DETERMINATION OF MOMENT OF INERTIA
PHYSICAL PENDULUM
Goal of the work: determine the moment of inertia of a physical pendulum in the form of a rod with weights based on the period of its own oscillations.
Equipment: pendulum, stopwatch.
THEORETICAL INTRODUCTION
The moment of inertia of a rigid body is a measure of the inertia of a body during its rotational motion. In this sense, it is an analogue of body mass, which is a measure of the inertia of a body during translational motion. According to the definition, the moment of inertia of the body equal to the sum products of body particle masses m i by the squares of their distances to the axis of rotation r i 2:
,
or 
.
(1)
The moment of inertia depends not only on the mass, but also on its distribution relative to the axis of rotation. As you can see, the inertia during rotation of a body is greater, the further the particles of the body are located from the axis.
There are various experimental methods for determining the moment of inertia of bodies. The paper proposes a method for determining the moment of inertia from the period of natural oscillations of the body under study as a physical pendulum. A physical pendulum is a body of arbitrary shape, the point of suspension of which is located above the center of gravity. If in a gravitational field the pendulum is deflected from the equilibrium position and released, then under the influence of gravity the pendulum tends to the equilibrium position, but, having reached it, by inertia it continues to move and is deflected in the opposite direction. The movement process is then repeated in reverse direction. As a result, the pendulum will perform rotational oscillations of its own.
To derive the formula for the period of natural oscillations, we apply the basic law of the dynamics of rotational motion. The angular acceleration of a body is directly proportional to the moment of force and inversely proportional to the moment of inertia of the body relative to the axis of rotation:
= 
. (2)
M 
The torque of a force is by definition equal to the product of the force and the arm of the force. The arm of a force is a perpendicular lowered from the axis of rotation to the line of action of the force. For a pendulum (Fig. 1a), the gravity arm is equal to d= a sin
,
Where A– the distance between the axis of rotation and the center of gravity of the pendulum. For small oscillations of the pendulum, the angle of deflection
is relatively small, and the sines of small angles are equal to the angles themselves with sufficient accuracy. Then the moment of gravity can be determined by the formula M=−mga∙
. The minus sign is due to the fact that the moment of gravity counteracts the deflection of the pendulum.
Because angular acceleration is the second derivative of the angle of rotation with respect to time, then the basic law of the dynamics of rotational motion (1) takes the form
.
(3)
This is a second order differential equation. Its solution must be a function that turns the equation into an identity. Such a function could be the sine function
= 0 sin ( t + ). (4)
In this case, the cyclic frequency is equal to 
. The cyclic frequency is related to the period of oscillation, that is, the time of one oscillation, the ratio T =
2
/
.
From here
. (5)
Oscillation period T and the distance from the axis of rotation to the center of gravity pendulum A can be measured. Then from (5) the moment of inertia of the pendulum relative to the axis of rotation WITH can be determined experimentally using the formula
. (6)
The pendulum, the moment of inertia of which is determined in the work, is a rod with two disks placed on it. Theoretically, the moment of inertia of a pendulum can be defined as the sum of the moments of inertia of the individual parts. The moment of inertia of the disks can be calculated using the formula for the moment of inertia of a material point, since they are small compared to the distance to the axis of rotation: 
,
.
Moment of inertia of the rod relative to an axis located at a distance b from the middle of the rod, can be determined by Steiner’s theorem 
. As a result, the total moment of inertia of the pendulum can be calculated using the formula
. (7)
Here m 1 , m 2 and m 0 – masses of the first, second disks and rod, l 1 , l 2 – distances from the middle of the disks to the axis of rotation, l 0 – length of the rod.
Distance from the suspension point to the center of gravity of the pendulum A, necessary to determine the moment of inertia in formula (6), can be determined experimentally using the concept of the center of gravity. The center of gravity of a body is the point to which the resultant force of gravity is applied. Therefore, if the pendulum is placed horizontally on a supporting prism located under the center of gravity, then the pendulum will be in equilibrium. Then just measure the distance from the axis WITH to the supporting prism.
But you can determine the distance A by calculation. From the equilibrium condition of the pendulum on the prism (Fig. 1b) it follows that the moment of the resulting force of gravity relative to the axis WITH equal to the sum of the moments of gravity of the loads and the rod: ( m 1 + m 2 + m 0)ga =m 1 gl 1 + m 2 gl 2 + m 0 gb. Where do we get it from?
. (8)
COMPLETING OF THE WORK
1. By weighing on a scale, determine the masses of the disks and the rod. Place the discs on the rod and secure them. Measure the distances from the axis of rotation to the middle of the disks l 1 , l 2 and to the middle of the rod b, rod length l 0 according to centimeter divisions on the rod. Record the measurement results in the table. 1.
Turn on the installation to a 220 V network, press the “Network” button.
Table 1
Weight of 1 disk m 1 kg | |
Weight of the 2nd disk m 2, kg | |
Rod mass m 0 , kg | |
Distance l 1 , cm | |
Distance l 2 , cm | |
Rod length l 0 , cm | |
Axle distance b, cm |
Turn off the installation.
3. Perform calculations in the SI system. Determine the average value<T> oscillation period. Determine distance A from the axis to the center of gravity of the pendulum according to formula (8), or place the pendulum on a supporting prism so that it is in equilibrium, and measure the distance using the divisions on the rod A.
4. Determine the experimental average value of the moment of inertia of the pendulum<J the ex> according to formula (6) according to the average value of the oscillation period<T>.
table 2
T 1 , With | T 2 , s | T 3, s | <T>,s | J theory , kg∙m 2 |
||
5. Determine the theoretical value of the moment of inertia of the pendulum J theory according to formula (7).
6. Draw a conclusion by comparing the theoretical and experimental values of the moment of inertia of the pendulum. Estimate the measurement error J =< J exp > – J theory .
