Mechanics: laboratory practicum in physics 9786010421097

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

MECHANICS LABORATORY PRACTICUM IN PHYSICS

Almaty «Qazaq university» 2016

UDC 53 / (076) M 63 Recommended for publication by the Scientific Council of the Faculty of Physics and Technology and RISO of Al-Farabi Kazakh National University (Protocol №1 dated 02.11.2016) Reviewers: doctor of Physics and Mathematics sciences, Professor M.E. Abishev doctor of technical sciences, Professor A.B. Ustimenko Authors: A.S. Askarova, S.A. Bolegenova, S.S. Issatayev, V.V. Kashkarov, I.N. Korzun, G. Toleuov, Sh.B. Gumarova, L.E. Strautman, O.A. Lavryshev, M.S. Isatayev, Zh.K. Shortanbayeva

M 63 Mechanics: laboratory practicum in physics / A.S. Askarova, S.A. Bolegenova, S.S. Issatayev, V.V. Kashkarov, [et. al.]. – Almaty: Qazaq university, 2016. – 220 p. ISBN 978-601-04-2109-7 The manual drawn up in accordance with the university programs for the general physics course for students of physical and physical-technical specialties. The workshop contains descriptions of 17 laboratory works for the part «Mechanics». Each work includes a brief theoretical introduction, a description of the experimental setup, work assignment, as well as how to conduct experiments and processing the results. The manual can be used in higher education institutions for the physical and physical-technical specialties. Published in authorial release. Учебное пособие составлено в соответствии с действующими университетскими программами по общему курсу физики для студентов физических и физикотехнических специальностей. Практикум содержит описания 17 лабораторных работ по разделу «Механика». Каждая работа включает краткое теоретическое введение, описание экспериментальной установки, рабочее задание, а также методику проведения экспериментов и обработки их результатов. Пособие может использовано в высших учебных заведениях для физических и физико-технических специальностей. Издается в авторской редакции.

UDC 53 / (076) © Askarova A.S., Bolegenova S.A., Issatayev S.S., Kashkarov V.V., [et. al.], 2016 © Al-Farabi KazNU, 2016

ISBN 978-601-04-2109-7

2

CONTENTS

PREFACE.................................................................................................. 3 INTRODUCTION ..................................................................................... 4 Laboratory work № 1. Statistical regularity in measurements ................... 20 Laboratory work № 2. Measurement of length, area and volume .............. 31 Laboratory work № 3. Measurement of the moment of inertia of the disk .................................................................................................. 41 Laboratory work № 4. Study of the fundamental law of dynamics of rotational motion ................................................................................... 56 Laboratory work № 5. Study of the law of oscillations of a physical pendulum .................................................................................. 70 Laboratory work № 6. Study of the stationary flow of liquid in the tube of variable cross section. Verification of Bernoulli's theorem........................................................... 82 Laboratory work № 7. Inclined pendulum ................................................. 98 Laboratory work № 8. Measurement of the moments of inertia of bodies by torsion pendulum .................................................................. 110 Laboratory work № 9. Studying of elastic collisions of balls and determination of young's modulus ...................................................... 124 Laboratory work № 10. Measurement of the speed of flight of the body with a torsion-ballistic pendulum ............................................ 133 Laboratory work № 11. Determination of the viscosity of liquids by the Stokes method ................................................................................. 142 3

Laboratory work № 12. Determination of the gravitational acceleration using a reversible pendulum .................................................. 153 Laboratory work № 13. Study of oscillations of coupled systems ............. 162 Laboratory work № 14. Study of the laws of translational rectilinear motion of bodies in the field of gravity on the Atwood machine ............................................................................. 178 Laboratory work № 15. Gyroscope............................................................ 187 Laboratory work № 16. Studying of dynamics of plane motion on Maxwell pendulum ............................................................................... 197 Laboratory work № 17. Free falling bodies in the gravitational field of the earth ......................................................................................... 205 APPENDIX ............................................................................................... 213

4

INTRODUCTION

The main objective of physical laboratory classes is quantitative study of physical phenomena, development of skills of independent research work and correct measurement of physical quantities. During laboratory classes students independently conduct physical experiments, get acquainted with the most important measuring instruments, acquire skills for carrying out precise measurements and correct processing of obtained results. Physical measurements During the experiments various types of measurements are made. Measurement is finding the value of a physical quantity empirically using special instruments. In the laboratory classes measurements are made to establish a functional relationship between two or more variables. There are several types of measurements. Their classification is based on the type of dependence of measured quantities on time, conditions determining the accuracy of measurements and methods of presentation of their results. According to the nature of dependence of measured quantities on the time of measurement, they are divided into static, when the measured value remains constant, for example, measurement of geometric dimensions of the body, constant pressure, and dynamic, when the measured value changes, for example, measurement of the amplitude of damped oscillations, pulsation rate, etc. According to the method of obtaining the results of measurements, they are divided into direct, indirect, cumulative and joint. Direct measurements are measurements where the sought value of a physical quantity is found directly using the measuring instrument. 5

In this case, the value of the physical quantity is determined by direct contact with the measuring device, i.e. the quantitative value of the unknown quantity is determined by the indications of the device, immediately, without further processing. For example, it is measuring of the size of the body using a micrometer, measuring time using stopwatch, etc. Direct measurements can be expressed by the formula x = Q, where x is the unknown quantity, and Q is the value directly derived from the experimental data. Direct measurements are widely used in mechanical engineering, industrial processes, and other spheres. Indirect measurements are measurements, in which the sought value is determined based on the known relationship between this value and the values subjected to direct measurements. Indirect measurements can be described by the formulas

x  f (Q1 , Q2 , Q3 ,...) , where х is the value of the unknown quantity; Q1 , Q2 , Q3 ,... are the values of quantities measured directly. Examples of indirect measurements are determination of the volume of the body from direct measurements of its geometric dimensions, finding of a specific electrical resistance of the conductor using its resistance, length and cross-section. Indirect measurements are widely used in cases when it is impossible or very difficult to find the unknown value as a result of direct measurements. Indirect measurements play an important role in measurements of quantities that cannot be measured by direct experimental comparison, for example, sizes of astronomical or intra-atomic level. Cumulative measurements are simultaneous measurements of several one-type quantities, in which the sought value is determined by solving the system of equations obtained by direct measurements of various combinations of these quantities. An example of cumulative measurements is determination of the mass of individual set of weights (calibration) by the known mass of one of them and by the results of direct comparisons of different combinations of weights. 6

For example, it is necessary to calibrate a set of weights, consisting of weights 1, 2, 2*, 5, 10 and 20 kg (asterisk denotes the weight having the same nominal value). Calibration is determination of the mass of each weight using a reference weight, for example, weight of mass of 1 kg. To do this, we make measurements, each time changing the combination of weights (numbers show masses of individual weights, 1ref means the mass of the reference weight of 1 kg) 1 = 1ref + a, 1 + 1ref = 2 + b, 2* = 2 + c, 1 + 2 + 2* = 5 + d , etc. Letters А, в, с and d denote weights that have to be added or subtracted from the weight indicated in the right-hand side of the equation. Solving this system of equations, we can determine the value of the mass of each weight. Joint measurements are simultaneous measurements of two or more different quantities aimed at finding dependencies between them. As an example, we can take simultaneous measurements of electrical resistance at 20ºC and temperature coefficients of the measuring resistor based on the direct measurements of its resistance at various temperatures. By the conditions determining the accuracy of the result, measurements are divided into three classes: 1. Measurements with maximum possible accuracy attainable at the current level of measuring techniques. These include reference measurements and measurements of physical constants; 2. Control-and-calibration measurements, the error of which cannot exceed a predetermined value. These include measurements carried out by state supervision laboratories to monitor the observance of state standards; 3. Technical measurements in which the error of the result is determined by the characteristics of measuring instruments. These include measurements made in scientific research, production, and others. 7

Units of measurement of physical quantities As physical quantities are in a functional relationship, it is impossible to establish their units arbitrary. There are seven units of physical quantities, called the basic units in SI, which are established arbitrarily. These are units of the following physical quantities: length (meter), mass (kilogram), time (sec), temperature (Kelvin), current (Amps), intensity (candela), the amount of substance (mole). International System of Units includes two units additional to the basic units of measurement. They are units for plane (radians) and solid (steradian) angles. The units of other physical quantities are found from the relations determining these values. Therefore they are called derived units. Measurement errors It is impossible to measure the value of a physical quantity with absolute accuracy. Firstly, there is no such a precision instrument, and secondly, the possibilities of human senses are limited. Therefore, when it comes to the measured value, it is necessary to specify the error of measurement. By the nature of occurrence the errors are divided into systematic, random and gross errors. Systematic errors are values constant in time or varying regularly during the measuring experiment, for example, the initial deflection of the arrow of the device from zero. Systematic errors are subdivided into instrumental and methodical errors. Errors of measuring instruments Accuracy of measurement of any measuring instrument is limited. Errors of instruments intended for industrial purposes (ammeters, voltmeters, potentiometers, etc.) are determined by their measurement accuracy. Measurement accuracy of the instrument is usually expressed in percentage form. For example, an ammeter with an accuracy class of 0.2% determines the current value corresponding to the full-scale, with an error not exceeding 0.2%. The accuracy class of the device, regardless of the position of the arrow, indicates the minimum relative error, and relative error of 8

the measured value increases as the arrow moves to the origin of coordinates. Therefore, during measurements we have to try to work in the second half of the scale. It is important at which angle we look at the arrow when you register the readings. To minimize the errors, we must look at the arrow of the device at the right angle, i.e. perpendicular to the scale. Methodological errors Such errors are, for example, errors of measurement techniques. For example, it is not taking into account the friction of the shaft in the laboratory work No. 3 in determining the moment of inertia of the disc by dynamic method or using the real liquid (water) to check the Bernoulli equation obtained for an ideal fluid. Errors in the methods of data processing They arise as a result of incorrect or imperfect methodology used to determine the sought value by computing and determining the errors. As the number of measurements increases, systematic errors decrease. To reduce them, more precise tools and more advanced measurement and data processing techniques are used. Random errors Random errors are components of the measurement error, changing randomly in repeated measurements of the same quantity. These include not only random variations in the mains voltage and the temperature in the laboratory but also factors associated with the imperfection of human senses and skills. In the process of measuring, the values and signs of random errors do not remain constant. To reduce the random error, it is necessary: 1) to reduce the influence of external factors; 2) to measure carefully and thoroughly; 3) to increase the number of tests. Gross errors Gross errors are errors of measurement, significantly exceeding expectations under given conditions. They are caused by the 9

negligence of the experimenter or by faulty equipment. When gross errors are found, such measurements are deleted. Methods of finding errors of direct measurements are described in detail in the laboratory work No. 1, so we do not consider them here, and continue with the method of finding errors of indirect measurements Finding errors of indirect measurements Let us consider some special cases of relations between two or more variables functionally connected with each other. 1. Let us consider the sum or difference of two quantities:

A BC.

(1)

The most probable values of quantities B and C are their arithmetic mean values  B  and  C  . Here and below the angular brackets denote the arithmetic mean values. The best value of A is equal to the sum (or difference) of the best values of terms: A  B    C  .

(2)

The arithmetic mean error S A , if В and С are independent, is found using the formula: S A  S B2  S C2 ,

(3)

in other words, not the errors are added but their squares. 2. Let the unknown quantity be a product or a quotient of two other quantities: А=В·С or А=В/С.

(4)

The most probable values of A are given by the formulas:

A  B    C 

or 10

А

B . C 

(5)

Relative standard errors of the product and quotient of independent variables are found using the formula: 2

SA  S   S    B   c  A  B   C 

2

.

(6)

3. Let us consider the product of several variables having different powers:

A  B   C   E  ...

(7)

The relative standard error of the value of A for independent B, C, E,… can be obtained using the formula 2

2

2

2

 SA  2 S  2 S  2 S       B     C     E   ... A B C       E

(8)

4. Let us derive the general formula for calculations. Let A  f ( B, C , E ,...) ,

(9)

where f is an arbitrary function of В, С, Е etc. Then the best value of А is:

Aнаил  f ( B ,  C ,  E ,...) .

(10)

The mean-square error of A is found as: 2

2

2

 f  2  f  2  f  S A2    S B2    S C    S E  ...  E   C   B 

(11)

The notation f has the usual meaning of the partial derivative B

of function f with respect to B, i.e. the derivative to find which we 11

assume all arguments except B constant. Partial derivatives should be calculated for the best values of arguments Вbest Сbest, Еbest, etc. Let us consider some consequences that can be derived from the analysis of the above formulas. 1. One should avoid measurements in which the sought value is found as the difference of two large numbers. Thus, we should not dedtermine the thickness of the pipe wall by subtracting its inner diameter from the outer (and dividing the result by two). The relative error of measurement in this case is greatly increased as the measured value – in this case, the wall thickness is small, and the error in its definition is found by adding the errors of measurements of both diameters and therefore it increases. 2. One should determine all measured values with approximately the same relative error. For example, if the volume of the body is measured with an accuracy of 1%, there is no sense to spend time and efforts on measurement of the body weight with an accuracy of 0.01% in order to determine its density. 3. In calculating the error of the product of power functions it is necessary to pay attention to the member of the highest power. According to the formula (8), the relative error of the factor is directly proportional to the square of its power. Therefore, the errors of individual members can be neglected. In real experiments, systematic and random errors are present simultaneously. Typically, they are independent, so the total error can be determined by the following formula: S2tot = S2 ran +S2sys.

(12)

Approximate calculations and rounding rules In the laboratory work it is often necessary to measure several quantities with different accuracy. If the quantities in the formula have different errors, it is necessary to pay attention to the value of the maximum error. Then, when determining the total accuracy it is not necessary to calculate errors of quantities, the accuracy of which is many times smaller than the value of the maximum error. Sometimes, they are simply rejected. 12

The accuracy of calculation of the required quantity must be within the measurement accuracy. When the accuracy of calculations is much higher than the measurement accuracy, it is considered a drawback of the work, rather than its advantage. Therefore, it is important to know the basic rules of rounding. Rounding rules and approximate calculations Rounding the number, we have to delete the last digits, and if the first of the deleted digits is less than 5, we do not change the remaining number, but if the first of the deleted digits is more than 5, we must add one to the previous number. If the first digit of the deleted part is equal to 5, there are two options. If the preceding digit is odd, then it is increased by one, if it is even – nothing is added. For example: А = 21.314  21.31;  = 3.141593  3.1416,  = 3.141593  3.14; l = 12,15 m  12,2 м; t = 5.25 s  5.2 s. The value of the physical quantity carries information about the amount of error. In fact, if all of the numbers are correct, then the price of last digit gives the accuracy of the measurement or calculation. For example: t = 5.021000 s, in this record of time the error can be estimated approximately as 1 microsecond. Significant digits. Significant digits are all correct digits except zero, standing in front of the number. For example, the number 0.000205 has three significant digits. The first four zeros to the left are insignificant digits, zero between two and five is a significant digit. In the number 2300 there are four significant digits. Recording 6.1 · 103 means that the number has only two significant digits (six and one). In the final recording the measurement error is usually expressed by the number with one or two significant digits. Two digits are used for more precise measurements, and if the higher digit of the number expressing error is three or less than three. For the least significant digit only 5 is used. It should be noted that in the intermediate calculations, in order the rounding errors not to distort the results too much, it is recommended to preserve three to four significant digits, depending on the computations. 13

Some recommendations to the fulfillment of the laboratory work The laboratory work can be divided into four stages. 1. Preparatory stage. Each lab work is devoted to the study of a concrete physical phenomenon or process, the specific features of which are to be identified. On the first day of physical practicum all students are given the theme of the laboratory work, which they are to do at the next lesson. Before the lab work, they should at home (in the library) get familiar with the theory of this work, the construction of the experimental device (drawings) or the experimental model and the experimental part of laboratory work. The main source of information is a methodological description of the lab. From it, the student writes down the name and purpose of the work, a brief theory of the problem, describes the setup, the process of t work fulfillment and methods of mathematical processing of the results. To the laboratory work the student brings his drawing tools, graph paper, multi-functional calculator, etc. 2. Experimental part. This stage of physical practicum is fulfilled directly in the laboratory. The students who showed good preparation to the performance of the work are allowed to do it. The necessary work equipment, tools and details are given to the students by the laboratory assistant. The students make measurements without assistance. The students write table values of physical constants, parameters of the setup and the data of the experiment into the lab journal. At the end of the experiment the teacher checks the correctness of measurements, If everything is correct, then the teacher signs in the lab journal. After the signature of the teacher it is prohibited to make any changes in the data table. 3. Mathematical processing of experimental results. In this part of the laboratory work the sought value is calculated, graphs of the processes are plotted, measurement errors are calculated, and the results are analyzed. In the analysis of the results it is necessary to pay attention to the factors influencing the results of the experiments. 4. Defense of the results of laboratory work. In this final part the student justifies his results making an oral presentation to the teacher. He tells about the theory of the subject, derives the final formula, and answers the questions given at the end of the description. 14

Graphic method of data processing In the experimental physics graphs are used for different purposes. For example: – to determine values; – to present visual description of the process; – to find an empirical relationship between the values; – to compare the experimental results with the theoretical data and the data of other authors, etc. Graphs are plotted on linear and logarithmic graph paper. To determine the functional dependence of the value it is accepted to put the value of the argument along the x-axis, and the values of the function – along the y-axis. Before plotting it is necessary to specify the minimum and maximum values for the abscissa and ordinate axes and to choose the most appropriate scale. The experimental results are shown on the graph as points and theoretical results – as a solid line. If several lines, corresponding to the different modes of the same process, are drawn on the same plot, it gives a visual representation of the dynamics of the process, depending on its parameters. Optimally choosing the scale, it is necessary to find such a scale that the data points on the graph cover all the picture plane. Typically, the points on the graph are located in a specific pattern. If any point does not follow the dependence (lies aside), you should pay attention to it: either there is an interesting effect or a gross error. If the measurement is correct, then the vicinity of the point is investigated in more detail in order to find out the effect. By the size of the symbols on the graph, we can judge about the measurement errors. Dimensions of the graphic characters are defined by the expression:

1 l  S . 2 Here l is the size of linear characters. In the construction of graphs the students should try to draw a linear dependence as it is easy «to read» a linear functional 15

dependence of variables and based on such a dependence it is not difficult to get an empirical relationship. Let us consider several ways of linearization of dependencies. 1. Let the process be described by a quadratic function. Consider the law of falling bodies described by the formula:

S

gt 2 . 2

If we put S and t along the axes, the point will lie on the parabola, and it is almost impossible to draw it by eye. But if we put S and t2 or S and t or ln S and ln t along the axes, the graph will become a straight line. In general, transforming the quadratic dependence to the form y  ax  b , it is possible to determine the unknown constants А and b. n

2. Let the process be described by a power dependence y  x . Let us take logarithms of both sides of this formula:

ln y  n ln x . If we put variables ln y and ln x along the axes, we will get a straight line. Least square method In general, many complex expressions can be reduced to the linear form:

y  ax  b .

(13)

The parameters a and b in this relationship can be found analytically by using the experimental data. Let in the experiment we got the value y i ( i  1,2,..., n ), corresponding to xi . Then the values of a and b in formula (13) corresponding to the most precise 16

dependence between y i and xi can be found by the least square method. If the experimental values xi and y i are substituted into (13), due to the errors in measurements the equation (13) is usually not valid, i.e. y i  (axi  b)  0 .

