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FAQ

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1. 5 .

2. . .

3. 10 , L = 35 .

4. , .

5. ³ L t 7.3.

 

7.3

r, L, t, c r2t, 2 h × Dh × [∆ηi]2 2×2
               
               
               
               
               
            Σ[∆ηi]2

 

6. L 5 , .

7. , 4÷6, 4 , 쳭 L. .

8. r2×t. 7.3.

9. r2×t L. (7.7) , . , (7.7), (7.3) .

10. (7.7) , 5- . Dh /h.

 

Ͳ

 

1. ?

2. .

3. ?

4. .

5. ?

 

 

 

1. .. . -VI.: . 1989. - . .

2. .. . -.: . ., 1990.

3. .., .. . - .: . 1985.

 

..

³, ..

3 01.12.2009 .

 

 

8 Laboratory work 3. MEASURING THE COEFFICIENT of internal friction BY STOCKS METHOD

 

The aim: to determine the internal friction coefficient of a liquid.

Equipment: a glass cylinder filled of oil, with a scale and two sliding rings AB and CD; metal balls; a vernier caliper; a stopwatch.

 

Theory

 

For a body falling in a liquid, the equation of motion is

. (8.1)

Here, is weight, is Archimedes force and is the force of viscous friction. Stocks established a resistance force experienced by a ball, which moves with low speed in a viscous medium. This force is

, (8.2)

where is the coefficient of viscous friction, is the radius of the ball, is the speed of motion. Criterion of smallness for the speed may be written as follows:

, (8.3)

where is the Reynolds number, is the density of the medium.

Projecting the forces onto z-axis pointed out vertically from the surface of the liquid, we obtain

, (8.4)

where = 7800 kg/m3 is density of the ball, = 960 kg/m3 is density of the liquid. Just after the ball has dipped into the liquid, it moves with acceleration, . The force of viscous friction increases until the moment of the force equilibrium

. (8.5)

From this moment, the ball falls with constant speed. Assuming in (8.4), we obtain

. . (8.6)

Experimental part

 

1. Select 5 balls of approximately equal radius.

2. Fix the upper ring at the point of 0.1 m under liquid level. Place the lower ring CD at the distance 0.35 m from the upper one.

3. Measure radius of a ball by the vernier caliper. Drop the ball into the cylinder. Switch on the stopwatch at the moment the ball passes by the upper ring, and switch off it when the ball moves near the lower ring. Put down time of crossing the distance between rings into the table 8.1.

 

Table 8.1

, mm , m , s , m/s , Pa s
    0.35      
    0.40  
    0.45  
    0.50  
    0.55  
  =      

 

5. Draw the graph versus . Make conclusions about character of balls motion. Determine the speed of motion as a slope of the line to the - axis.

6. Substituting speed and average radius into (8.6), calculate the coefficient of viscous friction .

7. Find out what is the main error in determination of . Estimate .

8. Examine condition (8.3).

 

Control questions

 

1. What is the physical origin of viscous friction in liquids and gases?

2. Write down Stocks formula for resistance force experienced by a ball moving in a viscous medium.

3. What is condition of validity for Stocks method?

4. Obtain (8.6) starting from the equation of balls motion.

5. Substituting into (6) reasonable values for radius and speed, estimate the coefficient of viscous friction of water ( = 1000 kg/m3).

 

Authors: V.P. Kurbatsky, the reader, candidate of physical and mathematical sciences.

Reviewer: S.P. Lushchin, the reader, candidate of physical and mathematical sciences.

Approved by the chair of physics. Protocol 3 from 01.12.2008.

 

4.1.

 

: 㳿 .

: , , 2-2.

 

9.1

 

䳿. , . , . ( ) , ( ). , 㳿.

9.1

 

. 㳿:

, (9.1)

. (9.2)

m1 m2 u1 u2 u1 u2 .

, (9.3)

(9.4)

(9.4) (9.3)

(9.5)

(9.5) m2 (9.3), :

, (3.6)

(9.5) m1 (9.3), :

. (9.7)

: m1 = m2 u2 = 0. (9.6) (9.7) u1 = u2; u2 = u1. , .

9.2. m1 = m2 = 0,6 . l = 0,72 . ³ j. E = mgh. ϳ :

,

. (9.8)

D , h = l(1-cosj) = 2l×sin2j / 2.

(9.9)

u2 = 0. u1 = 0. u2 = u1, .

9.2

p, 䳿 F, , :

(9.10)

(9.11)

(9.12)

m , t , j , , g , l .

t i j

, (9.13)

, i , k = 0,3÷0,5 . i lnt = f(lnj) ii

(9.14)

k.

t , . 9.3. - . ( ) N . , t = ×N. N, t

.

. (9.15)

9.3

 

 

1. 2-2 .

2. N t, 1, 3 ..

3. , . Ⳮ No, t . t No 9.1.