7. Write the answer in the form: J exp = < J > J .
CONTROL QUESTIONS
1. Give the definition of a physical pendulum, explain why natural oscillations of a pendulum are possible.
2. Write down the basic law of the dynamics of rotational motion for a physical pendulum.
3. In what form are they looking for a function that is a solution? differential equation speakers for a physical pendulum. Check if this feature is a solution.
4. Write down the formula for the period of oscillation of a physical pendulum. How will the period of oscillation change if the lower disk is moved even lower?
5. Define moment of inertia. Derive a formula to determine the theoretical value of the moment of inertia of the pendulum.
6. Define the center of gravity. Derive a formula to calculate the position of the center of mass. How can you experimentally determine the position of the center of mass of a pendulum?
DETERMINING THE SPEED OF SOUND IN AIR
Goal of the work: determine the speed of sound in air and wavelength using the Lissajous figure method, determine the adiabatic index.
Equipment: sound generator, handset with telephone and microphone, oscilloscope, heater.
THEORETICAL INTRODUCTION
Sound is waves in an elastic medium. In gases, sound waves are the process of propagation of areas of compression - rarefaction.
Let us consider the propagation of a sound wave in a gas. Let the telephone membrane located at the base of an imaginary tube with cross-sectional area S, began moving at subsonic speed U. Gas particles adjacent to the membrane move at the same speed. The air in front of the membrane is compressed and compresses subsequent layers of gas. The boundary between compressed and undisturbed gas, called the front, moves at the speed of sound V (Fig. 1).
To determine the speed of sound, let us apply the equation of Newton’s second law for a moving mass of gas: the change in the momentum of the gas is equal to the impulse of force from the membrane:
dm
U =
F
dt. We define the gas mass as the product of density and volume: dm =
dL∙
S,
A
the force of membrane pressure on the gas as an increase in pressure per area: F =
dp∙
S.
Let us assume that the ratio of the membrane and front velocities is proportional to the ratio of the distances they travel: 
, which, in turn, is equal to relative change gas density. Substituting the resulting transformations into the equation of Newton’s second law, making the replacement dL=
Vdt, we get the equation 
. Due to the short duration, the processes of compression and rarefaction of gas in a sound wave occur adiabatically, without heat exchange between the heated compression region and the cooled rarefaction region. Therefore, we apply the Poisson equation 
. Differentiating 
and substituting, we get
. (1)
Here R= 8.31 J/mol∙K – gas constant, T – absolute temperature, M= 28.9 10 –3 kg/mol is the mass of a mole of air, = 1.4 is the adiabatic index for diatomic gases.
Let's write down the wave equation. This is the parameter dependence equation ψ
(pressure, displacement, etc.)
at some point in space from time and distance Z to the source. If the source oscillates according to the equation 
, then the particles of the medium begin to oscillate later than the source, during the propagation of the wave 
. Then the wave equation takes the form
. (2)
D 
To experimentally determine the speed of sound in air, this work uses the method of Lissajous figures. A Lissajous figure is a repeating trajectory of a point participating in two mutually perpendicular oscillations. It occurs if the frequency ratio is equal to the ratio of integers.
In a laboratory setup, on the oscilloscope screen one observes the addition of electrical oscillations of the same frequency from the telephone as a sound source, and from the receiver - a microphone, which are fed respectively to the horizontal x and vertical y oscilloscope inputs (Fig. 2).
Let us consider special cases of the addition of two mutually perpendicular oscillations of the same frequency.
Example 1. Let the phase difference be a multiple of the integer 2
radians, so the vibrations occur according to the equations: x =
A
1
cos 2
t,
y =
A
2
cos
(2
t+
2πк) =
A
2
cos 2
t.
To obtain the trajectory equation (Lissajous figure) in explicit form y(x) exclude time t,
for example, dividing equations. As a result we get 
. This is the equation of a straight line (Fig. 3) passing through 1−3 quadrants in a rectangle with sides 2 A 2 –2A 1 .
Example 2. Let the phase difference be a multiple of an odd number 
radian, so x=A
1
cos
2
t,
y=
A
2
sin
2
t.
Let's exclude time t by ratio. As a result, we obtain the ellipse equation for the Lissajous figure: 
, inscribed in rectangle 2 A 2 – 2A 1 .
TO 
As can be seen, the Lissajous figure depends on the phase difference (Fig. 3).
With a constant distance between the microphone and the phone Z the phase difference of the components of the oscillations and the figure on the oscilloscope screen depends on the frequency
or 
. (3)
Transforming an ellipse back into an ellipse or a straight line
in the same straight line occurs if the phase difference increases by an integer 2
radians, that is 
, Where k =
0,1,2,3 –
an integer (it is equal to the increase in the number of wavelengths in the tube). Substituting the condition for repetition of the Lissajous figure into equation (3), we obtain
or 
(4)
COMPLETING OF THE WORK
Installation 1
3. Smoothly changing the frequency of the generator, observe the transformation of the Lissajous figure, as shown in Fig. 3. Get an image of the original figure. Write down the increasing number in the table k above the original k 0 and the corresponding generator frequency. Repeat the experiment at least five times.
k − k 0 | |||||
ν , Hz |
4. Plot the dependence of the generator frequency when repeating the figure on the number k− k 0 . The size of the chart is at least half a page. Apply a uniform scale on the axes. Draw a straight line near the points (Fig. 4).
5. Determine the average value of the speed of sound from the angular coefficient of the experimental straight line. To do this, on the experimental line, how to build on the hypotenuse right triangle(Fig. 4). Using the coordinates of the vertices of the triangle, determine the average speed
. (5)
6
. Estimate random measurement error 
. Record the result V=<
V>±
δV,
P=
0,9.