(14)

Let us square this difference. Then, find the sum of the squares of these differences for all measurements: 2

D    y i  (ax i  b) . n

i 1

(15)

Using the least square method, we get minimum values of parameters a and b corresponding to the most accurate approximation of the linear dependence between y and x in the formula (15). Therefore, to find the values of parameters a and b, from the equation (15) we find partial derivatives of D with respect to a and b, and equate the resulting expressions to zero. n D  2   y i  (axi  b)  0 , b i 1

n D  2  { y i  ( ax i  b)  xi }  0 . a i 1

(16) (17)

Then: n

n

i 1

i 1

 y i  nb  a  xi  0 ,

n

n

n

i 1

i 1

i 1

2  y i xi  b  xi  a  xi  0 .

17

(18) (19)

Solving the resulting system of equations, we can find the values of parameters a and b for the most accurate approximating linear dependence: n

a

n

i 1

i 1

i 1

2

 x   n  n x2  i  i  i 1  i 1  n

n

b

n

 y i   xi  n  xi y i

n

n

,

n

2  x i   xi y i   y i  x i

i 1

i 1

i 1

2

i 1

 x   n  x2  i  i  i 1  i 1  n

(20)

n

.

(21)

If the linear dependence passes through the origin of coordinates: y  ax ,

(22)

then n

a

 xi yi

i 1 n

2

.

(23)

 xi

i 1

The least square method allows us to find the errors of parameters in the linear dependence, which are determined by formulas (20) and (21): 2

2

n y  n n x y  n x n y   i    i i  i i i 1 i 1 i 1    i 1  , S 02  i 1   n  2 nn  2  n 2  n 2  nn  2n  xi    xi    i 1    i 1 n

2

 yi

S a2 

nS 02 n

n  xi2 i 1 18

n    xi   i 1 

2

,

(24)

(25)

n

S b2 

S 02  xi2 i 1

n

n  xi2 i 1

n    xi   i 1 

2

.

(26)

Here S 0 is the mean square error of an individual measurement,

S a and S b are the mean square errors of parameters in the linear dependence y  a x  b .

19

Laboratory work №
1 STATISTICAL REGULARITY IN MEASUREMENTS

1.1. The purpose of work: to get acquainted with the methods of processing the results of direct measurements. 1.2. Brief theoretical introduction 1.2.1. Measurement errors and their classification Let us assume that the true value of a quantity is x0 . As a rule, measuring this value we get the result different from x 0 . If the measurement is repeatedly performed, the results are not only different from x0 , but in most cases are different from each other. Let us denote the results of measurements as x1 , x 2 ,..., x n . Then the difference

xi  xi  x0 , где i  1, 2, ..., n,

(1.1)

is called the absolute error of measurement. It is expressed in terms of the measured value. In the classification of errors by their properties, the errors are subdivided into systematic, random and gross errors. Systematic error of the measurement is called a component of the measurement error, which remains constant or varies regularly with repeated measurements of the same quantity. For example, the scale of the measuring ruler is not uniform, the capillary of the thermometer has different diameters in different parts, weights have unequal shoulders, the arrow of the ammeter in the absence of current is not at zero, etc. 20

It is usually possible to take these errors into account, and therefore to eliminate by the introduction of corrections to the measured value (for example, to take into account the division, which the arrow of the ammeter shows in the absence of current and subtract it from the readings). Systematic error of the measurement can be detected experimentally or by comparing the obtained result with the result of measurement of the same quantity obtained by a different method or by using more accurate measuring instruments. However, systematic errors are usually estimated by theoretical analysis of the measurement conditions, based on the known properties of measuring. Gross errors are the consequences of wrong actions of the experimenter or equipment malfunctions. For example, it is a slip of the pen in the recording of the results of observation, incorrectly recorded reading of the device, etc. When such errors are detected, they must be excluded from the calculations. Random errors are errors that vary randomly. In this case the difference between the results of individual measurements is individually unpredictable, and dependencies appear only after a large number of measurements. 1.2.2. Methods of processing the results of direct measurements containing random errors. Let N measurements were made in the same conditions and x i is the result of the i-th measurement. The most probable value of the measured value is its arithmetic mean value:  x 

1 N xi . N i 1

(1.2)

The value  x  tends to the true value of the measured value x0 for при N   . The mean square error of individual measurements is the quantity N

SN 

 ( x   xi )

i 1

21

N 1

2

.

(1.3)

For N   S N tends to the constant limit  :

  lim S N .

(1.4)

N 

The value  2 is called variance of measurements. In practice, it is often necessary to determine the error of the arithmetic mean. Let x1 , x 2 ,..., xi ,..., x n be individual measurements, each of which is characterized by the same variance. The arithmetic mean of a series of measurements is determined by the formula:  x 

x x x 1 N  xi  1  2  ...  n . N i 1 N N N

(1.7)

Then, the variance of this value is determined as  2x  

2 N2



2 N2

 ... 

or



  x 

N

2 N2



2 ,

(1.8)

N

.

(1.9)

Similarly N

S  x 

SN N



2  ( x   x i )

i 1

N ( N  1)

.

(1.10)

The mean square error of the arithmetic mean is equal to the arithmetic mean error of the individual result divided by the square root of the number of measurements. This is a fundamental law of increasing accuracy with the growth of the number of observations. 22

The probability that the true value lies within a range from  x  x to  x   x is called confidence probability  (coefficient of reliability, reliability), and the interval is called the confidence interval. For sufficiently large N the confidence interval  x    x  corresponds to  = 0.68, the interval  x  2   x  corresponds to  = 0.95,  x  3  x  corresponds to  = 0.997. The quantity  of the approximation of the measured value of x to the true value x0 is determined by the physical essence of the measured value, as well as by physical and structural principles of the method of measurement, therefore, an infinite increase in the number of measurements does not give a noticeable increase in accuracy. As there is no sense to strive for a very large number of measurements, a limited number of tests is made in one experiment. However, for a given reliability  the confidence interval, measured in fractions of , is underestimated. The question arises how the reliability varies depending on the number of measurements? This dependence is complicated and cannot be expressed in terms of elementary functions. The multipliers determining the dimension of the interval in fractions of S x  as a function of  and N called Student coefficients, are denoted as t , N and are found from the tables of coefficients (Table P5, Appendix II). The confidence interval x is calculated as follows:

 x  t , N  S  x  .

(1.11)

The final result is presented as:

x  x   x

for   k % .

(1.12)

In is obvious that for  = 0.68 t , N  1 , but for N   t , N  1 . 23

In engineering calculations usually the confidence probability  = 0.95 is used. The error corresponding to  = 0.95 is called a standard error. For  = 0.95 t , N  2 , but for N   t , N  2 . To estimate the accuracy of the experiment the relative error of the result is calculated. The relative error is the error expressed as a fraction of the average value of the measured quantity:



x . x

It is often expressed as a percentage:



x  100% . x

(1.13)

Let us divide the whole set of measurements into intervals. From all N measurements, we select the minimum and maximum values xmin and xmax. The number of intervals K will be equal to the quotient of division of (xmin - xmax) per L, where L is the step of the interval:

K

xmax  xmin . L

The step of the interval in this work must be an integer and selected so that the number of intervals is not less than 8 and not more than 20. The intervals are numbered as follows: – 1-st interval x min   x min  L  , – 2-nd interval x min  L    x min  2 L  , – k-th interval x min  ( k  1) L   x min  kL  . If the number of intervals is put along the abscissa axis and the number of measurements ni is along the vertical axis, the results of which fall in this range, we get a graph of the empirical distribution of the number of measurements by intervals, called a histogram 24

(Figure 1.1.). For a large number of measurements the ratio ni N indicates the probability of appearance of the measured values in the interval with step L. If we divide ni N by the step of the interval L, n the value y i  i will characterize the relative number of favorable NL cases in the unit interval. The diagram constructed for yi shows the distribution of the probability density by intervals and is called the reduced histogram. It has the form shown in Fig. 1.2.

Fig. 1.1. Distribution of the number of impulses by intervals (histogram)

Now imagine that the measurements continue until the number of measured values is very large. The step of the interval L can be made very small (provided the measuring device has a sufficient sensitivity), and yet in each interval there will be a lot of measurements. In this case y i can be considered as a continuous function of x) If now, instead of the above histogram, we construct a curve of the dependence y  f (x) , which gives the percentage of measurements ni falling in the unit interval in case of continuous change in x, we will get a smooth curve called the distribution curve. 25

The function y  f (x) is called the distribution density.

Fig. 1.2. The distribution of the probability density by intervals: 1 – for a finite number of measurements (reduced histogram); 2 – Gaussian curve

The product f ( x )dx ( dx is the differential of the independent variable) gives the fraction of the total number of measurements ni / N in the interval from x to x  dx . In other words, f ( x )dx is the probability that a randomly selected value of the measured quantity will be in the range from x to x  dx . The shape of the reduced histogram obtained for a small number of experiments cannot be predicted beforehand. However, the probability theory allows us to calculate the shape of the smooth curve, which is the limit for the histogram in case of an unlimited increase in the number of experiments. This limit curve is called the Gaussian curve (Figure 1.2.). The distribution corresponding to the limit curve is called a normal (Gaussian) distribution. It is described by the distribution function: f ( x) 

1 2  26

e



( xi  x  ) 2 2 2

,

(1.5)

where  2 , as it was mentioned above, is a dispersion;  characterizes the dispersion of measurements with respect to the arithmetic average value and is called a standard deviation or RMS error. Gauss function is normalized, i.e. f (x) satisfies the relation 

 f ( x)dx  1 .

(1.6)



The integral has infinite limits. This means that the measured value with a probability 1 (or 100%) is in the range from   to   , or that the presence of the measured quantity within this range is a persistent event. The probability density function has the following properties (see. Fig. 1.2.): − it is symmetrical with respect to  x  , − it has maximum in the point  x  , − it quickly vanishes to zero when x i   x  becomes large with respect to . Figure 1.3 shows distribution curves corresponding to different values of . This figure shows that for small  the curve is more narrow and has higher maximum, which corresponds to more precise measurements.

Fig. 1.3. Gaussian distributions for different  values 1=10, 2=20 and 3 =30;  x  500 . 27

1.3. Experimental technique 1.3.1. Devices and accessories: counter SP-100, stopwatch. 1.3.2. Counter SP-100 enables us to measure the number of pulses at the input for any time that passes from the moment we press «Start» to the moment we release it. The pulses are sent to the counter SP-100 from the generator of AC voltage with a frequency of 100 Hz. Hence, the average number of pulses detected per one second ranges about 100. In the work the number of pulses for 5s is measured. Measurements are made 100 times. 1.4. The procedure of work performance 1.4.1. Switch on SP-100. Wait 15 min till it warms up. 1.4.2. Measure the number of pulses xi registered by the counter for t = 5 s. Repeat measurements 100 times. Write the measurements in Table 1.1. Table 1.1. The number of pulses registered by SP-100 for 5s

No. 1 2 . . . 33

xi

xi

xi2

№ 34 . . . . 66

xi

xi

xi2

№ 67 . . . . 100

xi

xi

xi2

1.4.3. Calculate the arithmetic mean  x  of all results. 1.4.4. Calculate deviations of individual measurements xi and their squares xi2 , write them into the table. 1.4.5. Calculate the mean square error of individual measurements by the formula (1.3). Calculate the mean square error of the arithmetic mean using the formula (1.10). 1.4.6. Calculate the confidence interval for the reliability given by the teacher and record the final result using formula (1.12.). Using formula (1.13) calculate the relative error of the experiment. 28

1.4.7. Determine xmin and xmax, divide the space between them into intervals and number them. 1.4.8. Determine into which interval the individual measurement falls. 1.4.9. Sum up the number of measurements (ni ) in each interval, record the numbers in Table 1.2. 1.4.10. Construct a histogram. 1.4.11. Construct a reduced histogram. 1.4.12. Calculate the value of the probability density function for xmax j  xmin j values   x  , where j is the number of the interval, 2 i.e. for the middle of each interval, and record these values in Table 1.2. Table 1.2. Data for the histogram and Gaussian curve

Interval number (j): j=1…k

n

ni NL

1 )L  2   x   х x min  ( j 

 

x 2 2 2

e- 

f(x)

1 2

1.4.12. Draw a Gaussian curve on the same graph, where we have drawn a reduced histogram. Questions for self-assessment

1. 2. 3. 4. 5.

Give the definition of absolute and relative error. How are errors classified by properties? What properties does the normal distribution of random errors have? What is the reliability of the standard deviation? Which  of the two ( 1   2 ) corresponds to good-quality measurements?

29

References

1. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main pbdlishing house of Physical and mathematical literature, 1983. – p. 704. 2. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106. 3. Kubyshkina V.D. The main methods of mathematical processing of results of physical experiments / Kubyshkina V.D. – Almaty: Kazakh State University, 1974.

30

Laboratory work № 2 MEASUREMENT OF LENGTH, AREA AND VOLUME

2.1. The purpose of work: to study the methods of measurement of linear values, to evaluate the accuracy of these measurements, to master the technique of processing the results of indirect measurements. 2.2. Brief theoretical introduction 2.2.1. In most physical measurements (including laboratory works) the quantity we are interested in is not measured directly. Instead, we measure some other quantities and then calculate the value of y, which is a known function of measured values: y  f ( x1 , x 2 , x3 ,..) . The most probable value of the function is the value obtained by substituting into it the values of arithmetic means of direct measurements

 x1 ,  x 2 ,  x3 ,... . The estimation of errors of direct measurements was obtained in lab work No. 1. 2.2.2. Measurement errors are small values (their squares are beyond the measurement accuracy), so to calculate the measurement error, we can use the methods of differential calculus. If case of indirect measurements, it is possible to use the following methods. 1. у is a function of one variable, i.e. y  f (x) . In this case it is possible to write with a rather high accuracy:

 y  f ( x) x . 31

(2.1)

If we multiply both sides by the Student coefficient, we obtain:

y  f ( x)x ,

(2.2)

where x is a confidence interval for the given probability . 2. If y is a function of several variables x1 , x 2 , x3 ,...x n , which  1 ,  2 ,  3 ,...,  n are known,  f



2

 f



2

 f



for

2

 y    x1     x2   ...    xn   x1   x 2   x n 

or N

 f



2

 y     xi  , i 1  x i 

(2.3)

where f xi is a partial derivative with respect to xi ;  x are i

standard deviations of individual arguments. To calculate the confidence interval of the function we can use the formula 2

  f y    xi  , i 1  x i  N

(2.4)

if the confidence intervals of all direct measurements are determined with the same reliability. Then the reliability of the function is equal to the reliability with which the arguments are determined. The final result of measurements is recorded in the form

y  y   y

where  is the chosen reliability. 32

for   k % ,

2.2.3. In special cases, to determine the error of indirect measurements we can use a different formula instead of (2.4.). If the function y  f ( x1 , x 2 , x3 ,..) is a product, a ratio or a power function, it is more convenient to first determine the relative error. For example, if n

y

A  x1  x 2 x 3m

then y  y

 n  x1   x1

2

 x 2      x2 

,

2

  mx 3       x3

2

  . 

(2.5)

In the particular case, if all the exponents of the measured value are equal to 1, the formula (2.4) takes the form: y  y

  x1   x1

  

2

 x2    x2

  

2

 x3    x3

  

2

 ... ,

(2.6)

i.e. for such functions the relative error is equal to the square root of the sum of squares of relative errors of direct measurements. 2.2.4. We should remind that the confidence interval of the direct measurement can be determined by the formula: N

 x i  t , N 

2  ( x   xi )

i 1

N ( N  1)

,

where xi is the result of an individual measurement; N is the number of measurements; t , N is the Student coefficient for the given confidence level. At first, it might seem that if the number of measurements increases unlimitedly, the error can be arbitrarily small. This is 33

certainly not true. Only random measurement errors can be arbitrarily small, but not instrumental and systematic errors. Whatever the number of measurements with a ruler, the smallest division of which is 1 mm, we cannot obtain the measurement error less than 0.5 mm. Therefore, the question of the number of measurements should be considered in more detail. Do not limit by a single measurement. If for three measurements, the measured values are the same, you should stop measurements, and the measurement error will be equal to the instrumental error, which is defined by half of the smallest division or class of accuracy of the instrument. If the results of measurements differ, the number of measurements must be such that the random measurement error is less than the instrumental. Knowing its value, it is not difficult to find the necessary number of measurements, guaranteeing a small effect of random error on the accuracy of the results. When the reliability of 0.95 (confidence level) is required, i.e. the random error must be no more than the value of the standard deviation, the number of measurements must be 7. If in the mathematical processing of measurement results, the confidence interval is of the same order as the instrumental error, the total error is calculated by the formula 2 . xполн  xcл2 .  xпр

(2.7)

If one of the errors is three or more times smaller, it is not taken into account. 2.2.5. In some cases, the conditions of the experiment are nonreproducible (e.g., in the studies of cosmic radiation). In this case, the value of the function is calculated each time to get the results. Then, these results are used to calculate the arithmetic mean, and the error is calculated as for direct measurements. In this case we say that the results are processed by the method of direct measurements. In the laboratory of mechanics we also have irreproducible experiments. For example, in the work on measurement of the speed 34

of the bullet in a ballistic pendulum it is impossible to make the bullet hit the same target point. 2.2.6. In this work, we study the method of processing of indirect measurements on the example of determination of the volume of bodies of regular geometric shape. The volume is calculated from the results of linear measurements. 2.2.7. In science and technology to determine lengths and distances many devices with various values of measurement accuracy are used. For example, the vernier scale is widely used for measuring lengths. The vernier scale is used as an additional scale (linear or circular) to improve the accuracy of measurements 10, 20 times. The linear vernier is a small graduated ruler that can slide along the scale. The vernier divisions are made so that (m  1) divisions of the main scale correspond to m vernier divisions (see. Fig. 2.1.) Let the smallest division of the main scale be known and equal to А (usually a = 1 mm). Denote the vernier scale division as х, then x  m  a (m  1) , and x

m 1 a a a  . m m

Fig. 2.1. Linear vernier

The difference between the scale division of the main ruler and the vernier scale is called vernier accuracy: a a  a  x  a  a    . m m 

(2.8)

To measure the size of the object let us place the zero point of the object at the zero division of the scale ruler, and put vernier at the 35

end of the subject. Then the length of the object will be equal to: L  ka   L , where k is the number of the nearest division of the main scale to the left of "0" on the vernier (see. Fig. 2.2.) As the scale division is not equal to the division of the vernier scale interval, there must be the division of the vernier n, which is the closest to the scale division, then, as seen from Fig. 2.2

Hence,

a a  L  na  nx  na  n a     n . m m 

L  ka 

a n, m

(2.9)

i.e. the length of the interval measured by the vernier is equal to the number of whole divisions of the main scale, multiplied by the value of its division, plus the number of the division, which coincides with some division of the main scale, multiplied by the vernier scale interval. The accuracy of the vernier is usually indicated on the meter. The error of the results of measurements made using vernier is equal to the precision of the vernier.

Fig. 2.2. Linear vernier with a subject

2.2.8. Caliper. Linear verniers are used in calipers. The caliper (Figure 2.3.) consists of a millimeter-scale M (instrument scale) rigidly attached to the leg LA. A movable vernier N, rigidly connected to the second leg LB and rack F of the instrument may be attached along the scale. The movable part of the caliper is provided with a clamping screw C. 36

When there is no gap between legs A and B, zero divisions of the vernier and the scale coincide.

Fig. 2.3. Caliper: LA – fixed leg, LB – movable leg, С – clamping screw, М – scale, N – vernier.