 

9.1

t, c No, . DNo, . (DNo)2, 2
         
       
       
       
       
    = å(DNo)2=

 

4. 3.

5. ³ . , j = 14. , , . ³ . ³ b . ˳ N, . b N 9.2.

 

9.2

j, b N, t, ln t ln j F, Po / P / Eko Ek
                     
                     
                     
                     
                     
                     
                     

 

6. 5 12, 10, 8, 6, 4, 2. 9.2. .

7. t (9.15), No; F (9.11); (9.10); (9.12). 㳿 (9.10) (9.12) j b.

8 ln τ lnj ln τ ln j. , , , , k

(9.16)

9. No . 1 , .

10. k , 㳿 , .

 

Ͳ

 

1. ?

2. ?

3. ?

4. .

5. 㳿.

6. .

7. 㳿 .

8. .

 

 

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4.2.

 

㳿 .

- .

 

10.1

 

, .

ϳ . ʳ . . . 㳿 , .

, , , .

, ', , .

, , . , .

, .

.

㳿.

, , .

㳿 - , - , , ( , ). .

, - , , ( , ). 䳭 䳿. .

:

)

,

, ,

, (10.1)

)

. (10.2)

㳿:

)

, (10.3)

)

, (10.4)

 

ΔW 㳿 , . , (, m2) , (10.1), (10.2), (10.3) (10.4) :

, (10.5)

, (10.6)

, (10.7)

. (10.8)

, . , . , .

(10.5), (10.6), (10.7) (10.8) :

, (10.9)

, (10.10)

, (10.11)

. (10.12)

 

: - ; - ; - 㳿 ; - 㳿 .

, .

䳺 ( , . .), "" 㳿 , , . 㳿 1. ( , m1 = m2 = 0..5. ' (10.10) (10.12), , =1). , , , 1, 1, (10.9), (10.10), (10.11) (10.12) ( ), 㳿 .

(10.9), (10.10), (10.11) (10.12) , . .10.1, m2 , m1 - α1 ( ୺䳿 α, 䳿 - β; : 1 , 2 - ).

1

,

.

,

.

α1 (< 10), ( ) . ϳ 2 m2 u2 β2.

.

1 β1.

.

ϳ v1, u1 2 (10.9), (10.10). (10.11) (10.12) ' ,

,

,

,

.

(m1 ≈ m2), β1, m1 (β1 = 0)

, (10.13)

, (10.14)

, (10.15)

. (10.16)

 

(10.13) ÷ (10.16), .

10.1

 

1. , .

2. ϳ . ³ , .

3. ³ α1, . α1 ( ) . , . β2 ( ) 10.1.

4. 5 . .

5. , , ("" ). ³ α1, . α1 .

 

10.1

α1, β2, Δα, Δβ, α1, β2, Δα, Δβ,
             
             
             
             
             
= =
  = =


7. . β2 ( ) .

8. ' , .

9. (10.13), (10.14), (10.15) (10.16) α1 β2.

10. . ( ).

11. 㳿 , . .

 

 

1 ?

2 ?

3 , .

4 㳿 , .

5 (10.9). (10.10), (10.11) (10.12)?

6 .

 

 

4. .. . -VI.: . 1989. - . .

5. .. . -.: . ., 1990.

6. .., .. . - .: . 1985.

 

..

- ..

,

3 01.12.2008 .

 

Laboratory work 4.

ELASTIC IMPACT OF BODIES

 

THE AIM: to study the laws of perfectly elastic impact.

INSTRUMENTATION AND APPLIANCES: two balls suspended to rods, a scale for the angles measurements, a device 02-2 to count electric pulses and a generator of electric pulses.

 

Task

1. Calculate a mean impact time .

2. Calculate a mean impact force .

3. Calculate the absolute and relative errors of τ and F.

4. Draw the conclusions.

 

Short theory

 

THE IMPACT is interaction of two or more bodies that takes place during a very short time. A magnitude of internal forces appearing in bodies system during the impact is relatively big, so the external forces acting on the bodies at this moment of time can be neglected. This enables to apply the law of conservation of momentum (law of conservation of impulse) for studying the phenomena of the bodys collision because a system of interacting bodies is isolated

PERFECTLY ELASTIC IMPACT is interaction of two bodies acting with elastic forces at each other, no mechanical energy of the system being transformed into thermal energy (heat). In this case we can apply both the law of conservation of momentum and mechanical energy.

PERFECTLY INELASTIC IMPACT is interaction of two bodies resulting in partial transformation of the mechanical energy into heat. The bodies move after collision as whole body. In this case we can not apply the law of conservation of mechanical energy, but we can apply the law of conservation of momentum and total energy of system. Hence we may write:

a) for perfectly elastic impact

,

,

where m1 and m2 are the masses of interacting bodies; V1 and V2 are the velocities of the bodies before the impact; and are the velocities of the same bodies after the impact.

b) for perfectly inelastic impact

,

,

where U is the velocity of the bodies system after the impact; Q = E1 E2 is the mechanical energy turned into heat (E2 and E1 are the mechanical energy of the system before and after the impact.