7. Compare with the theoretical value of the speed of sound in air, calculated using formula (1). Draw conclusions.
Installation 2
The operation is carried out in the same way as in installation 1. At a constant frequency of the generator, the distance between the telephone and the microphone changes. The speed of sound is determined by the formula 
.
CONTROL QUESTIONS
1. Explain the process of sound propagation in gases. Give the concept of a wave front.
2. Write down the formula for the speed of sound waves in gases. Explain why the process of compression and rarefaction of gas in a sound wave occurs adiabatically.
3. Write down the plane wave equation. Give the concept of phase.
4. Define a Lissajous figure. Derive the equation for the trajectory of a point participating in two mutually perpendicular oscillations of the same frequency, with a phase difference of 2 π k radian.
5. Derive the equation for the trajectory of a point participating in two mutually perpendicular oscillations of the same frequency, with a phase difference of /2 rad.
6. At what is the smallest change in the frequency of the generator, the Lissajous figure takes its original form.
DETERMINATION OF THE HEAT CAPACITY OF AIR
Goal of the work: get acquainted with the process of isobaric heating of air, determine the molar heat capacity of air during isobaric heating.
Equipment: heater, compressor, thermocouple with multimeter, power supply, ammeter and voltmeter.
THEORETICAL INTRODUCTION
Heat capacity is a thermophysical parameter of substances, defined as the amount of heat required to heat a certain mass of a substance by one Kelvin. If the mass of a substance is equal to one kilogram, then the heat capacity is called specific heat capacity; if the mass is equal to one mole, then it is called molar heat capacity. By definition, the molar heat capacity is equal to
. (1)
Here ν = 
– amount of substance per mole, m- weight, M– mass of one mole, dQ– the amount of heat sufficient to raise the temperature by dT. For gases, unlike solids and liquids, the heat capacity depends on the type of thermodynamic heating process occurring with the gas. This is due to the fact that, according to the first law of thermodynamics
, (2)
heat is spent not only to increase internal energy dU, that is, by increasing the temperature, but also by changing the volume of gas. Unlike solids and liquids, the change in volume can be relatively large and depends on the type of thermodynamic process. Therefore, the amount of work done by pressure forces and the amount of heat required to heat the gas also depends on the type of process.
Let's consider heating an ideal gas. An ideal gas is a gas whose own volume of molecules is negligible compared to the volume of the container, and there is no potential energy for the interaction of molecules. Air under normal conditions can be considered an ideal gas.
During isochoric heating of a gas, there is no change in volume, there is no work, and the heat goes only to increase the internal energy, dQ =
dU. For an ideal gas, according to molecular kinetic theory, internal energy is the kinetic energy of molecules 
. Whence the molar heat capacity during isochoric heating of an ideal gas is equal to 
.
During isobaric heating of a gas under constant pressure conditions, an additional part of the heat is spent on the work of volume change 
. Therefore, the resulting amount of heat ( dQ =
dU +
dA) will be equal 
. Comparing with formula (1), we find that the molar heat capacity during isobaric heating
In heat capacity formulas R– universal gas constant, i – number of degrees of freedom of a gas molecule. This is the number of independent coordinates needed to determine the position of a molecule in space. Or it is the number of energy components that a molecule has. For example, for a monatomic molecule these are the components of kinetic energy during translational motion relative to three coordinate axes, i= 3. For a diatomic molecule, kinetic energies of rotational motion relative to two axes are added, since relative to the third, passing through both atoms, there is no moment of inertia and energy. As a result, a diatomic molecule has 5 degrees of freedom. The same is true for air, which consists mainly of diatomic molecules of oxygen and nitrogen.
Experimental The molar heat capacity of air is measured using a calorimeter. In a calorimeter, air is heated at a constant pressure equal to atmospheric pressure. The heating temperature is measured using a thermocouple connected to a multimeter. To increase the accuracy of measurements, a larger mass of air should be heated. Therefore, with the help of a compressor, air is passed in a continuous stream through the calorimeter (Fig. 1).
The calorimeter heater is connected to the power supply. Power consumption is determined by voltmeter and ammeter readings N= JU. When, after turning on the installation, thermal equilibrium occurs and the temperature of the air leaving the calorimeter stops changing, supplied from the electric heater thermal power N is spent on heating the air entering the calorimeter and partially on heat transfer q through the walls of the calorimeter. Therefore, the heat balance equation has the form
. (3)
Here m– second air flow through the calorimeter, D T– increase in air temperature after passing through the calorimeter.
To eliminate unknown heat loss power q it is necessary to conduct experiments at different air flow rates, but at the same temperature increase. In this case, the power of heat loss will be the same, because heat transfer through the walls is proportional to the temperature difference. According to equation (3), the thermal power supplied to the calorimeter, with a constant increase in air temperature Δ T, depends linearly on the second air flow, and therefore the graph is a straight line. The slope of the line is equal to 
. It can be determined experimentally from the graph as the ratio of the legs of a right triangle constructed on the experimental line, according to the coordinates of its vertices A And IN. Where do we get it from?
. (4)
COMPLETING OF THE WORK
1. Measure the air temperature in the laboratory with a thermometer. Connect the calorimeter to a 220 V network and set the compressor variable resistor to a relatively high air flow rate.
2. Set the variable heater resistor to such a power that after thermal equilibrium has been established (3 minutes), the temperature of the air leaving the calorimeter would increase by 30–50 K. Measure the air temperature and determine the air flow rate using the compressor resistor scale. Record the temperature rise, air flow, ammeter and voltmeter readings in the tables.
3. Reduce the air flow by approximately one-fifth of the initial one and simultaneously reduce the heater power so that the air temperature at the outlet of the calorimeter remains the same. This part of the work requires patience and smooth adjustments. Record the results of measurements of air flow, current and voltage in a table. Carry out the experiment at least five times over the entire air flow range.