To measure the external dimensions, the object is placed between legs A and B, which are moved until they contact the object. Then the movable leg is fixed with the screw C and measurement is made. The number of whole millimeters is calculated directly on the scale of the instrument to the zero mark of the vernier, the number of fractions of a millimeter is determined by the vernier, as described above. When it is necessary to measure internal sizes, legs LL are used and to measure the depth, rail F is used. Calipers are made with verniers for n = 10, 20, 50 divisions. 2.2.9. Micrometer. For more precise measurements micrometers are used. They have several types: micrometers for external measurements, depth micrometers and micrometer internal gauges. Micrometer for outdoor measurements (Fig. 2.4.) consists of a hollow rod rigidly connected to the bracket. Micrometer screw A is screwed into the cavity of the rod. When measuring, the object is clamped between the fixed rod and the movable end of the micrometer screw A. The microscrew is rotated holding the ratchet B. The drum C rotates together with the microscrew and moves translationally with respect to the rod. The micrometer reading device has two scales. The horizontal scale of the rod is a dual scale with 1 mm divisions graded on both sides of the longitudinal line so that the upper scale is moved relative to the lower one by half of a division. 37

The scale interval of the drum can be set as follows. Let the number of divisions on the drum be n = 50. Step of the microscrew is h = 0.5 mm, i.e. one complete revolution of the microscrew (and drum) corresponds to the linear movement of the edge of the drum of 0.5 mm. The scale division of the round scale is:

a  h n  0,5 50  0,01 мм . The readings are taken in the following way: the size of the measured object is taken along the horizontal rod with an accuracy up to 0.5 mm. Hundredths of a millimeter are measured on the circular scale of the drum. The results are added. The number of hundredth fractions of the scale division corresponds to the division of the scale located against the line on the rod. The order of readings is the same for all types of micrometers.

Fig. 2.4. Micrometer: А – screw B – screw head, D – basic scale, C – drum with a scale

2.3. Procedure of work fulfillment 2.3.1. Devices and materials: caliper, micrometer, the body to be studied. 2.3.2. Get familiar with the caliper, determine the scale division of the main scale and accuracy of the vernier scale. 2.3.3. Get familiar with the micrometer. Determine the scale division of the main scale and of the drum. 2.3.4. Learn how to use these instruments. Choose measuring instruments that can be used to measure linear dimensions. When choosing instruments, you must remember that you must get the highest possible accuracy. 38

2.3.5. Measure the linear dimensions of the object, record the results in Table 2.1. Table 2.1. Linear sizes of the studied body

No. 1 2 3 4 5 6 7

ai, mm

bi, mm

=

=

ci, mm

=

2.3.6. Calculate the mean values of each linear size, record them in the same table. 2.3.7. Calculate the deviation of each measured value from the mean and record it in Table 2.2. A sample of the table is given below. Table 2.2. Data for error calculation

N o. 1 2 . . 7

ai, mm

ai2, mm2

bi2, mm2

bi, mm

ci, mm

ci2,mm2

2.3.8. Calculate the mean square error of the arithmetic mean

S  x  for each linear size using the formula N

S x 

 ( x   x ) . (see lab No. 1). i 1

2

i

N ( N  1)

39

2.3.9. Determine the confidence interval for the number of measurements to be made and the degree of reliability  = 0.95 for each linear dimension. 2.3.10. Determine the total error for each linear size, taking into account the instrument error of the measuring instrument. 2.3.11. Record the final result for each linear size in the form x  x   x for  = 0.95. 2.3.12. Calculate the relative error of each linear dimension. 2.3.13. Determine the absolute and relative errors of measurement of the volume choosing the most suitable formula (see Eq. (2.4) or (2.6)). 2.3.14. Record the final result for the volume. Questions for self-assessment

1. 2. 3. 4.

How is the accuracy of the vernier calculated? What is the instrument error of the micrometer and calipers? Give examples of direct and indirect measurements. What methods of calculating errors of indirect measurements do you know? 5. What is the absolute measurement error, if the random error is zero? References

1. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 2. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106. 3. Practicum in physics. Mechanics and molecular physics: Textbook / Belyankin A.G., Motulevich G.P. et al.; Ed. Iverenova V. I. – M.: Nauka, 1967. – p. 352. 4. Rabinovich S.G. Measurement error / S.G. Rabinovich – L., Energy, 1978. – p. 261.

40

Laboratory work № 3 MEASUREMENT OF THE MOMENT OF INERTIA OF THE DISK

3.1. Purpose of the work: using the law of conservation of mechanical energy, determine the moment of inertia of the disk by two methods: a dynamic method and a method of vibrations. 3.2. Brief theoretical introduction 3.2.1. The basic equation of dynamics of rotational motion. The basic equation of dynamics of rotational motion of a solid body around a fixed axis, in the projection on this axis, is written as: M I

d  I  , dt

(3.1)

where М is the projection of the moments of forces acting on the body on the axis of rotation; I is the moment of inertia of the body with respect to the same d axis;   is angular acceleration. dt As the rotation axis is fixed, the rotational angular velocity varies only in magnitude. At constant torque of acting forces the angular acceleration is constant, i.e., the rotation of the body is either uniformly accelerated or uniformly decelerated. 3.2.2. The moment of inertia of the body. The moment of inertia of the body is a physical quantity, similar to the weight of the body in translational movement, i.e., it characterizes the inertia of the body in rotational motion. The 41

moment of inertia of the body with respect to the axis depends on the distribution of the body mass with respect to this axis but does not depend on the movement of the body. The moment of inertia of the material point with respect to an axis is the product of its mass and the square of the distance from the axis: I  mr 2 . The moment of inertia about an axis is the sum of the moments of inertia of all elementary masses mi of the body with respect to the same axis: n

I   mi ri2 . i 1

For the body, the density of which is equal to  , the moment of inertia can be calculated by integrating: I   r 2 dm    r 2 dV . m

V

If the moment of inertia with respect to the axis passing through the center of mass of the body is equal I C , the moment of inertia I with respect to any axis parallel to the first one can be calculated using Huygens-Steiner's theorem: I  I C  md 2 ,

where d is the distance between the axes. Using the equation (3.1) and determining the moment of acting forces and angular acceleration, we can calculate the moment of inertia of the body. In experimental works the body is often made to rotate or swing with the help of a load falling from the height. The potential energy of the falling weight is converted into kinetic energy of the falling load and rotational kinetic energy of the body. Using 42

the law of conservation of mechanical energy under the condition of small friction losses, we can calculate the moment of inertia of the body. This method is used in this work. 3.3. Experimental setup and measurement procedure 3.3.1. Instruments and materials: – A disk mounted on the horizontal axis; – A stopwatch; – A caliper; – A set of loads; – A massive ball. 3.3.2. Determination of the moment of inertia of the disc by the dynamic method. We use the scheme of the setup shown in Figure 3.1. The disc with the shaft is placed on the horizontal axis OO with respect to which it rotates. The axis OO coincides with the axis of symmetry of the disc so that the disc is in the state of neutral equilibrium. The thread at the end of which there are loads, creating the torque, is attached to the shaft of the disc.

Fig. 3.1. The experimental setup for determining the moment of inertia of the disc by the dynamic method

If the thread is wound on the shaft, the load will rise to a certain height h, and the system will receive a potential energy equal to the product of the force of gravity on the height of rise. When the disc is released, the load starts to move down, causing the disc to rotate. The potential energy of the raised load is converted 43

into kinetic energy of translational motion of the load and rotational motion of the disk. The law of conservation of mechanical energy can be written as: mgh 

m 2 I 2 ,  2 2

(3.2)

where h is the distance traveled by the load from the starting point till falling on the platform;  is the linear speed of the translational motion of the load at the moment when it touches the platform; ω is the angular speed of rotation of the disk when the load touches the platform. The linear speed of translational motion of the load coincides with the linear speed of the rotational movement of points on the surface of the shaft (assuming there is no slippage of the thread). The angular velocity of rotation of these points is the angular velocity of disk rotation and is determined from the well-known relationship:    / r , where r is the radius of the shaft. Using equation (3.2), we can show that the movement of the load is uniformly accelerated. Let us rewrite equation (3.2), using the I notation   : mr 2

mgh 

m 2 (1   ) . 2

Then we obtain:

h

2 2g

(1   ) .

Comparing this expression with the formula for the path in

2 2 uniformly accelerated motion h    0 , we see that

2a

44

a

g  const .  1

Let us apply the laws of uniformly accelerated motion: h

2h at 2 ,   2h ,    r ,   , tr 2 t

(3.3)

with initial conditions:  = 0,  = 0 for t = 0. Substituting the values  and  in the formula (3.2), we obtain the expression for the moment of inertia of the disk (with the shaft): mgh  and

4m h 2 4 I h 2 2mh 2 2 I h 2  2 2  2  2 2 . 2t 2 2t r t t r

I

mr 2 ( gt 2  2h) . 2h

(3.4)

3.3.3. Determination of the moment of inertia of the disk by the method of vibrations. In this work we use the setup of the previous exercise, but on the rim of the disk a massive ball is placed (Fig. 3.2).

Fig. 3.2. The experimental setup for determining the moment of inertia of the disk by the method of oscillations

45

Initially, the disk with the ball is in the state of stable equilibrium. If this state of the system is violated (by turning the disk at a small angle (   8 0 ), it causes fluctuations of the system «disk – ball» with a period T around the horizontal axis. Neglecting the friction torque, we can write the equation of motion of the disk with the ball in the form:

I  I 1  d

2



dt 2

 m1 g ( R  R1 ) sin  ,

(3.5)

where I is moment of inertia of the disk with a shaft with respect to the axis of rotation OO, I 1 is the moment of inertia of the ball with respect to the axis of rotation OO, calculated by the theorem of Huygens – Steiner: I1 

2 m1 R12  m1 ( R  R1 ) 2 , 5

(3.6)

where R and R1 are radii of the disk and ball, respectively;; m1 is the mass of the ball;  is the angle of deflection from the equilibrium position of the disk. At small angles of deflection it can be assumed that sin    and the equation (3.5) takes the form: m g ( R  R1 ) d 2    02 ,  1 2 I  I1 dt

if we use the notation

(3.7)

m1 g ( R  R1 )   02 . I  I1

As it is known, the solution of this differential equation is a harmonic function

   o sin( 0 t   ) , 46

(3.8)

where  0

is the angular oscillation amplitude;  is the initial

phase;  0 is the cyclic frequency of oscillations:

0 

m1 g ( R  R1 ) 2 .  T0 I  I1

(3.9)

By measuring the period of oscillations of the disk with a ball

T0 and knowing the mass m1 and radius R 1 of the sphere, from

formulas (3.6) and (3.9) we can find the moment of inertia of the disk with respect to the axis of rotation:

 gT 2  2 I  m1  02 ( R  R1 )  R12  ( R  R1 ) 2  . 5  4 

(3.10)

3.4. Procedure of work fulfillment 3.4.1. Determination of the moment of inertia of the disk by the dynamic method. 3.4.1.1. Examine the setup. Tighten the bolt to the stop in the hole for ball fastening (to equilibrate the disk with respect to the axis of rotation). 3.4.1.2. Using the caliper measure the diameters of shafts 2r1 and 2r2 (large and small), the diameter of the disk 2R, the disk thickness a and the diameter of the ball 2R1 . The results of measurements must be recorded into Table 3.1. The density ρ of the material (steel) of the disk and the ball is given in the specifications of the setup. Table 3.1 Linear sizes of the setup and density of the material of the disk and the ball No. 1 2 3

2r1, mm

= =… mm

2r2, mm

= =… mm

2R, mm

А, mm

2R1, mm

= =… mm

= =…mm

= =… mm

47

, kg/m3

3.4.1.3. Attach one of the loads to the end of the thread (200 ÷ 500 g). At the moment of starting the stopwatch release the system and determine the time t of falling of the load from height h on the platform (height was measured from the lower base of the load). Make the experiment at least five times. Write the results of measurements in Table 3.2. Table 3.2 Time of falling of a load of mass m from height h = … m, holder mass m0  …kg.

№

m=

g

r1

1 2 . 5

m=

r2

r1

g

m=

r2

r1

g

m=

r2

r1

g

r2

t, s

t, s

t, s

t, s

t, s

t, s

t, s

t, s

t 

t 

t 

t 

t 

t 

t 

t 

3.4.1.4. Determine the numerical value of the moment of inertia of the disk by the formula (3.4). Write down the results of calculations into Table 3.3. 3.4.1.5. Calculate the error of the result by the method of direct measurements. Table 3.3 Results of calculation of the moment of inertia of the disk

r, m

m  m0 , kg

t , s

I, kg·m2

48

 I , kg·m2

I , kg·m2

I 2 ,

kg2·m4

3.4.1.6. Calculate the moment of inertia of the disk using its geometrical dimensions: the radius of the disk R, the disk thickness and density ρ of the disk material. 3.4.1.7. Compare the results. Explain the reasons for possible discrepancies. 3.4.2. Determine the moment of inertia of the disk by the method of vibrations. 3.4.2.1. Take the load from the thread. Attach the ball to the rim of the disc. 3.4.2.2. Remove the system from its equilibrium position and determine the time of 20 ÷ 30 oscillations for the same initial value of the oscillation amplitude  0 , given by the teacher. Repeat the experiment 5 ÷ 6 times. Write down the results of calculations into Table 3.4. 3.4.2.3. Calculate the moment of inertia of the disk using the formula (3.10). 3.4.2.4. Calculate the error by the method of direct measurements. 3.4.2.5. Compare the experimental results obtained by different methods. Explain the reasons for possible discrepancies. Table 3.4 Results of measurements and calculations. The number of oscillations n = ..., the initial amplitude of oscillation α0 = … №

t, с

T, с

I , êã  ì

2

 I , êã  ì

2

J , êã  ì

2

J 2 , êã2  ì

4

1 2 . 5 Questions for self-assessment

1. Write the equation of motion of bodies in translational and rotational motions and justify the validity of the use of the formula a   r . 2. What is called the moment of inertia with respect to the axis? What is the physical meaning of this value? What determines the moment of inertia? 49

3. Formulate the Huygens-Steiner theorem. 4. Why is the motion of the system uniformly accelerated when the moment of inertia of the disk is determined by the dynamic method? 5. The oscillation method for determining the moment of disk inertia uses the theory of small oscillations. What simplification is allowed in this case? 6. Name the reasons causing errors in the experiment. Will the oscillation and rotation of the moving load and the disbalance of the disk affect the accuracy of measurements? 7. Derive the formula for finding the moment of inertia of the disk by the dynamic method. 8. Derive the formula for finding the moment of inertia of the disk by the method of oscillations. References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989 – p. 576. 6. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 7. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

3.5. Assignment for Student research work 3.5.1. The purpose of research: – Further investigate the main factors affecting the accuracy of the experimental determination of the moment of inertia of the disk; – Check experimentally the dependence of the oscillation period of the pendulum on the amplitude of oscillations; – Determine more accurately the value of the moment of inertia of the disk taking into account the influence of 50

friction and the dependence of the oscillation period on the amplitude. 3.5.2. Investigation of the influence of the moment of resistance forces. When determining the moment of inertia of the disk by formula (3.4.) we did not take into account the moment of friction. It is recommended to carry out measurements and calculations similar to those in p.p.4.4.2 - 4.4.4 and 4.4.7 - 4.4.8 in work 4, «Studying of the fundamental law of dynamics of rotational motion» taking into account the effect of the moment of resistance forces in determining the moment of inertia of the disk by the dynamic method. 3.5.3. Study the influence of the oscillation period on the amplitude. 3.5.3.1. In case of not very small values of the amplitudes of the deflection angle (  0  5 0 ) in equation (3.5) the value sin cannot be replaced by the value of its argument . In this case it is necessary to solve the nonlinear equation: d 2 dt

2



m1 g ( R  R1 ) sin  . I  I1

(3.11)

As in formula (3.9)) we will use the notation:

 02 

m1 g ( R  R1 ) I  I1

and integrate the equation (3.11) with respect to the angle. In the integration it is convenient to count down from the maximum value

 0 when the speed of the pendulum is zero: d  0 ,    0 for dt

t = 0.  d 2 d   02  sin  d .  2  0 dt 0 

51

(3.12)

Transform the integrands: 2

2

1 d  d  1  d  d 2 d 2 d  d   dt   ,   dt  d  2 2 2 dt  dt  2  dt  dt dt dt

sin   d  d (cos  ) . Taking into account these transformations we obtain from (3.12): 2

 d  2    2 0 (cos   cos  0 ) .  dt 

(3.13)

We can rewrite it as: d cos   cos  0

 2 0 dt .

(3.14)

Let us use the transformation formula and introduce a new variable:   sin  2 . sin   0  sin   2 

(3.15)

Then equation (3.14) can be written as: d 1  sin 2

0 2

  0 dt .

(3.16)

 sin 2 

When integrating this equation, we assume that for t = 0, the pendulum moves to the position of equilibrium, i.e.,  = 0, in a 52

quarter period, t  T ,    0 , therefore, on the basis of (3.15) we 4

T   . 4 2 Taking this into account let us integrate (3.16) over time from 0  to t  T , and over  from zero to : can write that for t = 0  = 0, and for t 

2

4

T /4

 /2

0

0

d

 0  dt  

1  sin

2 0

2

.

(3.17)

sin 2 

From (3.17) taking (3.9) into account we get: T 4

 /2 I  I1   m1 g(R  R1 ) 0

d 2 0

1 sin

2

.

(3.18)

 sin2 

As seen from (3.18), in the general case the period of oscillation of any physical pendulum depends on the amplitude of oscillations  0 . Only if  0  0 we get: T0  4

 /2 I  I1 I  I1 .   d  2 m1 g ( R  R1 ) 0 m1 g ( R  R1 )

Here T0 is the period of oscillations of a physical pendulum with infinitesimally small amplitude of oscillations, for which the assumption sin    is valid. The integral in the right side of (3.18) is a full elliptic Legendre integral of the first kind. This integral may be written as a series: 53

2 2 2    1  2  1 3  4  1 3 5  6      1 k k k  ,         2   2  0 1 k2 sin2   2  4 6   2 4  

/2

d

(3.19)

where   k  sin  0  .  2 

(3.20)

When determining the moment of inertia of the disk by the method of oscillations, in formula (3.10) it is necessary to substitute the value of the oscillation period of the pendulum T0 with infinitesimal amplitude. In fact, we substitute in this formula the value of T, determined for fluctuations with finite amplitude  0 . Therefore, to determine the moment of inertia of the disk more accurately, we should take into account the effect of the amplitude of fluctuations on its period. 3.5.3.2. Procedure of work fulfillment. – Attach the ball to the rim of the disk and align the arrow of the pointer of the angle of disk rotation with zero in the position of stable equilibrium of the disk with the ball. – Deflect the disk from the equilibrium position at an angle  0 н  5 0 and release it without initial velocity. Find the period of oscillation by measuring the time of 30 fluctuations 35 times. Register and record the final oscillation amplitude  0 ê . Record the data in the table. – Repeat the experiments increasing the initial amplitude  0

by 50100, up to angles  0 = 700800. – Using the results of experiments calculate the average    0к amplitude of oscillations   0  0 н for each test 2

and the average value of the oscillation period for them 54

N

 T 

 Ti

i 1

N

and construct the graph of dependence of the oscillation period  T  on the mean amplitude of oscillations   0  .