Elastic impact

 

The experimental installation to study the perfectly elastic impact consist of two rods with suspended balls, the scale for measurement of angles, a generator of elastic pulses and a device to count the pulses. Electric pulses pase through the balls of there is a contact between those balls. It is possible to obtain the bodies interaction time (impact time) if we know the numbers N of electric pulses having passed the balls during impact time and the same numbers N having passed at a second. Evidently can be calculated by the formula

(11.1)

Mean balls interaction force (mean impact force) can be calculated by the formula, which shows change of momentum of the body and impulse of the force

,

where F is a mean impact force, Δτ is the mean impact time; FΔτ is the impulse of the force; Δp is the change of body momentum.

Hence

,

where v0 is the body velocity before the impact, v is the body velocity after the impact(v = 0 in our case). According to the law of conservation of mechanical energy

and

,

where h is a height of the lifted ball. If the rod length is equal to l then we can write (according to figure 11.1)

 

Figure 11.1

 

(11.2)

Here F is interaction force. You have to count the impact force in this laboratory work. To do this it is necessary to determine numbers N and numbers N0 of electric pulses at the beginning. For that you have to

1) switch on the device 02-2M;

2) wait for 5 - 7 minutes to warm up the device;

3) press the key "M" and "f";

4) connect the balls together (with you hand) and press them to each other for good electric contact and press the key "cpoc" and then "";

5) look at the indicator and write down the numbers of pulses passing during 1 second - N0;

6) repeat this measurement process four times (you must have five measurements of value N0).

Then you have to obtain the value of N. To do that you have to

1) press the key "100";

2) deviate one of the balls at the angle φ= 100;

3) release the ball and press the key "" simultaneously;

4) press the key "" after the first balls impact;

5) write out the value N from the indicator;

6) repeat your measurement process four times (you must have five measurements of N).

 

Experimental part

 

1. Measure the number of pulses N0 by the device 02-2M as in

was shown above.

2. Measure the number of pulses N the same way (φ = 100).

3. Fill in the table

 

n N0 ΔN0 (ΔN0)2 N ΔN τ, s v, m/s F, N
                 
           
           
           
           

 

4. Calculate mean impact time by the formula (11.1) using and .

5. Calculate mean impact force F by the formula (11.2) using

m = 0.6 kg, l = 0.72 m, φ = 10 and obtained above.

6. Calculate absolute and relative error for τ and F.

7. Draw the conclusion.

 

Control questions

1. Formulate the first Newton's law.

2. Formulate the second Newton's law.

3. Formulate the third Newton's law and the law of conservation of momentum (impulse).

4. What system is isolated (or closed)?

5. Write the relation between a force and an impulse.

6. Formulate the law of conservation of mechanical energy.

7. Give the determination of perfectly elastic and perfectly inelastic impact.

8. Write the laws of conservation for the impact of two balls.

 

 

Author: E.V. Rabotkina, the senior reader.

Reviewer: S.V. Loskutov, professor, doctor of physical and mathematical sciences.

Approved by the chair of physics. Protocol 3 from 01.12.2008.

 

4.3.

Ҳ

 

: 㳿 .

: , .

 

12.1

 

䳿. , 䳿 . ʳ . ( ) . , 㳿 , . 㳿

 

 

12.1

 

(. 12.1). ϳ , U . 㳿, , Q:

, (12.1)

 

. (12.2)

m1 m2 u1 u2 U Q.

, (12.3)

 

. (12.4)

: , , , U = 0,5u, , 㳿

. (12.5)

i , ii (. 12.2). . M m, L .

 

12.2

 

φ , /2.

.(12.6)

, , 㳿, ( ),

. (12.7)

J ML2 ( ) ; ; u .

(12.6) (12.7),

. (12.8)

ϳ . , U , (12.7) (12.8) m 2 2m, φ β

. (12.9)

 

. (12.10)

³ 㳿 (12.7) ÷ (12.10)

. (12.11)

 

1. φ = 15+3N . , φ = 3N . N .

2. .

3. ³ φ . β .

4. .3 ( 5), , . 12.1.

 

12.1

    φ β Δβ Δβ2
                 
       
       
       
       
    =    

 

5. , , Δβ β.

6. (12.11) 㳿, . 㳿 0,5 (. ).

7. 2 . , , (12.9) , .

8. 㳿.

1. ?

2. ?

3. ?

4. .

5. 㳿.

6. .

7. 㳿 .

8. .

 

 

1. .. ..1.- .: , 1982.- . 45-48, 134-136, 140-144, 151-152.

2. .. . - : , 1963.- .79.

3. .. . / , 2004.- 442 .

4. .. , .. . .1 / .: .-2003.-311 .

 

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3 01.12.2008 .

 

 

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