Temperature increase D T, TO | ||||||
Air flow m, g/s | ||||||
Current strength I, A | ||||||
Voltage U, IN | ||||||
Power N= IU, W | ||||||
Turn off the power to the multimeters. Turn off the installation.
4. Make calculations. Determine the power consumed by the electric heater, N= I U. Write it down in the table.
5. Plot a graph of power consumption versus air flow N (m). The size of the chart is at least half a page. Apply a uniform scale on the coordinate axes. Draw a straight line around the points so that the sum of the deviations of the points is minimal.
6. Construct a right triangle on the experimental line as on the hypotenuse (Fig. 2). Determine vertex coordinates A And IN triangle. Using formula (4), calculate the average value of the molar heat capacity<C P>. Take the value of the mass of a mole of air equal to 28.9 10 -3 kg/mol.
7. Rate graphical method random error in measuring molar heat capacity. To do this, draw two close lines on the graph parallel to the experimental line so that all points except misses are between them. Determine the distance between lines σ N. Calculate using the formula
. (5)
8. Write the result in the form WITH R = < C P > ± d C P , P = 90%. Compare with the theoretical value calculated using formula (3), with R= 8.31 J/mol K, i = 5.
Draw conclusions.
CONTROL QUESTIONS
1. Define the molar heat capacity of a substance.
2. Formulate the first law of thermodynamics. Write down the formulas for heat, work, internal energy of an ideal gas.
3. Derive formulas for the molar heat capacity of an ideal gas under isochoric and isobaric heating.
4. Write down the heat balance equation for the calorimeter.
5. Explain why heat loss through the walls of the calorimeter does not affect the heat capacity measurement.
6. Explain why in the installation air must flow through the calorimeter in a continuous stream.
DETERMINATION OF THE ADIABATH INDICATOR
Goal of the work: get acquainted with the adiabatic process, determine the adiabatic index for air.
Equipment: cylinder with valve, compressor, pressure gauge.
THEORETICAL INTRODUCTION
An adiabatic process is a process occurring in a thermodynamic system without heat exchange with environment. A thermodynamic system is a system containing a huge number of particles. For example, a gas whose number of molecules is comparable to the Avagadro number 6.02∙10 23 1/mol. Although the movement of each particle obeys Newton's laws, there are so many of them that the state of the system is characterized by macroscopic parameters such as pressure P, volume V, temperature T.
According to the first law of thermodynamics, which is the law of conservation of energy in thermodynamic processes, heat Q, supplied to the system, is spent on doing work A and the change in internal energy Δ U
Q = A + U. (1)
When applied to an ideal gas, heat added to the gas results in a temperature change: 
, Where
=
m/
M– the amount of gas equal to the ratio of mass to mass of one mole, WITH− molar heat capacity, depending on the type of process. The internal energy of an ideal gas is the kinetic energy of all molecules, it is equal to 
, Where C v
– molar heat capacity during isochoric heating. The work of an elementary volume change by pressure forces is equal to the product of pressure and volume change:
dA= PdV.
For an adiabatic process occurring without heat exchange ( Q= 0), work is done due to changes in internal energy, A = − U. During adiabatic expansion, the work done by the gas is positive, so the internal energy and temperature decrease. When compressed, the opposite is true. All rapidly occurring processes can be fairly accurately considered adiabatic.
Let us derive the equation
adiabatic process of an ideal gas. To do this, we apply the equation of the first law of thermodynamics for an elementary adiabatic process dA= −
dU,
which
takes the form
RdV =−
WITH v dT. Let us apply another equation obtained by differentiating the Mendeleev–Clapeyron equation ( PV=ν
RT)
: PdV +
VdP =
R
dT.
Excluding one of the parameters, for example, temperature, we obtain the relationship for the other two parameters 
. Integrating and potentiating, we obtain the adiabatic equation in terms of pressure and volume: P
V
=
const.
Similarly for other pairs of parameters:
TV -1 = const, P -1 T -- = const. (2)
Here 
– adiabatic index, equal to the ratio of the heat capacities of the gas during isobaric and isochoric heating. Let us obtain a formula for the adiabatic exponent in molecular kinetic theory. Molar heat capacity by definition is the amount of heat required to heat one mole of a substance by one Kelvin. 
. During isochoric heating, heat is spent to increase internal energy 
. Substituting heat, we get 
. Then the adiabatic exponent can be determined theoretically by the formula
. (3)
Here i– number of degrees of freedom of gas molecules. This is the number of coordinates sufficient to determine the position of the molecule in space or the number of constituent energy components of the molecule. For example, for a monatomic molecule, kinetic energy can be represented as the sum of three energy components corresponding to motion along three coordinate axes, i= 3. For a rigid diatomic molecule, two more components of the energy of rotational motion should be added, since there is no rotational energy about the third axis passing through the atoms. So, for diatomic molecules i= 5. For air as a diatomic gas, the theoretical value of the adiabatic index will be equal to = 1.4.