− For each value  0 in the experiment, calculate the relative change in the oscillation period of the pendulum from (3.18) and (3.19): f (0 ) 

  1 9 25 6 0 T 1  sin2 0  sin4 0  sin  ... (3.21) 4 2 64 2 256 2 T0

– Divide the experimental values of the oscillation period for each value  0 by f ( 0 ) , calculate the mean value of the period T0 from all the experiments. – Substituting the value of T0 determined above in the formula (3.10), calculate the moment of inertia of the disk and compare the obtained value I with its value calculated not taking into account the influence of the amplitude of oscillation on its period. Supplimentary reading

1. Korn G., Korn T. Reference Book of Mathematics / G. Korn, T. Korn. – M.: Nauka, 1974. – p. 832.

55

Laboratory work № 4 STUDY OF THE FUNDAMENTAL LAW OF DYNAMICS OF ROTATIONAL MOTION

4.1. The purpose of work: to check experimentally the fundamental law of dynamics of rotational motion on the X-shaped Oberbeck pendulum. 4.2. Brief theoretical introduction When you rotate a material point of mass mi around a circle of radius ri its momentum in the projection on the axis of rotation is equal to: Li  mi i ri . Linear speed of movement  i is related to the

angular velocity  by the relation, therefore Li  mi ri2 . If a system of material points with the same angular velocity  rotates around n

the axis 0, then L   mi ri2 where summing up is over all the i 1

material points of the system. The value of , the same for all material points, can be taken out from under the summation sign, and get: where

L  I ,

n

I   mi ri2 . i 1

(4.1)

(4.2)

The quantity I, equal to the sum of products of masses of material points to the squares of their distances from the axis of 56

rotation, is called the moment of inertia of the system with respect to this axis. The moment of inertia of a body is a physical quantity, similar to mass in the forward movement, i.e., it characterizes the inertia of the body in rotational motion. The moment of inertia of a solid body with respect to the axis depends on the distribution of the body mass with respect to this axis but does not depend on the movement of the body. In case of continuous mass distribution the sign of the sum is replaced by the integral sign, and the formula (4.2) is written as: I   r 2d m . m

Taking (4.1) into account the fundamental law of dynamics of rotational motion in the projection on the axis of rotation can be written as: dL d  I   M , dt dt

(4.3)

where М is the projection of the total moment of external forces on the axis of rotation. This equation is called the equation of moments. It is easy to see that in the rotational motion the role of acting   force F is played by the moment M :

 

   M  r,F ,  i.e. the vector product of theradius vector of the point r of  application of force on the force F . In the particular case of a solid body rotation around a fixed axis, the equation (4.3.) can be written as: I

d M dt 57

(4.4)

or

I  M .

(4.5)

The product of the moment of inertia of the body with respect to a fixed axis of rotation on angular acceleration  is equal to the moment of external forces with respect to the same axis. Comparing the equation (4.5) and Newton's second law, we see that the moment of inertia of the body during rotation plays the same role as the mass of the body during its translational motion. In this work you have to check that if for a given mass distribution with respect to the rotation axis the moment of force M changes, the angular acceleration changes so that these quantities remain proportional to each other. The proportionality coefficient is equal to the moment of inertia I. 4.3. Description of the setup and methods of work fulfillment 4.3.1. The setup is shown in Fig. 4.1. The Overbeck pendulum consists of four spokes, fixed on the sleeve at right angles to each other. On the same plug there are two pulleys of different radii r1

and r2 . This system may freely rotate with respect to the horizontal axis. The moment of inertia of the system can be changed by moving the cylinders m1 along the spokes. The moment of forces is generated by the load m attached to the thread H, which is wound on one of the pulleys. If the friction moment M òð applied to the axis of the pendulum is small in comparison with the moment of the tension force of the thread, it is not difficult to check the validity of the equation (4.5). Indeed, measuring the time t, within which the load falls from its initial position over a distance h, it is easy to find the acceleration a of the load in the projection on the coordinate axis coinciding with the direction of movement of the load: a

2h . t2

58

(4.6)

Acceleration is connected with the angular acceleration  (without thread slippage with respect to the rim of the pulley) by the relation: a  r ,

(4.7)

where r is the radius of the pulley.

Fig. 4.1. Overbeck pendulum

If Т is the tension force of the thread, then M  rT .

(4.8)

The force of the thread tension can be found from the equation of motion of the load: T  m( g  a ) .

ma  mg  T ,

Then

M  r  m g  a  . 59

(4.9)

4.3.2. The moment of friction forces M тр is quite large and depends on the speed of rotation. Therefore, M тр can significantly distort the results of the experiment. However, in the first approximation, we can take the moment of resistance constant and independent of speed. Then, taking into account the friction moment the equation (4.5) can be written as: M  M тр  I .

(4.10)

The task of taking into account the dependence of the resistance moment on speed was discussed in Section Student research work. 4.4. The procedure of work fulfillment 4.4.1. Before starting the experiment, it is recommended to put the pendulum into rotation several times, each time giving it time to stop, to be sure that it is its balanced. Measure the linear dimensions of the setup and write the results of measurements in table 4.1. Here, r is the radius of the pulley on which the thread is wound; h is the height, from which the load falls; l is the length of the generatrix of the hollow cylinder; R1 is the outer radius of the hollow cylinder;

R2 is its inner radius; R is the distance between the rotation axis and

the center of mass of the hollow cylinder.

Table 4.1 Results of measurements of linear dimensions of the setup

No. 1 2 3

2r, mm

h, mm

=

l, mm

2R1, mm

2R2, mm

R, mm

=



=

4.4.2. After attaching the load of mass m to the thread H, measure the time the load needs to fall from the height h. Repeat the experiment 5-7 times. The experimental results are tabulated 4.2. 60

4.4.3. Repeat p.4.4.2.for 4-5 different values of the load mass m. 4.4.4. Average the obtained values of t for each mass m. Determine the acceleration a of the mass using the formula (4.6.). Calculate the corresponding angular acceleration  and moments M by the formulas (4.7) and (4.9). Write the results in Table 4.2. 4.4.5. Fix the cylinders on the rods and repeat the experiments described in paragraphs 4.4.2. - 4.4.3., write the experimental results in Table 4.3. Table 4.2 The results of measurements and calculations for the pendulum without cylinders on the crosspiece, the weight of the holder m0 = kg m+m0, kg

t, s

, s

А, m/s2



, s2

T, Н

M, Н·m

m0 m+m0

4.4.6. Make the calculations for the pendulum with additional cylinders as in p. 4.4.4 and write the results in Table 4.3. Table 4.3 The results of measurements and calculations for the pendulum with cylinders on the crosspiece, the weight of the holder m0 = kg m+m0, kg

t, s

, s

А, m/s2



, s-2

T, Н

M, Н·m

m0 m+m0

4.4.7. Process the experimental results by the method of least squares. Taking into account the moment of the friction force, the basic law of dynamics of rotational motion can be written as: 61



M  M тр. I

.

It is more convenient to write this dependence as: M  I  M тр or y  Ax  B ,

where: y  M , x   , A  I , B  M тр . A and B coefficients are determined by the method of least squares using the following formulas: n

n

n

n

n

n

n

2 xi xi yi  xi yi

 yi  xi n xi yi

i1 i1 i1 i1 I  А  i1 i1 2 i1 , Mтр  В  . 2 n n n  x  n n x2  x   n x2  i  i   i  i  i1  i1 i1 i1 

It is convenient to make calculations using Table 4.4. Table 4.4 Data for calculating parameters of the linear dependence y  Ax  B No.

xi

yi



M

 xi 

 yi 

xi2

xi y i

1 2 .

n n

i 1

n

n

2  xi 

i 1

i 1

n

 xi y i 

i 1

Make calculations for the pendulum without cylinders and with additional cylinders. 4.4.8. Present the results of measurements and calculations as graphs (separately for the unloaded and loaded pendulums) plotting 62

 value along the abscissa and M value on the ordinate. Using the linear dependence of parameters A and B found by the least square method, draw straight lines that should best describe the experiment. 4.4.9. Knowing the value of the moment of inertia of the pendulum with cylinders I0, the moment of inertia of the pendulum without cylinders and using the additivity property of the moment of inertia, calculate the moment of inertia of one cylinder with respect to the axis of rotation: I cyl 

I  I0 . 4

(4.11)

4.4.10. Calculate the theoretical value of the moment of inertia of a hollow cylinder I C with respect to the axis passing through the center of gravity perpendicular to the generatrix using the formula: IC 

1 1 m1l 2  m1 ( R12  R22 ) , 12 4

(4.12)

where R1 is the outer radius of the hollow cylinder; R2 is its inner radius of the hollow cylinder; l is the length of the generatrix of the cylinder; m1 is the mass of the cylinder. 4.4.11. Using the Huygens – Steiner theorem, calculate the moment of inertia of the cylinder with respect to the axis of rotation of the pendulum: I cyl  I C  m1 R 2 ,

(4.13)

where R is the distance between the axis of rotation of the pendulum and the axis passing through the center of mass of the hollow cylinder parallel to the axis of rotation of the pendulum. 63

4.4.12. Compare the value of the moment of inertia of the cylinder I cyl relative to the axis of rotation of the pendulum calculated by formula (4.13) with the value obtained experimentally in Sec. 4.4.9 (Formula (4.11)). 4.4.13. Determine possible causes of errors in the experiment. Questions for self-assessment

1. Formulate and write down the fundamental law of dynamics of rotational motion. 2. What does the moment of inertia of the body with respect to the axis mean? 3. What formula simplifies calculation of the moment of inertia of the body with respect to the axis not passing through the center of mass? 4. Under what conditions do we have the right to consider the linear acceleration of the points on the rim of the pulley equal to the acceleration of translational movement of the load? 5. Show the direction of the moment of force, angular velocity and angular acceleration of the pendulum when it rotates. 6. Is it possible to determine the moment of inertia of the pendulum using the law of conservation of mechanical energy? 7. In the conditions of this experiment can we consider cylinders fixed on the rods as material points?

4.5. Assignment for Student research work 4.5.1. The purpose of research: determine a more precise value of the moment of inertia of the cross-shaped pendulum (Oberbeck) and cylinders, taking into account the dependence of the moment of friction on the speed. 4.5.2. When determining the moment of inertia of the crossshaped pendulum I 0 and the pendulum with weights I using the formula (4.10) it is assumed that the friction moment does not depend on the speed of rotation of the pendulum and is equal to the maximum moment of friction at rest. In fact, the friction moment depends on the speed of rotation of the pendulum and at high speeds it can reach values much larger than the values of the moment of friction. Therefore, a more accurate measurement is necessary to take into account the dependence of the moment of friction on the speed of the pendulum. 64

Suppose that in the first approximation, the moment of friction increases with increasing rotation speed according to the linear law: M тр  M 0  k ,

(4.14)

where M 0 is the maximum moment of friction; k is the coefficient of proportionality;  is the angular velocity of rotation of the pendulum. Then the equation of motion is written as:



d  m( g  a )r  M 0  k , dt

(4.15)

where r is the radius of the pulley; m is the mass of the load suspended on the thread. Linear acceleration of the falling load and angular d acceleration of the pendulum are related by (4.7) or a  r  dt and after substituting this expression into (4.15), it can be written as:

d dt  . mgr  M 0  k I  mr 2

(4.16)

Integrating this equation bearing in mind that  = 0 for t = 0 (at the initial moment the pendulum does not rotate), we obtain: kt   mgr M0  2 I   1  e mr  .   k  

65

(4.17)

It is seen that as the moment of friction depends on the speed of rotation, at t   the angular velocity of rotation tends to the maximum value

 max 

mgr  M 0 . k

(4.18)

Differentiating (4.17) with respect to time, we obtain the expression for the linear acceleration of the load: kt

 d mgr  M 0 2  re I  mr . a  r  2 dt I  mr

(4.19)

This expression shows that the acceleration of the load is maximal at the initial moment of time t = 0: a0 

mgr  M 0 I  mr 2

(4.20)

r

and decreases exponentially to zero for t   . The distance traveled by the falling weight during time t is found by the formula: mgr  M 0 h     rdt  k 0 t

Let us denote series:

kt    2  I   r t  1  e mr    

 I  mr 2   .  k   

(4.21)

kt  x and write the exponent e  x as a 2 I  mr

e x  1 

x x2 x3 x4 x5      . 1! 2! 3! 4! 5! 66

As in the conditions of our experiment

kt  1 , we confine I  mr 2

ourselves only by the first 4 terms of the series. Then (4.21) takes the form: h  a0

a 02 t2 kt 3 .   2 ( mgr  M 0 ) r 6

(4.22)

To determine the unknown quantities a0 and k for the same load m measure the time t1 and t 2 needed for the load to fall from two different heights h1 and h2 : h1 

a 0 t 12 a 02 kt 3   1 , 2 ( mgr  M 0 ) r 6

h2 

a 0 t 22 a 02 kt 3   2 . 2 ( mgr  M 0 ) r 6

Solving this system of equations, we get:

a0  2

k

h2 t13  h1t 23 , t 22 t13  t12 t 23

(h t 2  h2 t12 )(t12 t 23  t 22 t13 ) 3 ( mgr 2  M 0 r ) 1 2 . 2 ( h1 t 23  h2 t13 ) 2

(4.23)

(4.24)

Substituting the value of acceleration from the formula (4.23) into (4.20), we obtain the final formula for determining the moment of inertia of the cross-shaped pendulum (Oberbeck):

I  (mgr  M 0 )

rt12 t 22 t1  t 2  mr 2 . 2 h2 t13  h1t 23 67

(4.25)

4.6. Procedure of work fulfillment 4.6.1. By increasing the load on the thread H, find the minimum value m0 at which the pendulum starts to rotate. Estimate the value of the moment of friction at rest M 0 . 4.6.2. Fix the cylinder to the spokes of the pendulum at the same distance from the axis, as in exercise 4.4.5. Suspend one of the loads on the thread. 4.6.3. Measure the time t1 of load falling from height h1 and t 2 for falling from height h2 (h2  2h1 ) . 4.6.4. Determine the average time for  t1  and  t 2  by the results from p.4.7.3. 4.6.5. Substitute the obtained values t1 , t 2 , h1 , h2 and the value M 0 found by the data obtained in 4.7.1, in the formula (4.25) and find the moment of inertia of the pendulum I with four cylinders. 4.6.6. Remove the cylinders from the pendulum rod and repeat experiments similar to pp.4.7.3. - 4.7.4. According to the results of experiments using formula (4.25) find the moment of inertia of the pendulum I 0 without additional loads (take M 0 from 4.7.1). 4.6.7. Find the moment of inertia of the load with respect to the axis of rotation of the pendulum, using (4.11), and compare the results of experiments with theoretical calculations based on formula (4.13) and the results obtained without taking into account the dependence of the moment of friction on the rotation speed. 4.6.8. According to the formula (4.24), calculate the resistance coefficient k, determine the final angular velocity by the formula (4.17) and determine the value of the moment of resistance at the end of falling of loads. Compare the values of the moment of resistance with the value of the moment of friction at rest M 0 . 4.7. Questions for self-assessment 4.7.1. Try to answer the question: Why should the moment of resistance depend on speed? 68

4.7.2. How reasonable is the assumption of proportionality of the moment of resistance of the pendulum on the speed in the first approximation? 4.7.3. Try to deduce formulas for the case when Мтр=М0+k2. References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / |D.V. Sivukhin. – M.: Nauka, 1989. – p. 576. 6. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 7. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

69

LABORATORY WORK № 5 STUDY OF THE LAW OF OSCILLATIONS OF A PHYSICAL PENDULUM

5.1. The purpose of work: experimental test of the law of oscillations of a physical pendulum with small angular amplitude; development of the method of determining the acceleration of gravity with the help of a physical pendulum. 5.2. Brief theoretical introduction 5.2.1. The period of oscillations of a physical pendulum. Physical pendulum is a solid body of an arbitrary shape attached to a fixed horizontal axis (not passing through its center of mass) and able to oscillate under the influence of gravity with respect to this axis. In the state of stable equilibrium of the physical pendulum the line passing through the suspension point O and the center of mass C is directed vertically (see Figure 5.1). If the pendulum (Figure 5.1) deviates from the equilibrium position at an angle  and released, it will oscillate under the action of force of gravity relative to the axis of rotation. The projection onto the rotation axis of the moment of gravity force is:

M   mga sin  ,

(5.1)

where m is the mass of the pendulum; g is the acceleration of free all; a is the distance from the point of suspension O to the center of mass C. The Basic Law of dynamics of rotational motion for a physical pendulum in the projection to the axis of rotation is written as follows (see. Appendix): 70

I

d 2   mga sin  , dt 2

(5.2)

where I is the moment of inertia of the pendulum relative to the axis of rotation. If the deflection angle is small (   5 0 ), we can take approximately that sin   . Then, introducing the notation

mga   o2 , I

(5.3)

the equation of motion of the pendulum (5.2) is written as:

d 2   o2  0 . dt 2

(5.4)

The solution of this differential equation is a harmonic function:

   o sin(  o t   o ) ,

(5.5)

where  0 is the angular amplitude of oscillations (in radians);  0 is the initial phase of oscillations;    0 t   0 is the phase of oscillations at a given moment of time. Let us define the physical meaning of the quantity  0 . Since the function sin  is periodic with the period equal to  2  1  2 , the time T  t 2  t1 required to make a complete oscillation is found from the condition:

ot2  o   ot1  o   2 , hence

 o t 2  t1   oT  2 . 71

(5.6)

Thus, the time T of one complete oscillation, called the period of oscillations, is determined as:

T

2

o

 2

I . mga

(5.7)

The number of oscillations per unit of time is called the frequency and is denoted  . Then

 1 T . Angular (or cyclical) frequency is the number of oscillations per 2 seconds, that is, the angular frequency is equal to:

o 

2  2 . T

A comparison with formula (5.7) shows that the frequency  0 we introduced above is the angular frequency. Comparing the expression (5.7) with the formula for the period of oscillations of a mathematical pendulum T  2

lo , g

(5.8)

one can conclude that the mathematical pendulum, whose length is equal to I l0  , (5.9) ma will have the same period of oscillations as the physical pendulum. The value l0 is called the reduced length of the physical pendulum. 72

The reduced length of the physical pendulum is the length of the mathematical pendulum with a period of oscillations equal to the period of oscillations of the physical pendulum. 5.2.2. Center of oscillations. The point, which lies on the line passing through the point of suspension and the center of mass and is located at a distance of the reduced length from the suspension point is called the center of oscillation (Fig. 5.1., Point O  ). By the Huygens - Steiner theorem (see Appendix to lab No. 12) the moment of inertia of the body with respect to an arbitrary axis is equal to I  I C  ma 2 ,

(5.10)

where I C is the moment of inertia with respect to the axis passing through the center of mass and parallel to the given arbitrary axis; a is the distance between the axes. Substituting the expression (5.9) into this equation and solving it with respect to l 0 , we obtain l0

2  I C  ma  a  I C . ma ma

(5.11)

This formula shows that the reduced length is equal to the sum of the distance from the point of suspension to the center of mass I plus some distance C , it means that the center of oscillations O' ma lies below the center of mass C (see Figure 5.1). Denote the distance from the center of mass C to the center of oscillations O' by b, then b

Hence a 

IC . ma

IC . mb 73

Substituting these expressions in (5.11), we get: lo  b 

IC . mb

(5.12)

Fig. 5.1. Derivation of the equation of oscillations of a physical pendulum

Equations (5.11) and (5.12) show that if the pendulum is suspended by the center of oscillations, the oscillation period will not change. Therefore, when the point of suspension is moved to the center of oscillations, the former suspension point becomes the center of oscillations, i.e. the point of suspension and the center of oscillations are reversible. 5.2.3. The dependence of the oscillation period of the physical pendulum on the distance between the point of suspension and the center of mass. Substituting (5.10) in (5.7), we obtain the dependence of the period of oscillations of a physical pendulum T on the distance between the point of suspension O and the center of mass C: T  2

I a  C . g mga 74

(5.13)

Fig. 5.2. shows the dependence of T on a described by the formula (5.13). When a  0 (the suspension point is close to the center of mass), the period of oscillations T tends to infinity, which corresponds to neutral equilibrium. As the suspension point moves away from of the center of mass, the period of oscillations T of the pendulum first decreases and then increases again and with further increase in a tends to infinity. In case a   the second term under the square root in (5.13) can be neglected in comparison with the first and we get: T  2

a . g

The last formula corresponds to the period of oscillations of a simple pendulum. Indeed, the condition a   is the condition when it is possible to neglect the dimensions of the pendulum compared with the distance from the point of suspension to the center of mass, which is the definition of a simple pendulum. The period of oscillations of a physical pendulum has minimum at a certain distance А* from the point of suspension to the center of mass. The distance А* and the period of oscillations Tìèí corresponding to it can be found from the condition of extremum: dT da

a a

 0.

