic mean and describes all the collected data since it is the value for which the maximum entropy is obtained for our set of data.

To determine σ we can proceed in the same way

( )21

221 0

N

ii

x x

N

d ed

σ

σ σ

=

−

−⋅

⎡ ⎤∑⎢ ⎥

⋅ =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

from which:

( )2

1

1 N

ii

x xN

σ=

= −∑ (120)

But what should be use instead of x in equation (120)? If we use (119) then equation (120) is

slightly self-referential because 2

11

1 1 ( ... ... )N

i N ii

x x x xN N

σ=

⎛ ⎞= + + + + −⎜ ⎟⎝ ⎠

∑ and the i-esim term

appears two times. It is possible to show that the correct value of the standard deviation is:

( )2

1

11

N

ii

X xN

σ=

= −− ∑ (121)

Clearly σ is not defined for N=1 (we are assuming N greater of nearly 30, otherwise there are better distributions to consider). Suppose now we have a function Q of several variables ( , , ,...)Q f a b c= and we want to know how the experimental error on each variable contributes to Q. We can say that by varying the variables, the quantity Q varies of:

...Q Q QQ a b ca b c

∂ ∂ ∂∆ = ∆ + ∆ + ∆ +

∂ ∂ ∂

and if we identify our uncertainty ∆x with the standard deviation σx we can say that:

...Q a b cQ Q Qa b c

σ σ σ σ∂ ∂ ∂= + + +

∂ ∂ ∂ (122)

69

the modulus is due to the fact that errors could cancel each other and we want to consider the maximum error. We could do better, obtaining a smaller value, if the variables are normal and independent, by

starting from (121) ( )2

1

11

N

Q ii

QN

σ=

= −− ∑ where ( , , ,...)i i i iQ f a b c= is the i-esim value of Q by

taking the i-esim value of each variable of our set of data, ( , , ,...)f A B C= is the mean value of Q by taking the mean value of each variable of our set of data.

Since ( ) ( ) ( ) ( )2 2 2

2 2 2 2... ...i i i i i ii i i i

Q Q Q QQ Q a b a ba b a b

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂− = ∆ = ∆ + ∆ + ∆ + ∆ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(by

neglecting terms of higher order) we have:

( ) ( )2 2

2 22 2

1 1

1 1 ...1 1

a b

N N

Q i ii i

Q Qa ba N b N

σ σ

σ= =

∂ ∂⎛ ⎞ ⎛ ⎞= ∆ + ∆ +⎜ ⎟ ⎜ ⎟∂ − ∂ −⎝ ⎠ ⎝ ⎠∑ ∑

1442443 1442443

or

2 2

2 2 ...Q a bQ Qa b

σ σ σ∂ ∂⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ (123)

which is better of (122) since it’s always lower.

Suppose now that the function Q is just the arithmetic mean 1

1 N

ii

X xN =

= ∑ . By applying equation

(123) we get

1 2

2 22 2

1 2

...X x xX Xx x

σ σ σ⎛ ⎞ ⎛ ⎞∂ ∂

= + +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠ (124)

but 1

1 1N

iii i

X xx x N N=

∂ ∂ ⎛ ⎞= =⎜ ⎟∂ ∂ ⎝ ⎠∑ and

1 2...x xσ σ σ= = = and so

X Nσσ = (125)

which is called “standard deviation of the mean”. Analogously to the “standard deviation”, it tells us how good is the mean value X and we can assume it as the amount of uncertainty we have to allow for if we want to claim a roughly 68% chance of correctly predicting the result of any other mean value it is possible to obtain.

It is also useful to speak about relative error Q

Qσ

instead of absolute error Qσ . The relative error can

be expressed in percentage. For example let us return to the problem of determine the density of a cube.

Now, the function Q is the density which is function of the mass M and the side length L: 3

ML

ρ = .

If 60M g= and 54L mm= it’s easy to find that the mean value is 43 3 3

60 3,81 1054

g gcm cm

ρ −= = ⋅

By applying equation (122) we have that the relative error is:

70

3 3

3 4

1 1 1 3 1 3M L M L M L

L L MM L M L M L M L

ρσ ρ ρσ σ σ σ σ σρ ρ ρ

∂ ∂ −= + = + = +

∂ ∂.

If we can suppose the precision of the mass measurement is 2M gσ = and the precision of the length measurement is 1L mmσ = we have:

2 3 1 3,3% 5,6% 8,9%60 54

ρσρ

⋅= + + =

(this says it is more important to make a careful length measurement than a careful mass measurement). By applying equation (123) we get a better (lower) estimate of the density error:

2 2 2 2

2 2 2 2

9 2 9 1 6,5%60 54

M L

M Lρσ σ σρ

⋅ ⋅= + = + = .

This means that if we take another measurement of density there’s a probability of nearly 68% that

the new value will lie between 43(3,8 0, 2) 10 g

cm−± ⋅ .

It is important to note that since the standard deviation on density is 0,2 x 10-4 g/cm3 we can stop at the first decimal digit 3,8 x 10-4.

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4861.30 - Mechanics System 3