The adiabatic exponent can be determined experimentally by the Clément–Desormes method. Air is pumped into the balloon, compressing it to a certain pressure. R 1, a little more atmospheric. When compressed, the air heats up slightly. After thermal equilibrium has been established, the cylinder is opened for a short time. In this expansion process 1–2 the pressure drops to atmospheric R 2 =P atm, and the mass of gas under study, which previously occupied part of the volume of the cylinder V 1, expands, occupying the entire cylinder V 2 (Fig. 1). The air expansion process (1−2) occurs quite quickly; it can be considered adiabatic, occurring according to equation (2)
. (4)
In the adiabatic expansion process, the air cools. After closing the valve, the cooled air in the cylinder through the walls of the cylinder is heated to laboratory temperature T 3 = T 1 . This is an isochoric process 2–3
. (5)
Solving equations (4) and (5) together, excluding temperatures, we obtain an equation relating pressures: 
, from which the adiabatic index should be determined γ
. The pressure sensor does not measure the absolute pressure, which is written in the process equations, but the excess pressure above atmospheric pressure. That is R 1 = Δ R 1 +R 2, and R 3 =
Δ R 3 +R 2. Moving on to excess pressures, we get 
. Excess pressures are small compared to atmospheric pressure R 2. Let us expand the terms of the equation into a series according to the relation 
. After reduction by R 2 we get for the adiabatic exponent calculation formula
. (6)
Laboratory The installation (Fig. 2) consists of a glass cylinder, which communicates with the atmosphere through the “Atmosphere” valve. Air is pumped into the cylinder by a compressor with tap “K” open. After pumping, to avoid air leakage, close the tap.
COMPLETING OF THE WORK
1. Connect the installation to a 220 V network.
Open the cylinder tap. Turn on the compressor, pump air to excess pressure in the range of 4–11 kPa. Close the cylinder tap. Wait 1.5–2 minutes, record the pressure value Δ R 1 to the table.
Δ R 1, kPa | |||||
Δ R 3, kPa | |||||
Repeat the experiment at least five times, changing the initial pressure in the range of 3–11 kPa.
Turn off the installation.
3. Make calculations. Determine the adiabatic index in each experiment using formula (6). Write it down in the table. Determine the average value of the adiabatic index<γ >
4. Estimate the random measurement error using the formula for direct measurements
. (7)
5. Write the result in the form: = . R= 0.9. Compare the result with the theoretical value of the adiabatic index of a diatomic gas theory = 1,4.
Draw conclusions.
CONTROL QUESTIONS
1. Give the definition of an adiabatic process. Write down the first law of thermodynamics for an adiabatic process. Explain the change in gas temperature during adiabatic processes of compression and expansion.
2. Derive the equation of the adiabatic process for the parameters pressure – volume.
3. Derive the equation of the adiabatic process for the parameters pressure – temperature.
4. Define the number of degrees of freedom of molecules. How does the internal energy of an ideal gas depend on the type of molecules?
5. How are processes carried out with air in the Clément – Desormes cycle, how do pressures and temperatures change in the processes?
6. Derive a calculation formula for the experimental determination of the adiabatic index.
Introductory session 3
Work 1. Study of impact of bodies 13
Work 2. Determining bullet speed using the ballistic method 18
Work 3. Study of the movement of bodies in a gravity field 22
Work 4. Study of the dynamics of rotational motion 27
Work 5. Determining the speed of a bullet using a torsion pendulum 32
Work 6. Determination of the moment of inertia of bodies 37
Work 7. Study of gyroscope precession 42
Work 8. Study of plane motion when rolling bodies 47
Work 9. Study of the plane motion of Maxwell's pendulum 52
Work 10. Study damped oscillations 57
Work 11. Study forced oscillations 62
Work 12. Study of the addition of vibrations 67
Work 13. Determination of the moment of inertia of a physical pendulum 71
Work 14. Determining the speed of sound in air 76
Work 15. Determination of the heat capacity of air 81
Work 16. Determination of the adiabatic index 86
Mechanics
Educational and methodological manual
for laboratory classes
Compiled by Shusharin Anatoly Vasilievich
Editor L. L. Shigorina
The work is intended...
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6.11. Physical pendulum
A rigid body of arbitrary shape that oscillates freely around a fixed horizontal axis that does not pass through its center of mass is called physical pendulum .
According to the definition, a physical pendulum when oscillating has one degree of freedom, i.e. is indeed a one-dimensional harmonic classical oscillator (Fig. 6.14, where point 0 is called the oscillation axis, and point 0 * - the center of swing of the physical pendulum, point C is the center of mass).
For harmonic vibrations, the angle of deviation from the equilibrium position q small and amounts to no more than three to five degrees, which allows in some cases to assume sin q » q (if angle qtake in radians, and not in degrees), and consider the vibrations themselves to be harmonic and isochronous , those. their period or frequency does not depend on the amplitude of the oscillation.
First, let's write the differential equation of oscillations of a physical pendulum. To do this, let's consider what forces act on it. The friction force at the suspension point 0 (axis Z) we do not take into account the physical pendulum. When oscillating, a physical pendulum is affected by the force of gravity G and the normal reaction of the support F (Fig. 6.14). To find the resultant force, we decompose the force of gravity into two mutually perpendicular forces: G ^ = mg sin q and G | | = mg cos q(Fig. 6.14). Then strength normal reaction the supports and the parallel component of gravity cancel each other out (Newton's third law). Therefore, the force that forces a physical pendulum to continue to perform harmonic oscillations remains the perpendicular component of gravity, which is often called the restoring force.
The same result can be obtained if you add the gravity vector and the normal support reaction force vector according to the parallelogram rule. (We invite the reader to perform this operation independently.)
From the dynamics of rotational motion(5.17 ) should, that in this case the physical pendulum (like any rigid body) is affected by the moment of force M relative to the Z axis, equal to the product moment of inertia of the body I on angular acceleration e relative to the same axis:
| M = I×e, |
(6.33) |
Where
| . |
(6.34) |
The moment of force M is equal to the product of the gravity component G ^ on the shoulder ℓ :
where sin q » q, as noted above. Let's substitute the values of expressions (6.34) and (6.35) into formula (6.33):
Thus, we obtained a homogeneous second-order differential equation characterizing the oscillations of a physical pendulum.
Its solution is the function q = q 0 сos (w t + j o), Where q 0 - amplitude value of the angle qdeviation of a pendulum from its equilibrium position during its oscillations.