Using (5.13), we can get the following expressions: a 

IC , m

Tmin  2 4

4I C mg 2

.

(5.14)

5.2.4. Determination of the acceleration of free fall with the help of the rod pendulum. 75

The proposed method of determining the acceleration of gravity is based on the formula (5.8):

g

4 2 l 0 T2

.

(5.15)

The reduced length l0 is determined graphically using the experimentally obtained dependence of the oscillation period of the physical pendulum on the distance between the point of suspension and the center of mass, which is qualitatively shown in Figure 5.2. As seen from the figure, this curve is composed of two branches located symmetrically with respect to the vertical axis (axis T).

Fig. 5.2. The dependence of the oscillation period T of the physical pendulum on the distance a between the center of mass and the point of suspension

5.3. Description of the experimental setup Devices and accessories: rod pendulum, electronic stopwatch. A general view of a universal pendulum is shown in Figure 5.3. The base 1 is equipped with 2 adjustable feet that allow the alignment of the device. On the base the column 3 is mounted, on which the upper bracket 4 and the lower bracket 5 with the 76

photoelectric sensor 6 are fixed. On the one side of the bracket 4 is a physical pendulum 8, made in the form of a steel rod of length 590 mm and diameter 10 mm, on which a prism suspension 12 is fixed. On the rod, annular grooves are made every 10 mm for accurate fixation of the suspension, which can be moved along the rod and fixed in any groove with a screw. The lower bracket 5 with the photoelectric sensor 6 can also be moved along the column 3 and set in such a position that the end of the rod 8 crossing the light beam of the sensor does not collide with the bracket 5.

Fig. 5.3. Universal pendulum

The front panel has the control buttons: «Power» – a power switch; «Reset». The button Reset is used to reset the readings of the stopwatch and start measurements of a new timing and the number of periods of oscillation; «Stop». The button Stop stops timing and oscillation periods, as well as digital indicators of the number of periods and full time of oscillations. 77

Electronics of the device enables us to count the number of periods of oscillations from 0 to 100, and the digital electronic stopwatch measures the oscillation time from 0 to 100 s with an error  0.001 s. 5.4. Procedure of work fulfillment 5.4.1. Preparation for measurements: – Fix the lower bracket 5 in the lowest position, fix the top of the fourth prism on the fourth groove from the edge of the annular groove of the rod 8 so that the lower end of the rod was in the section of the photosensor 6; – Connect the device to the power; – Press the button «Power», check that the lamp of the photosensor 6 glows and zeros on the digital display flash. 5.4.2. Measurements. 5.4.2.1. Preliminary experiment. – Deflect the rod 8 from the vertical equilibrium position at a small angle and release. The rod must make oscillations freely, crossing the light beam of the photosensor 6. The initial angular deflection  0 must be made as little as possible, but it cannot be made arbitrarily small. It must be so small that during 10 ÷ 20 oscillation periods, oscillations do not attenuate so that the light beam sensor is not periodically interrupted by the oscillating rod. – Press and release the button «RESET». Visually inspect the stopwatch and the counter of the number of oscillations, which should record every second crossing of the beam of the photodetector. After counting 10 ÷ 20 oscillations, stop the measurement by pressing the button «STOP». – Repeat the experiment with different angles of deflection  0 and find such an angle, a further decrease in which does not change the period of oscillations. In further measurements you should use this angle. 5.4.2.2. Basic measurements. – Prepare a copybook to record the results of measurements with the following columns: 78

1) number of the i-th annular groove against which the suspension prism is set; 2) distance ai from the point of suspension to the center of mass; 3) measurements of time t i of 10 periods of oscillations and average value of  t i  ; 4) period of oscillations Ti ; The number of required measurements is 25. Table 5.1 The results of measurements of oscillations of the physical pendulum

№ 1 2 …

i

ai, cm

t1, s

t2, s

t3, s

, s

Ti, s

− Moving the suspension prism step-by-step by one groove (10 mm) to the center of mass, measure 10 periods of oscillations of the pendulum 3 times for each position of the suspension prism. Record the results in Table 5.1. The bracket (5) also must be moved up so that the lower end of the rod passes through the beam of the photosensor. 5.5. Mathematical processing of the results of measurements 5.5.1. Using the average values  t i  determine the period of oscillations Ti and record these values in Table 5.1. 5.5.2. Plot the graph of the dependence of the oscillation period Ti on the distance ai between the point of suspension and the center of mass, using all experimental points. 5.5.3. Plot in a larger scale (increasing the scale 2050 times) separately the area of function T  f (a ) , in which two values of

ai correspond to the same value of period Ti . a1i

5.5.4. After drawing 56 horizontal lines find the values of and a 2 i corresponding to the points of intersection of 79

these lines with the function T  f (a) and determine the reduced lengths l 0 i  a1i  a 2 i for each value. Record the data in Table 5.2. Table 5.2 Calculation of the gravitational acceleration

T, s

А1, m

А2, m

l0, m

g, m/s2

, m/s2

 g, 2

m/s

 g2,

m2/s4

5.5.5. Substituting the values of l0 i and Ti obtained in formula (5.15), find the experimental values of the acceleration of gravity. 5.5.6. Determine the average value  g  and the confidence interval g by the method of direct measurements for a given confidence level. 5.5.7. Compare the obtained value with the reference value for Almaty (g = 9.804 m / s2). 5.5.8. Using the formula for the moment of inertia of a thin rod with respect to the rotation axis perpendicular to the rod and passing through the center of mass IC 

ml 2 , 12

(where m is the mass of the rod, l is its length) and formula (5.14), calculate the theoretical value of the minimum period of oscillations Tmin and the corresponding distance А* from the point of suspension to the center of mass; and compare them with the data obtained experimentally on the graph. 80

Questions for self-assessment

1. Give definitions of the physical and mathematical pendulums. 2. Are the curves of the oscillation period and the distance between the point of suspension and the center of mass for the mathematical and physical pendulums, shown in the same graph, different? Explain why they are different. 3. What is the physical meaning of the term «reduced length of the physical pendulum»? 4. How can the weight and shape of a pendulum suspension affect its motion? Will the account for the presence of the suspension increase or decrease in the period of oscillations? 5. Explain why the curve of dependence of the period of oscillations T on the value a in Figure 5.2. is symmetric with respect to the center of mass, though the physical pendulum itself may not have such a symmetry. References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989. – p. 576. 6. Laboratory works in physics: textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 7. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

81

Laboratory work № 6 STUDY OF THE STATIONARY FLOW OF LIQUID IN THE TUBE OF VARIABLE CROSS SECTION. VERIFICATION OF BERNOULLI'S THEOREM

6.1. The purpose of work: to study the laws of stationary motion of an incompressible low-viscous fluid through a tube of variable cross section, check the applicability of the Bernoulli formula and study the effect of viscous friction on the decrease of the total pressure along the channel. 6.2. Brief theoretical introduction The movement of liquids or gases is called a flow, and the collection of particles in the moving fluid (or gas) is called a flux. In aerohydromechanics, liquid and gas are considered as a continuous medium, continuously distributed in space, without going into details of the molecular structure of matter. The motion of a liquid (or gas) can be described in two ways. We can track the movement of each individual fluid particle, i.e., indicate the position and velocity of the particle at any moment of time. This way of describing the fluid motion is called the Lagrange method. It can be also done in the other way, as it was done by Euler. We can specify the magnitude and direction of different velocities of liquid particles passing through the same point in space at different moments of time. If we set the value and direction of the velocity of particles in the whole space at any given moment of time, we obtain the field of velocities. Then at each point in space we will have a velocity vector 82

of the particle in fluid which passes through this point at a given moment of time. The flow is called stationary if its speed at each point in space does not change with time. If the velocity vector of the fluid depends on coordinates and time, i.e., if the speed of the fluid particles varies in space and time, such a fluid motion is called non-stationary (unsteady). If the fluid flows in layers, without mixing, this flow is called laminar. If the layers are mixed, the flow is called turbulent or vortex. The turbulent flow is always unsteady. In hydrodynamics for the convenience of calculations the concept of a streamline and a stream (flow) tube are introduced. The line whose tangent at any point indicates the direction of the velocity of the particle of the liquid passing at any moment of time through the point of contact, is called the stream line. In the stationary motion of a fluid, the stream line does not change with time and coincides with the trajectories of motion of fluid particles. Streamlines in a fluid can be made visible by dropping into it a paint or admixing any visible suspended particles. In the moving fluid let us take an arbitrary closed contour C and draw flow lines through its points (Fig. 6.1.). The collection of flow lines drawn through all the points of the contour C forms a tubular surface, called the stream tube. Velocities of the liquid particles are directed tangentially to the surface of the stream tube, so the fluid particles do not intersect the side surface. The stream tube behaves like a side surface of a solid tube within which the fluid flows. If the cross section of the flow tube is small enough, it can be assumed that the velocity of the fluid is the same across the crosssection and is directed along the axis of the flow tube. The mass of the fluid flowing during the time dt through the cross section of the tube, is determined by the expression: dm   S u dt ,

(6.1)

where  is the fluid density; S is the area of the normal section of the flow tube; u is the velocity of fluid flow. 83

Fig. 6.1. A stream tube

In case of stationary flow of the fluid the mass flowing per unit of time through any cross section of the flow tube is the same. If we take two sections of the pipe, the areas of which are equal S1 and S 2 , we can write:

1u1 S1   2 u 2 S 2 .

(6.2)

This formula is one of the basic laws of thermodynamics – the law of conservation of mass or the continuity equation. If the fluid is incompressible, i.e. 1   2   , the law of mass conservation gives the relation: or

u1 S1  u 2 S 2

(6.3)

u1 S 2 .  u 2 S1

(6.4)

The flow rate of fluid in the same flow tube is the greater, the more narrow is the cross section of the tube. Study of the real movement of liquids and gases is a difficult task. For simplicity, the model of ideal, i.e, non-viscous liquid in which internal friction forces are neglected, is considered. The only force that can act on the surface of the selected volume of the ideal fluid is the force of normal pressure. Consider a steady flow of ideal incompressible fluid in a gravitational field. To this flow we can apply the law of conservation 84

of mechanical energy. In this case we neglect the heat exchange, which may occur between the flow of the liquid and the environment. Let us select a narrow flow tube in the fluid and consider the part of the fluid occupying the volume MNDC (Fig. 6.2). During time t this part of the fluid travels at a short distance l1 and occupies the volume M1N1D1C1. Let us calculate the work A done by the forces of pressure to move the selected volume of liquid from the position MNDC to the position M1N1D1C1. The pressure forces acting on the lateral surface of the tube perpendicularly to the flow movement do not do any work. The pressure p1 acts on the left section of the flow tube of the area S1 and when the cross section moves from the position of MN to the position M1N1 the pressure force performs positive work A1  p1 S1 l1  p1 S1u1 t ,

(6.5)

where l1  u1 t is the distance at which the left cross-section moved during time t . The pressure forces p2 acting on the cross section CD are directed upward the flow and to move the right cross section with area S2 from the CD position to the position C1D1 perform negative work:

A2   p 2 S 2 l 2   p 2 S 2 u 2 t .

(6.6)

The total work done by the pressure forces to move the selected volume of fluid in the tube from the position MNDC to the position M1N1D1C1, is equal to:

A  A1  A2  p1 S1u1 t  p 2 S 2 u 2 t .

(6.7)

According to the law of conservation of mechanical energy, this work must be equal to the increment of the total mechanical energy E of the selected portion of liquid during movement: 85

A  E .

(6.8)

As we consider the steady motion of the fluid, the mechanical energy of the unshaded part of the volume of the liquid does not change. Therefore, E is equal to the difference between energies of fluid in the volumes CDD1C1 and MNN1M1, which corresponds to the mass of the fluid flowing through the section CD during t :

m2   u 2 S 2 t and flowing through the section MN during the same time t :

m1   u1 S1 t . The mechanical energy of the mass m of the fluid is the sum of mu 2 and potential energy in the gravity its kinetic energy 2 field mgh . Therefore, the increment of the mechanical energy of the selected volume of liquid is equal to: E 

2  m u 2  m2u2  m2 gh 2   1 1  m1 gh1  .  2  2  

(6.9)

Here h1 and h2 are heights of centers of mass m1 and m 2 , counted from a certain initial level (Figure 6.2.). Based on the law of mass conservation: m1  m2  m ,

 u1 S1 t   u 2 S 2 t   uS t .

(6.10)

Substituting A and E values from (6.7) and (6.9) into (6.8), we get: 86

 u2   u2   m2  2  gh2   m1 1  gh1  p1 S 1 u1  t  p2 S2 u2  t . 2  2 

(6.11)

Fig. 6.2. To the derivation of Bernoulli's equation

Taking into account the expression (6.10), canceling m and transferring members with the same indexes to the same side of the equality, we finally have: p1 

 u 12 2

  gh1  p 2 

 u 22 2

  gh 2 .

(6.12)

This formula was first obtained by Petersburg academician Daniel Bernoulli in 1738 and is called the Bernoulli equation. Equation (6.12) was derived for two arbitrary cross sections of the flow tube. This is possible if the sum of the three quantities in the Bernoulli's equation remains constant throughout the length of the flow tube. Therefore, for any section of the flow tube in a stationary flow of an ideal fluid the equality p

 u2 2

  gh  const . 87

(6.13)

is valid. The first term р is called the static pressure, the second  u2 is called the dynamic pressure or dynamic head, the third term 2 term  g h is called the leveling pressure. The Bernoulli equation can be considered as the law of conservation of mechanical energy in the moving liquid or gas. When the fluid moves along the horizontal channel, the leveling pressure does not change, and the Bernoulli equation is written as: p1 

 u12 2

 p2 

 u 22 2

.

(6.14)

When a liquid moves along a horizontal tube, the sum of static and dynamic pressures is constant for any section of the flow tube. In case of an ideal fluid, the flow tube can be replaced by a tube from a solid material. Hence, for an ideal fluid moving in a tube with a variable cross section, the Bernoulli equation is also valid. However, in case of motion in a real (viscous) fluid, forces of viscous friction appear in the tube acting from the walls of the tube on the fluid flow. Therefore, in case of motion of a viscous fluid the total pressure along the pipe will decrease. The Bernoulli equation can be considered approximately valid only for low-viscous fluids (water, gasoline, etc.) and for small sections of the pipe length. On the basis of the continuity equation (6.3) and (6.14) we conclude that when a liquid flows through the tube of a variable cross section, the liquid flows slowly in wide sections and quickly in narrow sections, and the static liquid pressure acting on the pipe wall is higher in wide cross-sections and smaller in narrow cross-sections of the pipeline. 6.3. Description of the setup and work fulfillment 6.3.1. The scheme of the laboratory setup is shown in Fig. 6.3 (a). It consists of a tank 1 providing a constant level of liquid poured into the tank; a tube of variable cross section 2; a block of valves 3, used to regulate the flow of liquid through the tube 2 and to change the direction of flow along this tube; a battery of gauges 4 for 88

measuring the static pressure in the three cross-sections of the tube and the measuring flask 5, which is used to measure the flow of fluid through the tube 2. The scheme of valves is shown in Fig. 6.3 (b). When the handle 6 of the valve is turned counterclockwise up to the end position, the fluid flows through the tube 2 from left to right. Turning the handle in the opposite direction reverses the direction of flow. The lower handle 7 of the block of valves is designed to control the flow of liquid. The diameters of cross sections, in which measurements of the static pressure of the liquid are made, are shown in Fig. 6.4. Numbering of the tube sections is made in the direction of fluid flow in the tube 2. Thus, if the current is directed from the maximum to the minimum cross-section, d1 = 12 mm, d2 = 4 mm, d3 = 6 mm. In case of the opposite direction d1 = 6 mm, d2 = 4 mm, d3 = 12 mm. Experiments begin when the flow in the pipe moves from the maximum to the minimum cross section. 6.3.2. Starting the work, open the tap 8 of the water pipe and fill in the tank 1. The valve 7 must be closed. When the water level in the tank reaches the edge of the discharge funnel in the tank, the water begins to flow out through the drain pipe 9. Then, gradually turn the handle 7, so that the height of the liquid in the middle manometer tube is in the range of 50 ÷ 100 mm. Tap 7 is to be regulated so that the water flows out from the drain hose 9 into the sink, and the float 10 is at a constant height corresponding to the lower mark on its stock. Under these conditions, the movement of fluid in the tube of variable cross section is stationary, and the levels of liquid in the manometer tubes will not change over time. Before starting the experiments, prepare table 6.1 to record the measurements. 6.3.3. Using the graduated flask 5 and a stopwatch, determine the time  needed for a predetermined volume of water Q to flow out and write down recordings of the manometric tubes hi . Repeat the experiment 3 ÷ 5 times (write down the data in Table 6.1) and 89

determine the average value of the flow rate of the liquid according to the formula:  q   Q .  

6.3.4. Using the average value of the flow rate, calculate the average flow speed in each section of the tube by the formula: ui 

where S i 

 d i2 4

q , Si

are cross sections of the tube.

6.3.5. Static pressures in the cross-sections of the tube of variable cross sections are determined by the height of the liquid column in the manometer tube 4, measured from the axis of the tube 2 of variable cross sections:

pi   g hi , where indices i = 1, 2, 3 correspond to the numbers of cross sections;

 = 1000 kg/m3 is the water density; g = 9.804 m/s2 – is the

gravitational acceleration. If we measure the value of hi in meters, we will get the pressure in Pascals.

Fig. 6.3 (А). Scheme of the setup

90

Table 6.1 Results of measurements

№

h1, mm

h2, mm

h3, mm

 1 ,s

 2 ,s

 3 ,s

 

q, cm3/s

,s

1 2 3

Fig. 6.3 (Б). Scheme of the block of taps

3-rd section 2-nd section 1-st section 2-nd section

1-st section 3-rd section

 u    u  

Fig. 6.4. A channel of variable cross section

6.3.6. For each section of the tube calculate the dynamic  ui2 . pressure 2 u 2 6.3.7. Calculate the total pressure for each section p пi  p i i . 2 The results of calculations are recorded in Table 6.2. Compare the 91

total pressure for three cross-sections of the tube 2, thereby verifying the applicability of the Bernoulli formula (6.14) for our setup. Table 6.2 The results of calculations verifying the Bernoulli equation

Modes

1 mode

q,

1 cross section

м3 с

2 cross section

3 cross section

h, mm p, Pa u, m/s  u 2 , Pa 2

2 mode

pп, Pa h, mm p, Pa u, m/s  u 2 , Pa 2

3 mode

pп, Pa h, mm p, Pa u, m/s  u 2 , Pa 2

pп, Pa

6.3.8. Repeat exercises in 6.3.3. - 6.3.7. for two other modes of fluid flow (with other water discharges). 6.3.9. Explain the results. Questions for self-assessment

1. Write and explain physical meaning of the continuity equation and the Bernoulli equation. 2. How can we measure the static pressure of a moving fluid in the tube? 3. How can we find the average flow velocity of the fluid in a section of the channel? 4. If the difference between static pressures in two sections and diameters of these sections are known, is it possible to determine the speed of fluid flow 92

in one of these sections? Derive the formula for calculating the speed in this case, using the Bernoulli equation and continuity equation. 5. What pressure is measured by the manometric tubes 4: absolute pressure or an excess over the atmospheric pressure? 6. Calculate the pressure of the water column of 1 meter high in Pascals. References

1. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 2. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 3. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989. – p. 576.

6.4. Task for Student research work 6.4.1. Purpose of Student research work. Using the measurements of the total pressure in the crosssections of the tube with variable cross sections, study the dependence of hydraulic resistance of confusers (convergent section of the tube) and diffusers (expanding section of the tube) on the degree of expansion and contraction of cross-sectional area and the Reynolds number. 6.4.2. Brief theory. As stated in the first part of the work, the Bernoulli equation is strictly valid only for the motion of a perfect fluid. In the motion of a viscous fluid flux, some part of the flux energy is spent on work against the friction force, thus, the total pressure of the flow decreases in the direction of the fluid flow. Therefore, instead of formula (6.14) for fluid motion in the horizontal tube of variable section the Bernoulli equation must be written in the form: p1 

p2 

 u12 2

 u 22 2

 p2 

 p3  93

 u 22 2

 u 32 2

 p12

,  p 23

(6.15)

where p12 is the loss of the total pressure between sections 1 and 2 due to resistance; p 23 is the loss of the total pressure due to the resistance between sections 2 and 3 in the tube of variable section. Here we suppose that the flow is directed from the maximum cross-section to the minimum cross-section. The loss of the total pressure over the length of the channel depends on many factors. When a viscous liquid moves through the tube of constant cross section, the pressure loss is caused only by the friction on the channel walls. When the liquid moves in the convergent channel (confuser), losses are also mainly due to the viscous friction. When the liquid moves in the expanding area of the tube (diffuser), the total pressure loss is caused not only by viscous friction, but also by the loss of energy on the formation of a vortex of chaotic (turbulent) fluid motion due to partial flow separation from the walls of the channel (formation of a vortex flow field). The area of the vortex flow and its intensity is the higher, the larger is the expansion of the diffuser: nД 

S exit , S ent

( S exit  S ent ) ,

where S exit and S ent are the cross-sectional areas at the input and output of the diffuser. The confuser resistance mainly depends on the area of its lateral surface and slightly depends on the degree of convergence:

nK 

S ent , ( S ent  S exit ) . S exit

In hydrodynamics, the resistance of diffusers and confusers is usually expressed through their hydraulic resistance coefficient and the dynamic pressure of flow in the narrow section: 94

p K   K p Д   Д

2  u nar

2

2  u nar

2

,

(6.16)

,

(6.17)

where  K and  Д are hydraulic resistance coefficients of confuser and diffuser; unar is the mean flow rate in the narrow section of the channel of variable section. For our experiments at the setup, if the flow is directed from the 1st section of the 3rd, the pipe section between the sections 1 and 2 corresponds to a convergent flow and the pipe section between sections 2 and 3 corresponds to a diffuser flow. Then, taking into account (6.16) and (6.17), from (6.15) we get:  u 22

 p12   K

 p 23   Д

2

 u 22 2

 p1 

 p2 

 u12

 p2 

2

 u 22 2

 u 22

 p3 

2

,

 u 32 2

(6.18) .

(6.19)

Introducing notations of full pressure p1п  p1 

 u12 2

; p2п  p2 

 u22 2

; p3п  p3 

 u32 2

,

(6.20)

from (6.18) and (6.19) we can find the coefficients of hydraulic resistance of confuser: p 1п  p 2 п

K 

and diffusor: Д 

 u 22 / 2 p 2 п  p 3п .  u 22 / 2

95

(6.21) (6.22)

In the general case  К and  Д depend on the Reynolds number (determining flow regime) and the degree of compressing (confuser) or expanding (diffuser). In this work we will study this dependence. 6.4.3. Procedure of work fulfillment. 6.4.3.1. To determine the hydraulic resistance coefficient of a confuser  К 1 and a diffuser  Д 1 as a function of the Reynolds number of the liquid, when it moves from section 1 to section 3 of variable cross section, it is necessary to use the experimental data of the first part from p.p. 6.4.1.-6.4.3. The degree of narrowing of confuser is n K 1 = 9.00; the degree of widening of the diffuser is

n Д 1 = 2.25. 6.4.3.2. Using the block of valves change the direction of fluid flow through the tube of variable section on the opposite direction and repeat the experiment described in paragraphs 6.4.1.-6.4.3. Here indices 1, 2, 3, correspond to sections d1 = 6.0 mm, d2 = 4.0 mm, d3 = 12 mm. The results of measurements and calculations must be recorded in table 6.2. 6.4.3.3. Using 6.6.3.2. calculate hydraulic resistance coefficients:

K2 

p  p p1п  p 2 п and  Д 2  2 п 2 3п . 2  u 2 / 2  u 2 / 2

Here the degree of widening of the diffusor is n Д 2 = 9.00; the degree of narrowing of the confuser is n К 2 = 2.25. Write down the results of calculations in 6.6.3.1.-6.6.3.3. in Table 6.3. 6.4.3.4. Using the experimental data, on the same graph draw two curves:

 K  f n K , Re  ,  Д  f n Д , Re  and make conclusions. 96

The Reynolds number is calculated by the diameter and velocity in the narrow section u 2 . Table 6.3 Results of calculations

Mode

Confuser

S n 1 S2

p p  K 1П 2 2П  u2 2

ud Re 2 2 

Diffusor S n 2 S3

Д 

p2П  p3П  u22 2

Re

 u2d2 

1 2 3 4 5 References

1. Povkh I.L. Aerodynamic experiment in engineering / I.L. Povkh. – L.: Mechanical Engineering, 1974. 2. Yablonsky V.S. A brief course of technical fluid mechanics / V.S. Yablonsky. – M.: Fizmatgiz, 1961.

97

Laboratory work № 7 INCLINED PENDULUM

7.1. The purpose of work: To study manifestations of friction forces for rolling bodies; to experimentally determine the coefficient of the moment of rolling friction for the pair of a ball and a plane made of different materials. 7.2. Brief theoretical introduction In engineering a special attention is paid to the forces arising in a cylinder (wheel) or a ball rolling on the plane. These forces, called the friction forces, can be divided into three types: the rolling friction, sliding friction and static friction (or adhesion friction). The static friction (or adhesion) is different from the other two types of friction in that it is not connected with the obligatory transfer of mechanical energy into heat. Sliding friction occurs in the forward movement of the body (slip) with respect to the plane. The work of this force acting against the vector of relative velocity is always negative and leads to the transition of the mechanical energy into heat. However, the experience shows that rolling of the cylinder (ball) over the plane without sliding is also accompanied by losses of mechanical energy. These losses are caused by the work of the force of rolling friction. During rolling the cylinder (ball) and the plane are deformed under the action of normal pressure. If these deformations are elastic, the force of interaction between the body and the plane must be completely symmetrical with respect to the vertical plane AB passing through the axis of the cylinder. The resultant of all forces of elastic deformation of the contacting surface is directed vertically, 98

and the moment of the forces with respect to the axis of the cylinder is equal to zero (Figure 7.1). In this case there are no rolling friction forces.

Fig. 7.1.

Fig. 7.2.

Fig. 7.4.

Fig. 7.3.

In the case of inelastic deformation (as it always happens in practice), the forces acting on the cylinder from the contacting surface are no longer symmetrical relative to the plane AB due to the residual strain (Figure 7.2). Therefore, the resultant of these forces always has a horizontal component, directed against movement. The moment of these forces with respect to the axis of the cylinder is not equal to zero, and slows down the rotational movement. Because of the character of decelerated (both translational and rotational) motion of the cylinder on the horizontal plane it can be 99



concluded that the resultant Q of deformation forces applied to the



cylinder must have a horizontal component Fk directed backwards. Moreover, the point of application of this force must be ahead of the  plane AB (Fig.7.3), the line of action of the force Q must be above the axis of the cylinder, otherwise the moment of this force with respect to the axis would lead to an acceleration of the rotational motion, which is impossible in this case. The horizontal component   of the force Q is the force of rolling friction Fk , which slows down (if there are no other external forces) the translational motion of the center of mass of the cylinder.  Usually, the connection between the force of rolling friction Fk and other quantities is determined experimentally.  Suppose that an external horizontal force F balancing the force    of rolling friction Fk : F + Fk = 0, is applied to the axis of the cylinder rolling on a horizontal plane without slipping (Figure 7.4). In this case the movement of the cylinder, both translational and rotational, will be uniform, without acceleration. This is possible only when the sum of all the external forces is zero, and the sum of the moments of all these forces is zero. The condition that the resultant of all external forces is equal to zero:

   Q  mg  F = 0;

(7.1)

can be written for components as follows: Q  cos   mg ,

(7.2)

Q  sin   Fk ,

(7.3)

where  is the angle of inclination of force vector Q to the vertical; Fk is the force of rolling friction. 100

Let us add the condition of zero moment of external forces with respect to the axis of the cylinder to equations (7.2) and (7.3):

RFk  r Q cos  ,

(7.4)

where R is the radius of the cylinder; r is the distance from the point of force application Q to the vertical plane AB passing through the axis of the cylinder.   Forces F and m g do not create moments with respect to the axis of the cylinder, and therefore are not included in the equation (7.4). Taking into account small values of deformations for r  R and sin   r R  1   , cos   1 ,equations (7.2) − (7.4) can be written as: Q  mg , Q    Fk , Fk  R  r  Q , hence

Fk  Q  r R  m  g  r R .

(7.5)

Tables usually give the values of r, but do not give the values of the rolling friction forces Fk (7.5), however, they give the values of the moment of the rolling friction forces:

M  Fk R  N  r .

(7.6)

Here М is the rolling friction moment equal to the product of the normal pressure N and r. In this case N  Q  mg . The value of r is called the coefficient of rolling friction moment. Note that r has the dimension of length! The experiments show that within certain limits the r value practically does not depend on the speed and radius of the cylinder, and is determined by the materials from which the cylinder and the plane are made. 101

Knowing the value of r for a given pair of materials, it is always possible to determine the force of rolling friction

Fk  r  N R

(7.7)

for the known radius R of the cylinder and the normal pressure N on the plane. Let us now consider one of the methods of experimental determination of the coefficient of rolling friction moment.  Suppose the external potential force F directed parallel to the plane acts on the ball rolling on a plane without slipping (Figure 7.4). We can write the system of equations of motion of the ball:

ma  F  Fk ,

(7.8)

I  Fk R  rN .

(7.9)

Here m is the mass of the ball; R is the radius of the ball; I is the moment of inertia of the ball with respect to the axis passing through the center of mass; a is the acceleration of translational motion of the ball;  is the angular acceleration of rotation around the center of mass; Fk is the horizontal component of the reaction force of the support (the force of rolling friction); N is the vertical component of the reaction force of the support, r – is the coefficient of the rolling friction moment. Equation (7.8) is the second Newton's law for the translational motion of the ball, and (7.9) is the equation for the rotational motion around the axis passing through the center of mass. It is necessary to add to the system of equations (7.8) - (7.9) the condition of absence of ball slipping with respect to the surface:

 R,

(7.10)

where  is the v Velocity of the center of mass of the ball;  is the angular velocity of the ball rotation. 102

From this equation we can get the relation for accelerations: a  R.

(7.11)

Using this formula, we can exclude angular acceleration  from the system of equations (7.8) - (7.9): ma  F  Fk ,

(7.12)

 ma  Fk  N  r R  .

(7.13)

Here   I (mR 2 ) . Adding equations (7.12) and (7.13), we obtain:

m(1   )

d r FN . dt R

(7.14)

Here we write acceleration a as a time derivative of the velocity

 of the center of mass of the ball.

Multiplying the last equation by , after some transformations we obtain: d dt

  m  2 I 2 r dx r    2  2  U    N  R    R N dt .  

(7.15)

It was taken into account that m

d d  m 2  ,   dt dt  2 

m

2 2



I 2 , 2

dU  Fdx,

U is the potential energy of the ball in the field of the conservative force F. The expression under the sign of the time derivative in the equation (7.15) is the mechanical energy of the ball including the kinetic energy of translational movement 103

Ek n 

m 2 , 2

the kinetic energy of rotation

Ek в 

I 2 2

and the potential energy of the ball in the field of the conservative force F:

U    F  dx. The right side of (7.15) is the power developed by the force of rolling friction. Thus, we can write that the change in the total mechanical energy of the ball is equal to the work of friction forces: E  r R  N S .

(7.16)

where S is the distance covered by the ball. The formula (7.16) can be used for the experimental determination of the coefficient of rolling friction moment r. The total mechanical energy of the ball in the setup with an inclined pendulum can be written in terms of the angle  of the maximum deviation from the equilibrium position: E  mgl (1  cos  ) cos  ,

(7.17)

where g is the gravitational acceleration; l is the length of the suspension;  is the angle of the plane of the pendulum with the vertical. Expression (7.17) represents the maximum potential energy of the pendulum in the gravitational field. 104

Using trigonometric formulas and the condition of small angles of deviation of the pendulum from the equilibrium position (   1 ), we can rewrite (7.17) as: E  2mgl sin 2 ( 2)  cos   mgl ( 2 2)  cos  .

(7.18)

Let the pendulum make n full oscillation. In this case the force of rolling friction does the work: A  (r R)m g S sin   (r R)

where S 

k  н

k  н 2

4l n  sin  ,

(7.19)

4l n is the total distance covered by the 2 pendulum during n periods. The amplitude of oscillations of the pendulum changes during this period of time from the initial angle  н to the final angle  k . The formula (7.16) can be written as: mgl

 k2   н2 2

cos   2

r nl ( k   н ) sin  . R

After mathematical transformations we can obtain the final expression for calculation of the coefficient r of the moment of rolling friction using the known values of initial angle  н and the final angle  k of the pendulum and the number of full oscillations n: r  R

н к 4n

ctg  .

(7.20)

7.3. Description of the experimental setup General view of the setup «Inclined pendulum FPM-07» is shown in Fig. 7.5. The electronic stopwatch 1 measuring the number and time of oscillations is attached to the base 2 equipped with four legs with 105

adjustable height. The tube 3, on which the case 4 with a worm gear is fixed, is mounted on the base. The worm gear is connected to the bracket 5 on which the scales 6 and 7 are mounted. The column 8 is fixed in the bracket on which the ball with a carrier 9 is suspended. Plates 10 from different materials are inserted in the guide tracks of bracket 5. To incline the pendulum the handle 11 is used. A photoelectric sensor 12 is fixed on the bracket 5. To replace the ball it is necessary to unscrew it from the carrier and to screw a new one.

Fig. 7.5. Experimental setup

7.4. Procedure of work fulfillment 7.4.1. Experimental part. 7.4.1.1. Preparations for measurements. To prepare the setup it is necessary to: – Check the grounding; – Make such a horizontal alignment of the device that when the angle of inclination of the pendulum to the vertical plane is   0 , the ball slightly touches the insert 10, and the position of the thread of the pendulum corresponds to the zero division on the scale 6; – Connect the device to the power supply; 106

– Press the bottom «POWER», check the glow of the lamp of the photoelectric sensor and flashing zeros on the digital display of the number of oscillations and the stopwatch; – Check the operation of the stopwatch and counter of oscillations, deflecting the pendulum at a slight angle and starting the system by the button «Reset» (stopping of counting by the button «Stop»). 7.4.1.2. Measurements. Set the angle of the pendulum with the vertical ε = 30º. Deflect the ball from the equilibrium position at an angle   5 0  10 0 and release it. The ball starts to oscillate about the equilibrium position. When the maximum angular deviation coincides with one of the lines of the scale 6, press the «RESET», recording the initial deflection angle on the scale 6. Write down the reading in Table 7. Observe the decrease in the amplitude of oscillations of the pendulum on the same side of the scale 6 where the initial reading was taken. When the maximum deflection angle decreases approximately two times, press the «STOP» (it is desirable to do it at a time when the thread suspension of the pendulum with the maximum deviation is again located directly against one of the divisions on the scale of angles). Record the final deflection angle of the pendulum and the number of full oscillations in Table 7. Repeat these measurements 5  7 times. Make the previous measurements for the inclination angles of the pendulum to the vertical ε = 45º and ε = 60º. Table 7 Results of measurements and calculations Ball radius R = …m

 1 30º

н 2

k 3

r,m

n 4

5

107

r , m

r , m

6

7

r 2 , m2 8

1 45º

2

3

4

5

6

7

8

60º

Write the results of measurements in Table 7. 7.4.2. Mathematical processing of obtained results. For each experiment calculate the coefficient r of the friction force moment using (7.20). To determine the angle of inclination of the pendulum to the vertical, find the average value of r and the confidence interval using the method of direct measurements separately for each series of experiments. What conclusions can be made from these results? Questions for self-assessment

1. 2. 3. 4. 5. 6. 7. 8.

What kinds of friction forces do you know? What is the main feature of static friction? What is the mechanism of rolling friction? Can the force of (static, sliding, rolling) friction be directed along the movement of the body? Give examples. When can we use the law of conservation of mechanical energy, despite the fact that there is the force of friction? Why are locomotives (diesel locomotives, electric) made massive? What determines the coefficient of rolling friction moment? What are the main factors that can limit the accuracy of your experiment?

References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 108

2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V.Sivukhin. – M.: Nauka, 1989. – p. 576. 6. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 7. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

109

Laboratory work № 8 MEASUREMENT OF THE MOMENTS OF INERTIA OF BODIES BY TORSION PENDULUM

8.1. The purpose of work: to study the experimental method of determining the moment of inertia of bodies with respect to different axes; to understand the use of the concepts of the tensor of inertia and ellipsoid of inertia in theoretical calculations. 8.2. Brief theoretical introduction 8.2.1. Moment of inertia. The moment of inertia of the material point with respect to any axis is the quantity equal to the product of its mass m by the square of the distance R to the axis: I  mR2 .

(8.1)

The moment of inertia of the mechanical system with respect to an axis is determined by the sum of the moments of inertia of material points of the system with respect to this axis: I 

m R i

i

2 i

.

(8.2)

In solid bodies mass is distributed continuously, so the sum of (8.2) must be replaced by the integral: I   R 2 dm   R 2  dV , m

V

110

(8.3)

where  is the density of the solid body; V is the volume of the body. The integration should be made all over entire mass or volume of the solid body. As seen from (8.3), the moment of inertia of the mechanical system depends not only on the mass m of the system, but also on the distribution of mass with respect to the selected axis. The physical meaning of the moment of inertia is that it characterizes the inertia of the body to the change of the angular velocity of rotation. Thus, in the rotational motion the moment of inertia plays the same role as the mass of the body in the translational movement. Analytical calculation of integrals of the type (8.3) is possible only in the simplest cases of bodies of regular geometric shape. For bodies of irregular geometric shape such integrals can be found by numerical methods. Calculation of the moments of inertia in many cases can be simplified by using considerations of similarity and symmetry, Huygens - Steiner theorem, the concept of inertia tensor, etc. 8.2.2. The tensor of inertia. The concept of the tensor of inertia can be introduced on the basis of the following example. Let us calculate the moment of inertia of the body with respect to an arbitrary axis AA' (Figure 8.1). Without loss of generality it can be assumed that the axis AA' passes  through the origin of coordinates. We expand the radius vector r  of the mass element dm into components d along the axis AA' and  along R perpendicular to it.    r  d  R. From the definition (8.3) of the moment of inertia we get:





I   R 2 dm   r 2  d 2 dm .