It is not difficult to show that any motion of a rigid body (for example, the motion of an astronaut in training centrifuges, etc.) can be represented as a superposition of two simple types of motion: translational and rotational.
During translational motion, all points of the body receive, over equal periods of time, movements equal in magnitude and direction, as a result of which the velocities and accelerations of all points at each moment of time are the same.
During rotational motion, all points of a rigid body move in circles, the centers of which lie on the same straight line, called the axis of rotation. For rotational motion, you need to set the position in space of the rotation axis and the angular velocity of the body at each moment of time.
It is of interest to compare the basic quantities and formulas of the mechanics of a rotating rigid body and the translational motion of a material point. For the convenience of such a comparison, Table 1 shows on the left the values and basic relationships for translational motion, and on the right - similar ones for rotational motion.
Table 1
| Forward movement | Rotational movement |
| S- path - linear speed - linear acceleration m- body mass - body impulse - force Basic law of dynamics: Kinetic energy: - Job | - turn - angular velocity- angular acceleration J- moment of inertia - moment of impulse - moment of force Basic law of dynamics: Kinetic energy: - work |
The table shows that the transition in the relationships from translational motion to rotational motion is carried out by replacing speed with angular velocity, acceleration with angular acceleration, etc.
In this work, plane motion is considered, i.e. one in which, under the influence of external forces, all points of the body move in parallel planes. An example of plane motion is the rolling of a cylinder along a plane.
This movement can be represented as the sum of two movements - translational with speed and rotational with angular speed.
Having named the frame of reference relative to which we are considering complex movement rigid body, motionless, the movement of the body can be represented as rotation with angular velocity. In a reference system that moves relative to a stationary frame translationally with speed .
Thus, the acceleration of each point of the body is the sum of the acceleration of translational motion and the acceleration during rotation around an axis passing through the center of mass. The acceleration of translational motion is the same for all points of the body and is equal to
where is the moment of all external forces relative to the axis passing through the center of mass of the body,
- moment of inertia of the body relative to the same axis.
In this work, the plane motion of a body is studied using the example of the motion of a Maxwell pendulum.
Maxwell's pendulum consists of a flat metal rod - axis AB with disk C symmetrically fixed to it (Fig. 1). Attached to the ends of the axle are two threads that are pre-wound around the axle. The opposite ends of the threads are fixed to the upper bracket. The disc is lowered by gravity on the threads, which unwind to their full length. The disk, continuing its rotational movement in the same direction, winds the threads around the axis, as a result of which it rises up, while slowing down its rotation. Having reached the top point, the disk will go down again, etc. The disk will oscillate up and down, which is why such a device is called a pendulum. The essence of the work is to measure the moment of inertia of the pendulum and compare the results obtained with those theoretically calculated using known formulas.
Let's create an equation for the translational motion of the pendulum without taking into account the forces of friction with the air (see Fig. 1)
where is the radius of the axis;
Tension force of one thread.
Translational and rotational accelerations are related by the relation
From equations (4.3), (4.4), (4.5) and (4.6) we express the moment of inertia of the Maxwell pendulum:
where is the moment of inertia of the pendulum axis;
m o - axle mass;
Moment of inertia of the pendulum disk;
Outer radius of the disk;
m D - disk mass;
Moment of inertia of the replacement ring only;
Outer radius of the ring;
m k is the mass of the ring.
DESCRIPTION OF THE EXPERIMENTAL INSTALLATION
A general view of the installation is shown in Fig. 2.
Two brackets are attached to the vertical post of the base 1: the top 2 and the bottom 3. The top bracket is equipped with electromagnets and a device 4 for fastening and adjusting the bifilar suspension 5. The pendulum is a disk 6 mounted on an axis 7 suspended on the bifilar suspension. Replaceable rings 8 are attached to the disk. The pendulum with replaceable rings is fixed in the upper initial position using an electromagnet.
There is a millimeter scale on the vertical stand, which is used to determine the stroke of the pendulum.
Photoelectric sensor 9 is a separate assembly, fixed using bracket 3 at the bottom of the vertical stand. The bracket provides the ability to move the photo sensor along a vertical post and fix it in any position within the scale of 0 - 420 mm.
Photosensor 9 is designed to output electrical signals to physical millisecond watch 10. The millisecond watch is made as an independent device with a digital time display. It is rigidly fixed to base 1.
EXPERIMENTAL METHOD AND PROCESSING OF RESULTS
Exercise 1 . Determine the parameters of Maxwell's pendulum.
1. Draw a table. 1.
Table 1
| Pendulum axis | Pendulum disk | Rings | |||||||
| № | R o, m | L o, m | R D, m | L D, m | R k1, m | R k2, m | R k3, m | ||
| Average values | |||||||||
| V o = | m o = | V D = | m D = | ||||||
2. Using a caliper, measure R And L, calculate the volumes of the axle and disk V o and V D.
3. Using tabulated values of the density of the metal (aluminum) from which the axle and disk are made, calculate the mass values m o and m D. Enter the results obtained in the table. 1.
4. Measure the values with a caliper R k (for three rings) and enter into the table. 1. Determine the average values.
Task 2. Determine the moment of inertia of the pendulum
1. Draw a table. 2.
2. Using the scale, using the indicator of bracket 3, determine the stroke of the pendulum h.
table 2
| m k1 = kg; | h= m; | |||||||
| t, With | t Wed, s | |||||||
| m k 2 = kg; | ||||||||
| t, With | t Wed, s | |||||||
| m k 3 = kg; | ||||||||
| t, With | t Wed, s | |||||||
3. Press the “Network” button located on the front panel of the millisecond watch; the light of the photosensor and the digital indicators of the millisecond watch should light up.