 Let s be a unit vector directed along АА' axis, then   d  r , s   x  s x  y  s y  z  s z , 111

(8.4)

where s x , s y , s z 2

2

2

are Cartesian components of the unit vector, 2

and r  x  y  z .

Fig. 8.1. To the notion of the tensor of inertia

Taking into account these relationships, as well as the equality

s x2  s 2у  s z2  1 , we obtain:











I   r 2  d 2 dm   x 2  y 2  z 2  xs x  ys y  zs z dm   I xx s x2  I yy s y2  I zz s z2  2 I xy s x s y  2 I xz s x s z  2 I yz s y s z   I xx cos 2   I yy cos 2   I zz cos 2   2 I xy cos   cos  

(8.5)

 2 I xz cos   cos   2 I yz cos   cos  . where  ,  ,  are angles formed by АА/ axis with Cartesian axes ох, oy, oz; are constants I xx , I yy , I zz ; I xy  I yx , I xz  I zx , I yz  I zy determined as:





I xy  I yx    xy dm ,

I xx   y 2  z 2 dm , 112





I xz  I zx    xz dm ,





I yz  I zy    yz dm .

I yy   x 2  z 2 dm , I zz   x 2  y 2 dm ,

(8.6)

The set of nine quantities

I ij

I xx  I yx I zx

I xy I yy I zy

I xz I yz I zz

(8.7)

forms a second-rank tensor and is called the tensor of inertia with respect to the point O. As it is seen from (8.6), the tensor of inertia is symmetrical, i.e., I ij  I ji (i, j  x, y, z ) , so it is fully determined by its six components. Expression (8.5) can be written in a more compact and symmetrical form: I    I ij s i s j . i

j

(8.8)

If for a coordinate system all six components of the inertia tensor are known, then using (8.5) and (8.8) we can calculate the moment of inertia with respect to any axis passing through the origin O. This shows the important role played by the tensor of inertia. Consider a specific example. Let us choose a uniform parallelepiped with sides a, b, c as a body for which we have to determine the moment of inertia. Let us orientate the Cartesian coordinate system as shown in Figure 8.2. With this choice of the coordinate system, its origin coincides with the center of mass of the parallelepiped. Let us calculate all components of the inertia tensor of the system. Using formulas (8.6) we obtain: 113













I xx   y 2  z 2 d m   y 2  z 2   dV   y 2  z 2 dx dy  dz  m

V

a/2 b/2 c/2 c / 2 b3     dx  dy  dz y 2  z 2    a    bz 2 dz  a / 2 b / 2 c / 2 c / 2  12  3 3 b c   a  b  c  2 m 2    a  c  b    b  c2  b  c 2 . (8.9) 12 12 12 12  













Fig. 8.2. The main central axes of inertia of the parallelepiped

Other components of the tensor are calculated in the same way (to determine them it is useful to use their symmetry). They are written as: I

yy







m a2  c2 ; 12

I xy  I

yx

I zz 





m a2  b2 ; 12

 I xz  I zx  I yz  I zy  0 .

(8.10)

Now using the formula (8.5) we can calculate the moment of inertia I of the parallelepiped with respect to any axis passing through the origin O. In this case, we must know the orientation of the axis with respect to our coordinate system, i.e. components of the  unit vector s directed along the selected axis or, which is equivalent, cosines of direction angles α, β, γ. For example, direct the axis AA' through the opposite vertices of the parallelepiped (Fig. 8.3.). It can be shown (using your knowledge of trigonometry) that 114

s x  cos   s y  cos  

s z  cos  

a a  b2  c2 b

,

2

a2  b2  c2 c

a2  b2  c2

(8.11)

,

.

From the expression (8.5) and formulas (8.9) − (8.11), we obtain: I  I xx  s x2  I yy  s y2  I zz  s z2  









m 2 a2 b  c2 2  12 a  b2  c2

m b2 m c2 a2  c2   a2  b2 2 2 2 12 12 a b c a2  b2  c2











m 2 2 1 . a b  a 2c 2  b 2c 2 2 6 a  b2  c2

(8.12)

We can calculate the moments of inertia of the parallelepiped with respect to the other axes in a similar way. A parallelepiped with a square base and a cube are particular cases of the parallelepiped considered above. Formulas for calculating moments of inertia of such bodies are simplified by the conditions А = b and А = b = с, respectively.

Fig. 8.3. To the definition of the tensor inertia of the parallelepiped.

115

8.2.3. The ellipsoid of inertia. Formula (8.5) allows us to get a clear geometrical interpretation. Let us find the equation of the surface, which is the locus of points whose distance from the origin O is defined by the equality r  1 I where I is the moment of inertia of the solid body with respect to the axis connecting the origin with the selected point on the surface. According to the construction, the radius vector of an arbitrary 

point on the surface is determined by the expression r 

 s

I

, which

taking into account (8.8) leads to the equation of the sought surface:   I ij  ri  r j  1 i

or

j

I xx x 2  I yy y 2  I zz x 2  2 I xy xy  2 I xz xz  2 I yz yz  1 .

(8.13)

This second-order surface is an ellipsoid. It is called the ellipsoid of inertia of the body with respect to the point O. If the point O coincides with the center of mass of the body, such an ellipsoid of inertia is called central. The principal axes of the central ellipsoid of inertia are called the principal central axes of the body. If the coordinate axes are directed along the principal axes of the ellipsoid of inertia, the non-diagonal elements of the tensor of inertia are equal to zero and the tensor is diagonalized. These coordinate axes are also called the principal axes of the tensor of inertia. They are rigidly connected with the body. The ellipsoid of inertia is also rigidly connected to the rigid body. An important factor for the dynamics of rotational motion of a solid body is not symmetry of the body, but symmetry of the corresponding ellipsoid of inertia. All the bodies with the same ellipsoid of inertia are dynamically equivalent. 8.3. Description of the experimental setup and methods of work fulfillment The experimental setup «torsion pendulum FPM – 05» is shown in Fig. 8.4. As it can be seen from this figure, a stopwatch 1 is fixed 116

on the base 2 equipped with four legs of adjustable height. Column 3, on the base on which brackets 4, 5 and 6 are fixed by means of clamping screws, is mounted on the base. The construction of the frame allows us to fix solids 12 with different external dimensions. Bodies are fixed with the help of a movable beam, which moves along the guides between two stationary rods. The movable beam is fixed by nuts on the clamping sleeve, located on it. This setup allows us to determine the moments of inertia of different bodies experimentally by torsional oscillations. The essence of this method is as follows. The period of oscillation T of the torsion pendulum with a studied body is determined by the formula: T  2

I , k

(8.14)

where I = I+ Iр; Iр is the moment of inertia of the empty frame of the pendulum; I is the moment of inertia of the studied body relative to the axis of rotation of the pendulum; k is the torsion modulus of the wire, on which the frame with the body is suspended. The moment of inertia I of the studied body can be determined from the expression: I

k 4

2

T 2  I p ,

(8.15)

if k and Iр values are known. These values can be determined experimentally if we use a set of parallelepipeds from the installation kit. The set of bodies for the installation includes a cube with a side a, and a parallelepiped of height 2a, the base of which is a square with the side a. Brackets 4 and 6 have clamps serving for fixing a steel wire, on which the bracket frame 7 is suspended. To the bracket 5 is attached a steel plate 8, which serves as a base for the photoelectric transducer 117

9, the electromagnet 10 and the angle scale 11. The electromagnet 10 can change its position on the plate, and the arrow, attached to the electromagnet, indicates its position with respect to the photoelectric sensor.

Fig. 8.4. Torsion pendulum

We can imagine that this parallelepiped consists of two cubes with sides a, therefore the moment of inertia InС of such a parallelepiped with respect to the axis passing through its center of mass and directed perpendicularly to the plane of the large face is equal to the double moment of inertia Ik of the cube with respect to the axis parallel to the edges of the cube and located at a distance of half a rib А/2 from the center the mass. 2  a  I nC  2  I k  2 I k C  m     ,  2   

(8.16)

where IkС is the moment of inertia of the cube with respect to the axis passing through its center of mass; m is the mass of the cube with the side a. 118

To derive the formula (8.16) we used the Huygens - Steiner theorem connecting the moment of inertia IC of a solid body with respect to the axis passing through the center of mass with the moment of inertia the body with respect to the axis parallel to the previous one and located at a distance R from it: (8.17)

I  I C  mR 2 .

Let us write expressions for the period of oscillations of the torsion pendulum: without additional bodies (an empty frame): T p  2

Ip k

,

(8.18)

with a cube the rotation axis of which is perpendicular to the sides:  I  I kC   Tk  2  P , k  

(8.19)

with a parallelepiped whose axis of rotation passes through the center of gravity and is parallel to the major faces:





 I p  I nC  Tn  2  . k  

(8.20)

If we make measurements of these oscillation periods Тр, Тk, Тn on the experimental setup, then, using a system of equations (8.16) (8.20), we can determine the desired expression for the moment of inertia of the free frame Ip and the torsion modulus k of the wire on which the pendulum is suspended. T p2 ma 2 , Ip   2 Tn2  T p2  2Tk2 119

(8.21)

k

Tn2

2 2 ma 2 .  T p2  2Tk2

(8.22)

Now, to determine the moment of inertia I of the body using a torsional pendulum it is sufficient to measure its period of oscillations T with a body fixed in the fixed frame and to calculate the moment of inertia I of the body using the formula: I

T 2  Tp2 ma 2 .  2 2 Tn  Tp2  2Tk2

(8.23)

8.4. Procedure of work fulfillment 8.4.1. Experimental part. 8.4.1.1. Preparations for measurements. To prepare the setup it is necessary: – To check the grounding; – To make a horizontal alignment of the setup using the adjustment screws; – To connect the setup to the power supply; – To press the button «POWER» to check whether all the indicators light number zero, and the lamp of the photoelectric sensor glows. 8.4.1.2. Procedure of work fulfillment Release the button «Start» and rotating the frame move the arrow to the electromagnet so that the electromagnetic force fixes the position of the frame. Press the button «RESET». Press the button «START». Frame will start to oscillate. After 9 periods of oscillations, press "STOP". Write the results of measurement of the time of 10 oscillations into Table 8.1. Repeat the measurements 5 ÷ 7 times. Measure the time of 10 oscillations of the cube and the parallelepiped with a square at the base with respect to the axes of rotation given in section 8.3 to determine the constants of the installation k and Ip. 120

Table 8.1 Results of measurements and calculations

№

t, c

 t , c

T,c

I экс , кг  м 2

I теор , кг  м 2

An empty frame

Cube

A parallelepiped with a square at the base

A parallelepiped with a rectangle at the base

Measure the time of oscillations of the pendulum with bodies fixed on its frame. Measure the time for all bodies for all different axes of rotation using the scheme given for measurement of the time of oscillations of the empty frame. Write the results into Table 8.1, write down the parameters of the body, with which you are making the experiment, and the axis with respect to which you observe torsional oscillations. 8.5. Mathematical processing of the results of measurements 8.5.1. From these measurements, determine the average periods of oscillation of the pendulum  T p  ,  Tk  and  Tп  without

a body, with a cube and with a parallelepiped with respect to the axes given earlier. These values will be used to calculate the moment of inertia by the formula (8.23). 121

Estimate the error of measurements for the periods of oscillations by the method of processing the results of direct measurements. 8.5.2. For all the bodies and axes of rotation calculate the average periods  T  of oscillations of the torsion pendulum. 8.5.3. Using the formula (8.23) calculate the corresponding moments of inertia  I  . 8.5.4. Using the formulas given in the brief theoretical introduction, calculate theoretical values of the moments of inertia with respect to the axes of rotation for the bodies used in the experiments. Use parameters of the bodies from Table 8.2. Compare the results of theoretical calculations with the experimental values. Table 8.2 Parameters of studied bodies

mass, g

Dimensions А, mm

b, mm

с, mm

Body 1 Body 2

1884 1962

40 50

60 50

100 100

Body 3

980

50

50

50

8.5.5. For the system of coordinates given in Fig. 8.2. calculate the components of the inertia tensor of the studied body. Try to qualitatively draw the ellipsoids of inertia for these bodies. Questions for self-assessment

1. What properties of a physical body characterize its moment of inertia? 2. Does it make sense to speak about the moment of inertia, without specifying the axis with respect to which it is calculated? 3. What quantity is called the tensor of inertia of a physical body with respect to the point? 4. What is the practical use of the concept of the tensor of inertia? 5. What characterizes the ellipsoid of inertia of a solid body? Are there different physical bodies with the same ellipsoids of inertia and vice versa? 122

References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989. – p. 576. 6. Laboratory works in physics: textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 7. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

123

Laboratory work № 9 STUDYING OF ELASTIC COLLISIONS OF BALLS AND DETERMINATION OF YOUNG'S MODULUS

9.1. The purpose of work: measuring the time of collision of balls, and using the laws of conservation of momentum and mechanical energy determine the Young's modulus of the material of balls, the maximum impact force, the maximum pressure in the contact «patch», the radius of the contact area and maximum radial deformation of balls. 9.2. Brief theoretical introduction 9.2.1. Central absolutely elastic collision of balls. In the central absolutely elastic collision of balls, the total momentum and the total kinetic energy of the ball do not change after the collision. If we denote the masses of the balls m1 and m2 ,   the speed before the collision 1 and  2 , and after the collision   1



and и   , we can write 2

    m11  m2 2  m11  m2 2 , 1 1 1 1 2 m1 12  m2 22  m11 2  m2 2 . 2 2 2 2

 

Solving this system of equations, we can find the speed of the balls after the collision: 

1 

  2m2 2  (m2  m1 )1 , m1  m2 124





 2 

  2m11  (m1  m2 ) 2 . m1  m2



When two balls of equal mass ( m1  m2  m ) collide, they exchange velocities: 







1   2 ,  2  1 . In a special case, when before the collision the second ball is at rest, the first ball after the collision will stop, and the second will move at the speed of the first ball before the collision. 9.2.2. Speed of ball-to-ball collision. The speed, with which the right ball hits the left ball at rest, can be found from the law of conservation of mechanical energy. The center of mass of the ball at the angle of deflection of the suspension a rises to a height h:

Fig. 9.1. To the derivation of the formula (9.5)

From Fig 9.1 it is seen that h  l 1  cos    2l sin 2

 2

.

The potential energy of the ball increases by the value: 125



E п  mgh  2mgl sin 2

2

,

which transforms into its kinetic energy before its collision with a stationary ball: m 2   2mgl sin 2 . 2 2

From this expression we can find the speed of the ball:   2 gl sin

 2

.

(9.5)

9.2.3. Dependence of the time of collision and the force of collision of balls on the properties of their material and sizes. In the elastic collision of two balls from the moment of their contact with each other, they experience elastic compression deformations in the points of contact. Elastic forces arising in such cases begin to reduce the relative speed of colliding bodies. At the moment, when the relative speed difference between the balls decreases to zero, the force of elastic deformation of the body reaches the highest value, which also corresponds to the closest approach of the balls and the largest contact area between them. The kinetic energy of the relative motion of the balls is completely transformed into the energy of elastic deformation. After reaching the closest approach, the balls pushing each other under the action of elastic deformation again fly apart in opposite directions at a speed determined by formulas (9.3) and (9.4). In the case of a perfectly elastic central collision of the balls, based on the condition of equality of the kinetic energy of the relative motion before the collision and the potential energy of the elastic deformation at the closest approach of the balls, A.N. Dinnik derived the following formulas for impact parameters: 126

F  1, 2827 5

 m1  m 2   m1  m 2 1     E 1 

2 1



2

1 12 1  22    1,2356 5   E 2   E1

1  E2

2 2

3

    6     

2

 1 1     R R 2   1

,

2

 1 1   m 1 m 2  4   ,      R 1 R 2   m 1 m 2 

2

 R1 R 2   m1 m2  1  12 1  22  2    R  R   m  m   E  E   , 2   1 2  1 2 1

a  0,9872 5 

p max  0 ,6285 5

 1 1   R R 2  1

  

3

 m 1 m 2   m 1 m 2

 1   12 1      E E 2 1  2

 1   12 1   22     5 E E 2  1  t  2,8662

2 2

   

4

 2   

6

7

8

9

,

2

 1 1   m 1 m 2        R 1 R 2   m 1 m 2  .



10

In these formulas, F is the maximum impact force of the balls at the closest approach;  is the absolute value of the maximum linear deformation of the balls; a is the radius of the contact circle with the greatest deformation; рmax is the maximum pressure in the center of the contact circle; t is the duration of the collision; R1 and R2, m1 and m2 are radii and masses of the colliding balls; E1 and E2  1 and  2 are Young's moduli and Poisson's coefficients for materials of colliding balls;  is a relative impact velocity of balls. If the balls are made of the same material and have equal radii, formulas (9.6) - (9.10) are considerably simplified and reduced to the form: 127

F  7,705 

 R 3 ,

1

t

  0,4308  t ,

2

a  0,4122    t  R ,

3

 R2 ,

4

p max  21,66 

t  5,8432 R

5

t2

1     2 2

2

E 2 

,

5

where  is the density of the material of the ball. Poisson's coefficients for all materials range from 0 to 0.5; for example, for steel  = 0.29. Knowing the density  of the material, the radius R of the balls, their relative velocity  at the moment of impact, and the time t of collision of balls, from the formula (9.15) we can determine the Young's modulus of material of the balls: E  82,53 

1     2 2 5

t

2

R5

(9.16)

Knowing the time of collision between balls, using formulas (9.11)-(9.14) we can find all other quantities characterizing the elastic interaction of balls. 9.3. Description of the experimental setup and experimental procedure 9.3.1. Experimental setup. The experimental setup shown in Fig. 9.1. consists of the base 1, the vertical rod 2, the upper bracket 3, the casing 4, the electromagnet 5, threads 6 for hanging metal balls 12, wires 7 to provide electrical contact of the ball with the terminals 10. 128

The base 1 has three adjustable legs 8 and a thumbscrew 9 to fix the vertical rod 2. The vertical rod 2 is made of a metal pipe. On the upper bracket 3, designed for hanging balls, adjustment nodes providing their central impact, and terminals 10 are mounted. Casing 4 is designed for attachment of the scale 11 for angular movements. The electromagnet 5 is designed for fixing the starting position of one of the balls 12. The metal balls 12 are formed in pairs from aluminum, brass and steel. To measure the time of collision of balls the electronic stopwatch is used.

Fig. 9.1. General view of the setup

9.3.2. Experimental procedure. 9.3.2.1. Preparation of the setup for measurements. – Place the balls in the brackets of suspenders. – With the help of regulating legs set the base 1 so that the lower arrows on the lower brackets of suspenders pointed to zero on scale 11. Adjust the position of the balls in the vertical and horizontal planes to superpose upper arrows on 129

– – – – –

bracket hangers. The adjustment is made by varying the length of threads of ball suspenders and changing the position of the points of attachment to the bracket. Connect terminals 10 of the upper bracket and the electromagnet 5 of the setup to the electronic stopwatch with cables. 9.3.2.2. Measurement of the time of ball-to-ball collision. Switch on the stopwatch (the switch is on the back panel of the stopwatch). By switching the stopwatch, the indication display and the electromagnet are switched. Fix the right ball with the help of the electromagnet. Press the START button, the ball will start moving. Record the stopwatch readings of time t of ball collisions in the table. Press the STOP button.