4. While rotating the pendulum, fix it in the upper position using an electromagnet, while ensuring that the thread is wound onto the axis, turn to turn.
5. Press the “Reset” button to make sure that the indicators are set to zero.
6. When you press the “Start” button on the millisecond watch, the electromagnet should de-energize, the pendulum should begin to unwind, the millisecond watch should count down the time, and at the moment the pendulum crosses the optical axis of the photosensor, the counting of time should stop.
7. Carry out tests according to points 4 - 6 at least five times and determine the average time value t.
8. Determine the moment of inertia of the pendulum using formula (4.7).
9. Carry out tests according to points 4 - 6 for three replacement rings.
10. Enter all the results obtained into the table. Determine the average values.
12. Compare the theoretical values of the moment of inertia of the pendulum (4.8) with the experimental values.
1. What is called plane-parallel motion?
2. What two movements make up the complex movement of a pendulum? Describe them.
3. Prove that the pendulum moves with constant acceleration of the center of mass.
4. Define moment of inertia. Write down the expression for the moment of inertia of the disk or ring.
5. Formulate the law of conservation of mechanical energy. Write it down as applied to Maxwell's pendulum.
While gravity R, applied at the center of mass WITH, directed along the axis of the rod (Fig. 5.1, A), the system is in equilibrium. If you deflect the rod by a certain small angle (Fig. 5.1, b), then the center of mass WITH rises to a small height and the body acquires a reserve of potential energy. On the pendulum relative to the axis ABOUT, the direction of which we choose “towards us”, the moment of gravity will act, the projection of which on this axis is equal to
Where 
; L– distance between the axis of rotation ABOUT and center of mass WITH.
Torque M, created by force R, at small angles is equal to
It causes acceleration during the rotational movement of the pendulum. The relationship between this acceleration and the torque is given by the basic equation of the dynamics of rotational motion
, (5.2)
Where J– moment of inertia of the pendulum relative to the axis ABOUT.
Let's denote
Then from equation (5.2) we obtain
Equation (5.4) describes an oscillatory process with a cyclic frequency.
The period of oscillation is therefore equal to
From formula (5.5) we express the moment of inertia
If the position of the center of mass of the system does not change, then the value L is constant and can be introduced into formula (5.6) constant coefficient
. (5.7)
Measuring time t, during which it occurs n complete oscillations, we find the period . Substituting T And K in (5.6), we obtain the working formula
Using formula (5.8), indirect measurements are made of the moment of inertia of a physical pendulum relative to the axis ABOUT.
On the other hand, the moment of inertia J depends on the position of the weights on the rod. Let us move the weights along the rod so that they are located symmetrically relative to a certain point A. This mathematical point is chosen arbitrarily near the middle of the rod. The center of mass of the system retains its location. We will consider the size of the loads to be small compared to and (see Fig. 5.1). Then they can be considered as material points. In this case, the moment of inertia of the system is determined by the expression
where is the moment of inertia of the system without loads; x– distance of the load to the point A; l– point distance A to the axis of rotation of the pendulum ABOUT.
Transforming formula (5.9), we obtain
where is the moment of inertia of the pendulum when the loads are positioned at the point A.
We will check dependence (5.10) by obtaining the quantities J And J A experimentally using formula (5.8).
Assignment for work
1. When preparing for laboratory work, obtain a calculation formula for the error of indirect measurements D J moment of inertia (see Introduction). Please note that the moment of inertia is determined using the working formula (5.8). To simplify calculations, we can assume that the coefficient K measured exactly in this formula: D K= 0.
2. Prepare a sketch of the table. 1 for statistical processing of direct fivefold time measurements t(for a sample, see Introduction to Table B.1).
3. Prepare a sketch of the table. 2 for dependence research J from x 2 .
4. Turn on the electronic stopwatch. By pressing the “Mode” button, set mode No. 3 (the “Mode 3” indicator lights up), and the braking device holding the body will turn off.
5. When starting work, place both weights at the point A(its position is indicated in the table of initial data located in the Appendix and near the laboratory installation on which you will work).
6. Deflect the pendulum by hand at a small angle, and at the moment the pendulum is released, turn on the stopwatch by pressing the “Start” button. After counting 10 full swings of the pendulum, stop the stopwatch by pressing the “Stop” button. Record the obtained time in the measurement table.
7. Take five time measurements t ten complete oscillations of a physical pendulum without changing the position of the weights.
8. Calculate the average time and determine the confidence error of measurement D t.
9. Using the working formula (5.8), determine the value of the moment of inertia J A, and using the formula obtained in step 1 of this task, determine the measurement error of this value D J. Write the result in the form and enter it into the table. 2 for value.
10. Spread the weights symmetrically relative to the point A to a distance (see Fig. 5.1). It is recommended to take a distance equal to that the value that was used in the individual task. Take one-time time measurements t ten complete oscillations of a physical pendulum.
11. Repeat experiment step 7 at five different distances x.
12. Determine the moment of inertia of the pendulum using formula (5.8) at various distances x. Enter the results in the table. 2.
13. Plot a graph of the moment of inertia of the pendulum
from x 2, using table. 2. Plot the expected time on the same graph.
dependence (5.10). Compare and analyze the results obtained
tatov.
Control questions
1. What is the purpose of this work?
2. What is the moment of inertia of a body? What is its physical meaning?
3. Formulate and apply to this work the basic law of the dynamics of rotational motion.
4. What is the center of mass of the system?
5. Why does the location of the center of mass of the pendulum not change when the position of the weights changes?
6. Find the moment of inertia of the system relative to the center of mass by setting or measuring the quantities necessary for this.