9.4. Procedure of work fulfillment 9.4.1. Measure the length of the suspension l, and the diameter 2R of balls 3 ÷ 5 times. Write the results in Table 9.1. 9.4.2. Take from the reference book the values of the density  of ball materials and Poisson's coefficients  and write in Table 9.1. Table 9.1 Linear dimensions of the setup and the reference data

№

Material

l, m

, m

2R, m

, m

, kg/m3



9.4.3. Place the electromagnet in the right end position. Note the angle of deflection of the ball on the scale 11 and record it in Table 9.2. Measure the time of collision of balls in accordance with paragraph 9.3.2.2. Repeat measurements 3 ÷ 5 times. Record the results in Table 9.2. 130

9.4.4. Moving the electromagnet 11 along the scale, repeat the measurement of time of collision of balls for the other two primary angles of deflection of the right ball. Write the results in Table 9.2. Make similar measurements for balls made of other materials (as instructed by the teacher). Table 9.2 Results of measurements and calculations

№

Material

,0

t, s

, s

υ, m/s

E, Pa

, Pa

9.4.4. Using formula (9.5), determine the speed of the ball  in the collision and write it down in Table 9.2. 9:45. Using formula (9.16), calculate the value of Young's modulus E for each angle  of deflection of the ball, and write the results in the Table. 9.4.6. Find the average value of Young's modulus for the material of the ball. 9.4.7. Using formulas (9.11) - (9.14) determine the maximum force of impact F, the value of the largest linear deformation , the radius a of the contact circle for the largest deformation of balls and the maximum pressure рmax. Write the results in Table 9.3. Table 9.3 Results of measurements

№

Material

,0

 ,m

F, Н

131

a, m

рmax, Pa

Questions for self-assessment

1. 2. 3. 4.

What collision is called an absolutely elastic collision? What conservation laws are fulfilled in this case? Why does the moving ball stop after the collision? Why are forces arising in the collision so large?

References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989. – p. 576. 6. Laboratory works in physics: textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704.

132

Laboratory work № 10 MEASUREMENT OF THE SPEED OF FLIGHT OF THE BODY WITH A TORSION-BALLISTIC PENDULUM

10.1. The purpose of work: using the laws of conservation of mechanical energy and angular momentum, determine the speed of the body (bullet) in the setup of a torsion-ballistic pendulum. 10.2. Brief theoretical introduction Direct measurement of the speed of the flight of shells is a difficult experimental task, as this speed reaches high values (e.g. 800 ÷ 1000 m / s for a combat rifle). Therefore indirect methods of measurement are widely used. In one of these methods a flying bullet gets into the stationary body of much greater mass, experiencing absolutely inelastic collision. After the collision, the body begins to move, and its speed of movement is much less than the speed of the bullet. If we measure a low speed of the body, it is possible to calculate the speed of the bullet using the conservation laws for an isolated system of interacting bodies. This method is used in this work, in which the experimental problem of determining the speed of flight of the projectile is modeled. The scheme of the experimental setup is shown in Figure 10. The torsion-ballistic pendulum is made in the form of the cross. On the horizontal bar 12, the target 10 and two movable loads 11 are fixed. The cross is suspended on a steel wire to the brackets 4 and 5. In order to provide absolutely inelastic collisions, the target is covered with a layer of plasticine. The deflection angle of the pendulum after the collision is measured on the scale 8. A horizontally flying body of mass m hits the target and gets stuck in it. As a result of the impact the pendulum rotates around its 133

axis at an angle  0 . As the thread of the pendulum is twisted, the kinetic energy of the pendulum transforms into the potential energy of the elastic deformation of the suspension thread. Then the reverse process of transformation of potential energy into kinetic energy starts, etc. The pendulum oscillates, its period T is significantly longer than the stopping time t of the bullets in the target:

T  t .

(10.1)

Under the condition (10.1) the law of conservation of the angular momentum is fulfilled, and in the projection on the axis of rotation of the pendulum it is written for the pendulum-body system as: m r  I  I 0 ,

(10.2)

where m is the mass of the body;  is the speed of the body before the collision; r is the distance from the line of flight of the body to the axis of rotation of the pendulum (the impact parameter); I is the moment of inertia of the pendulum; I  is the moment of inertia with respect to the axis of the pendulum;  0 is the angular velocity of rotation of the pendulum immediately after the collision of the body with the target. In our setup I  I  , therefore formula (10.2) is simplified: m r  I 0 .

(10.3)

Neglecting friction losses in the setup, the law of conservation of mechanical energy of the pendulum, after hitting by the projectile, can be written as: I  02 k  02 ,  2 2

(10.4)

2 where I  0 is the kinetic energy of the pendulum immediately

2

2 after the collision; k  0

2

is the potential energy of the elastic 134

deformation at the maximum angle of rotation of the pendulum; k is the torsion modulus of the thread;  0 is the maximum angle of rotation of the pendulum. In real cases, the pendulum oscillations are always damping (due to air friction and other causes). Therefore, the formula (10.4) can be used if the energy loss for a quarter of the period is small as compared to the energy of the pendulum after the collision, i.e. k 02 1 , Eп  4 2

(10.5)

where Еп is the energy loss for the period. To verify the validity of condition (10.5), it is sufficient to measure the number N of full oscillations of the pendulum, for which the initial amplitude of oscillations of the pendulum decreases by half. If (10.6) N  1 , the oscillations are weakly damping, and we can use the formula (10.4). Solving (10.3) and (10.4), we obtain the formula for determining the speed of the body flight:



 o kI  1/ 2 mr

.

(10.7)

The unknown product kI can be found from the formula for the period of oscillations of the torsion pendulum, measuring it for two values of the moment of inertia of the torsion pendulum. The formula for the period of oscillations of the torsion pendulum is written as:

T  2 I k .

(10.8)

The moments of inertia of the pendulum for two different positions of loads ( R and R1 ) are calculated by the formulas: 135





I  I 0  2 I 0  MR 2 ,

(10.9)



(10.10)



I1  I 0  2 I 0  MR12 ,

where I 0 is the moment of inertia of the cross without loads: – I 0  MR 2 is the moment of inertia of the load 11, calculated using the Huygens-Steiner theorem (see. Appendix to the lab No. 12); – M is the mass of the load 11. From formulas (10.9) and (10.10) we can express I1 through I:





I1  I  2M R12  R 2 .

For the position of the load 11 corresponding to the distance R1 , the period of oscillations T1 is equal to: T1  2

I1 . k

Raising the last expression in the square and substituting the value for I1 , we obtain: T 12 4 2





I 1 4 2 8 2  I  2M R 12 R 2  T 2 M R 12 R2 , k k k



or T12  T 2 











8 2 M R12  R 2 . k

Using (10.8) we find: T12  T 2 

4 M R 2 R2 T , kI  1 / 2 1 136



(10.11)

and kI 





4 M R 12  R 2 T . T 12  T 2

(10.12)

Substituting (10.12) in (10.7), we obtain the final formula for the speed of flight of the body: 

 



 0 4 M R 12  R 2 T r



m T 12  T 2



.

(10.13)

10.3. Description of the experimental setup General view of the torsion-ballistic pendulum ARM-09 is shown in Figure 10. The base 1 is equipped with 2 adjustable legs which allow leveling of the setup. On the base the column 3 is mounted, on which the upper bracket 4, the lower bracket 5 and the middle bracket 6 are mounted. The middle bracket holds the firing device 7, a transparent casing with an angular scale 8 and a photoelectric sensor 9. The brackets 4 and 5 have clamps for fastening the steel wire 13, to which the pendulum consisting of the target 10 with plasticine on it, two moving loads 11, the rod 12 and the suspender 14 is attached.

Fig. 10. Torsion-ballistic pendulum

137

15.

A photoelectric sensor is connected with an electronic stopwatch

On the front panel there are buttons: «POWER» – to connect the power supply; «RESET» – to remove the previous reading and to start the stopwatch; «STOP» – to stop the measurement of time. 10.4. Procedure of work fulfillment 10.4.1. Experimental part. 10.4.1.1. Preparations for measurements. To prepare the setup it is necessary: – To check the grounding; – To make a horizontal alignment of the setup using the adjustment screws; – To check that the initial position of the torsion pendulum is within the scale 8 for measuring rotation angles – To check the tension of the suspension (it should not sag under the crosspiece) and free movement of the pendulum in the range – 900 to + 900 ; – To check the position of the driver 14 with respect to the photoelectric sensor: in the initial position of the pendulum the driver must be in the section of the sensor 9; – To connect the device to the power line; – To press the button «POWER», to check the glow of the lamp of the photoelectric sensor 9, flashing of zeros on the digital display; – To test the electronic stopwatch and counter of oscillations, deflecting the pendulum by an angle of 200 ÷ 300 and starting the scheme by the button «Reset» (stop of counting – «Stop» button). 10.4.1.2. Preliminary measurements. Measurement of the periods of oscillations of the pendulum. Move apart loads 11 at the maximum distance R1 between the center of mass of the loads and the axis of rotation of the pendulum. Move the pendulum at an angle of 200 ÷ 300 and check the condition (10.6). Measure the total number N of oscillations of the 138

pendulum, for which the initial amplitude of oscillations of the pendulum decreases two times. Measure at least 5 times the time of (5 ÷ 10) oscillations of the pendulum at the initial angle of 200 ÷ 300 and write down the results in Table 10.1. Repeat the same measurements for closely located loads (at a distance from the center of mass equal to R), write down the results in Table 10.1. Table 10.1 Measurement of the period of oscillations of the pendulum The number of oscillations N =

№

t,s

t ,s

T ,s

t,s

R1 =

1 2 3 4 5

 t  ,s

T ,s

R=

Switch off the power supply of the setup by the button «POWER». 10.4.1.3. The main measurements. Table 10.2 Results of measurements and calculations

№

 0 ,0

 0 , rad

r,m

 , m/s

1 2 3 4 5 139

  , m/s

 , m/s

 2 , m2/s2

Measurements are made for closely located loads. Make 5 ÷ 10 shots from the setup, fixing for each the maximum deflection angle  0 of the pendulum on the scale 8 and the impact distance r from the point where the projectile hits the target to the axis of rotation of the pendulum. Write down the experimental results in Table 10.2. Write down the mass of the projectile and the mass of the load, in our setup they are: m = (2.148  0.015) g, М = (190.0  0.5) g. 10.4.2. Mathematical processing of the results. Using formula (10.13) calculate the speed of flight of the projectile in each experiment. Find the average speed of flight of the projectile, and the confidence interval for the method of direct measurements. Find the relative error of the result. Questions for self-assessment

1. What laws of conservation are valid for absolutely elastic and inelastic interaction of bodies? 2. What conservation laws are used in this work? 3. Why cannot we determine the speed of the projectile, equating its kinetic energy to the potential energy of the elastic deformation of the suspension for the deviation of the pendulum at the maximum angle? 4. What simplifying assumptions are used in this work? 5. What factors may affect the accuracy of the experiment? 6. Can we use the above theory, if the collision occurs with the target body at not a right angle? 7. How does the angular momentum of the projectile with respect to the axis of the pendulum change during the flight to the target? References

1. Matveev A.N. Mechanics and Relativity: A Textbook for special physical universities / A.N. Matveev. – 2nd ed., revised. and enlarged. – M.: Higher School, 1986. – p. 320. 2. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 3. Irodov I.E. Mechanics. Basic laws / I.E. Irodov. – M.: Laboratory of basic knowledge, 2002. – p. 312. 140

4. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 5. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989. – p. 576. 6. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 7. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

141

Laboratory work № 11 DETERMINATION OF THE VISCOSITY OF LIQUIDS BY THE STOKES METHOD

11.1. The purpose of work: experimental determination of the viscosity of the fluid by Stokes' formula used to calculate the friction force acting on the ball falling in the liquid. 11.2. Brief theoretical introduction A body moving in a liquid experiences resistance from the same liquid. The liquid particles adjacent to the surface «stick» to it due to the forces of attraction between molecules of the liquid and the material of the surface and therefore move with the body. Because of intermolecular interaction, the molecules adjacent to the «stuck» molecules are also set in motion. As the distance from the surface increases, the speed of the molecules decreases. This causes friction between the layers of the liquid. The value of the friction force between two adjacent layers of fluid, moving along the x-axis parallel to each other at different speeds, is determined by the Newton’s law for viscous friction: FТР  

d ( y ) , S dy

(11.1)

where S is the area of the liquid layer; d ( y ) is the module of the dy

gradient of the speed module (Fig. 11.1);  is the dynamic viscosity of the fluid. 142

As follows from (11.1), the viscosity  is numerically equal to the force of viscous friction between layers of liquid or gas, moving parallel to each other with different speeds if areas of the layers are equal to one, with a unit gradient of the speed module between the layers. Dimension of viscosity : ML1T 1 . Units of  measurement in SI: [] = kg/(ms) = Pas.     is called kinematic viscosity. Dimension of kinematic viscosity is  : L2T 1 . Units of measurement of  in SI: [] = m2/s The viscosity of the liquid depends on its properties and temperature.

Fig. 11.1. To the determination of the force of viscous friction

Consider the forces acting on a solid ball falling freely in the fluid (Fig. 11.2).

Fig. 11.2. Forces acting on the ball falling in the liquid

143

Three forces are acting on the ball falling in the liquid: P1 is the force of gravity, P is buoyancy, Ffr is the power of resistance of the medium. Let us introduce the following notations: r is the radius of the ball,  is the speed of the ball, 1 is the density of the ball 1 is the liquid density,  is viscosity, g is acceleration of free fall. The force of gravity of the ball:

 4  P1   r 3 1 g . 3

(11.2)

The buoyancy force is determined according to the law of Archimedes:   4 P    r3 g . (11.3) 3 If the ball is moving at a low speed (laminar flow) in a fluid not bounded by walls, then according to Stokes' law the friction force caused by the internal friction forces is equal to:   F   6 r .

(11.4) 



Let us write the second Newton’s law (m d   Fi ) in the projection on the direction of the ball motion: m

dt

d  P1  P  F . dt

The mass of the ball can be found by the formula: 144

i

(11.5)

4 m   r 3 1 . 3 as:

(11.6)

Equation (11.5), using (11.2) – (11.4) and (11.6), can be written d 9  dt 2r 2 1

  2( 1   ) gr 2  .  9  

(11.7)

Separating the variables, we obtain: d





 н  2( 1

 

   ) gr 2  9 



9 t .  dt 2 r 2 1 0

(11.8)

If at the initial time (t = 0), when the ball drops in the liquid, its speed is equal to   н, then integrating (11.8), we obtain:

 (t )   0  ( 0   н )e  t /  ,

(11.9)

where  0 is the terminal speed of the ball motion in the fluid;



2 1r 2 9

(11.10)

is called the relaxation time of the ball motion. If t  , the ball in its movement in the viscous liquid reaches its maximum speed:

0 

2 1   2 gr . 9 

 The condition  Fi  0 is fulfilled. i

145

(11.11)

Using (11.9), we find the dependence of the path of the ball in a viscous fluid on time: t

x    dt   0 t   ( 0   н )(1  e  t /  ) . 0

(11.12)

Fig. 11.3 shows the dependence of the speed and distance on time for the case when the initial velocity of the ball in the fluid is zero ( н  0).

Fig. 11.3

When the time t1  4,6 

1 r 2 

(11.13)

passes from the beginning of the movement, the speed of the ball is equal to  1 0,99  0 . During this time the ball covers the distance x1  3,6  0 . If t  t1, the ball will move into the liquid uniformly at a constant speed  0 . If the ball moves in a vessel with a finite cross section, it is necessary to take into account the effect of walls on the movement of 146

the ball. If the radius of the cylindrical vessel is R, then in the expression for the force of resistance to the movement of the ball an additional term appears: r  F  6  r 0 1  2,4  . R 

(11.14)

As a result  0

2 g r2 9

 1  r   1  2,4  R 

.

(11.15)

Using this formula and measuring the speed of the ball, we can find viscosity of the liquid:



2 2 gr 9

 1  r   0 1  2,4  R 

.

(11.16)

Stokes' formula (11.4) for determination of the friction force acting on the moving ball is valid only for Re  0.5 where

Re 

2 0 r 



is the Reynolds number.

(11.17)

For Re  0.5 the fluid resistance force does not obey Stokes' law. Brief information about the Reynolds number The properties of a viscous liquid depend on its density , dynamic viscosity or kinematic viscosity . They are determined by characteristic flow speed  and linear body size d = 2r. For example, when a ball moves in the liquid (or, conversely, when the liquid flows around a stationary ball) the characteristic 147

velocity is the velocity of the ball (or, respectively, the average fluid velocity), and the characteristic size is the ball diameter 2r. From the quantities , d, ,  (or ), using the analysis of dimensions, one can make a dimensionless complex, which is called the Reynolds number and is denoted by Re: Re 

2  r





2 r



.

The Reynolds number is one of the important characteristics of a viscous medium. The character of the flow, laminar or turbulent, depends on its value. 11.3. Description of the experimental setup The setup consists of two transparent cylindrical vessels with castor oil and glycerol. On the vessels there are horizontal marks at a distance of 10cm between them. They are used to determine the distance covered by the ball moving in the fluid. Steel balls of different sizes are used for measurements. 11.4. Procedure of work fulfillment 11.4.1. Write down the density of the fluids and the material, from which the balls are made, and the radii of vessels (specifications of the setup) in the laboratory log. 11.4.2. Using the micrometer, measure the diameter of each ball (5  7 balls) 3 times in different places. 11.4.3. Determine the depth of the liquid from which the movement of the ball becomes uniform. For this purpose it is necessary to determine the speed of the ball in the first interval (10 cm) and the second interval (10 cm). If these speeds are equal, then we can select the first line as the starting point for further measurements of the ball motion. 11.4.4. Measure with a stopwatch the time of passing the distance of 60 cm by the ball (or the other distance by the assignment of the teacher) for each fluid (castor oil and glycerin). Write down the measurements in Table 11.1. 148

Table 11.1 Results of measurements and calculations Distance l, which the ball passes in the liquid: l =… m.

2r , m

t, s

, m/s

, ПА·с

  ,

 ,

Pa·s

Pa·s

 2 , Pa2·s2

Castor oil

Glycerin

11.4.5. Using formula (11.16) calculate the viscosity and write down the results of calculations in Table 11.1. 11.4.6. Calculate the error of the result obtained by the method of direct measurements. 11.4.7. To estimate the influence of the vessel walls on the movement in the ball, calculate the viscosity for one of the experiments using the formula (11.11), and compare the results with those calculated using the formula (11.16) (for each fluid). 11.4.7. For one of the experiments, calculate the Reynolds number by the formula (11.17). Using (11.13), estimate the time during which the movement of the ball becomes uniform, and the distance x0     , covered by the ball during this time (for each fluid). Questions for self-assessment

1. What is viscosity? Give the definition of viscosity, name its unit in the SI. 2. What does the viscosity of the fluid depend on? 149

3. What forces act on the ball moving in the liquid? 4. Why does, from a certain moment of time, the movement of the ball become uniform? 5. How do the ball diameter and the diameter of the cylinder in which it falls affect its movement? References

1. Savelyev I.V. Course of general physics. Mechanics / I.V. Savelev. – M.: Astrel, 2003. – p. 336. 2. Strelkov S.P. Mechanics / S.P. Strelkov. – SPb.: Publisher «Lan», 2005. – p. 560. 3. Sivukhin D.V. General course of physics. Volume 1. Mechanics / D.V. Sivukhin. – M.: Nauka, 1989. – p. 576. 4. Laboratory works in physics: Textbook / Goldin L.L., Igoshin F.F., Kozel S.M., et al.; Ed. Goldin L.L. – M.: Nauka. Main publishing house of Physical and mathematical literature, 1983. – p. 704. 5. Zeidel A.N. Errors of measurements of physical quantities / Zeidel A.N. – SPb.: Lan, 2005. – p. 106.

11.6. Task for Student research work 11.6.1. A brief theory. As mentioned at the beginning, the Stokes formula   F   6  r 0

is valid only for Reynolds numbers Re