7. Formulate the law of conservation of energy and write it down in relation to a physical pendulum.
8. How to obtain the working formula (5.8) and dependence (5.10)?
9. How to obtain a formula for calculating the error of indirect measurements of the moment of inertia?
10. How is Steiner’s theorem formulated? How can it be applied to the system under study?
11. Why is it proposed to plot the dependence of the moment of inertia on the square of the value x?
12. What is moment of force, angular velocity, angular acceleration, angular displacement, how are these vectors directed?
Individual tasks for team members,
performing laboratory work on one installation
| Crew member number | Individual task |
| Calculate the moment of inertia of a pendulum consisting of a drum and a spoke with weights attached to the spoke close to the point A | |
| Calculate the moment of inertia of a pendulum consisting of a drum and a spoke with weights attached to the spoke at a distance from the point A. Take the numerical values of the masses, dimensions of the drum and the spokes in the table of initial data placed in the Appendix or near the laboratory installation on which you will perform the experiments | |
| Perform a task similar to the task for the second number, but with a different distance from the point A |
Literature
Savelyev I.V. General physics course. – M.: Nauka, 1982. – T. 1 (and subsequent editions of this course).
Laboratory work No. 6
DETERMINATION OF THE ADIABATH INDICATOR
BY THE METHOD OF CLEMENT AND DEZORMES
Goal of the work - study of equilibrium thermodynamic processes and heat capacity of ideal gases, measurement of the adiabatic exponent using the classical method of Clément and Desormes.
OUTPUT OF THE CALCULATION FORMULA
A physical pendulum is a rigid body that, under the influence of gravity, oscillates around a fixed horizontal axis. ABOUT, not passing through the masstel center point WITH(Fig. 2.1).
If the pendulum is moved out of its equilibrium position by a certain angle j, then the gravity component is balanced by the reaction force of the axis ABOUT, and the component tends to return the pendulum to the equilibrium position. All forces are applied to the center of mass of the body. Wherein
. (2.1)
The minus sign means that the angular displacement j and restoring force have opposite directions. At sufficiently small angles of deflection of the pendulum from the equilibrium position sinj » j, That's why F t » -mgj. Since the pendulum, in the process of oscillation, performs a rotational motion relative to the axis ABOUT, then it can be described by the basic law of the dynamics of rotational motion
Where M- moment of power Ft relative to the axis ABOUT, I– moment of inertia of the pendulum relative to the axis ABOUT, is the angular acceleration of the pendulum.
The moment of force in this case is equal to
M = F t×l =–mgj×l, (2.3)
Where l– the distance between the suspension point and the center of mass of the pendulum.
Taking (2.2) into account, equation (2.3) can be written
(2.4)
Where .
The solution to the differential equation (2.5) is a function that allows you to determine the position of the pendulum at any time t,
j=j 0 × cos(w 0 t+a 0). (2.6)
From expression (2.6) it follows that for small oscillations the physical pendulum performs harmonic oscillations with an oscillation amplitude j 0, cyclic frequency , initial phase a 0 and period determined by the formula
Where L=I/(mg)– reduced length of a physical pendulum, i.e. the length of such a mathematical pendulum, the period of which coincides with the period of the physical pendulum. Formula (2.7) allows you to determine the moment of inertia of a rigid body relative to any axis if the period of oscillation of this body relative to this axis is measured. If the physical pendulum has the correct geometric shape and its mass is uniformly distributed throughout the entire volume, the corresponding expression for the moment of inertia can be substituted into formula (2.7) (Appendix 1).
The experiment examines a physical pendulum called negotiable and representing a body oscillating around axes located at different distances from the center of gravity of the body.
The reversible pendulum consists of a metal rod on which support prisms are fixedly mounted O 1 And O 2 and two moving lentils A And B, which can be fixed in a certain position using screws (Fig. 2.2).
A physical pendulum performs harmonic oscillations at small angles of deviation from the equilibrium position. The period of such oscillations is determined by relation (2.7)
,
Where I– moment of inertia of the pendulum relative to the axis of rotation, m– mass of the pendulum, d– distance from the suspension point to the center of mass, g– acceleration of gravity.
The physical pendulum used in the work has two supporting prisms O 1 And O 2 for hanging. Such a pendulum is called a reversible pendulum.
First, the pendulum is suspended on a bracket using a supporting prism O 1 and determine the period of oscillation T 1 relative to this axis:
(2.8)
Then the pendulum is suspended by a prism O 2 and T 2 is determined:
Thus, the moments of inertia I 1 And I 2 O 1 And O 2, will be respectively equal to and . Pendulum mass m and periods of oscillation T 1 And T 2 can be measured from high degree accuracy.
According to Steiner's theorem
Where I 0– moment of inertia of the pendulum relative to the axis passing through the center of gravity. Thus, the moment of inertia I 0 can be determined by knowing the moments of inertia I 1 And I 2.
PROCEDURE FOR PERFORMANCE OF THE WORK
1. Remove the pendulum from the bracket, place it on a triangular prism so that the distances from the support to the prisms O 1 And O 2 were not equal to each other. Moving the lentil along the rod, set the pendulum to the equilibrium position, then secure the lentil with a screw.
2. Measure the distance d 1 from the equilibrium point (center of mass WITH) to the prism O 1 And d 2– from WITH to the prism O 2.
3. Suspending the pendulum with a supporting prism O 1, determine the oscillation period, where N– number of oscillations (no more 50 ).
4. Similarly, determine the period of oscillation T 2 relative to the axis passing through the edge of the prism O 2 .
5. Calculate moments of inertia I 1 And I 2 relative to the axes passing through the supporting prisms O 1 And O 2, using formulas and , measuring the mass of the pendulum m and periods of oscillation T 1 And T 2. From formulas (2.10) and (2.11), determine the moment of inertia of the pendulum relative to the axis passing through the center of gravity (mass) I 0. From two experiments, find the average < I 0 > .
