The body rolls down the mountain as the values \u200b\u200bare related. Study of bodies rolling down an inclined plane
V. M. Zrazhevsky
LAB No.
ROLLING A RIGID BODY FROM AN INCLINED PLANE
Objective: Verification of the law of conservation of mechanical energy when a rigid body rolls down an inclined plane.
Equipment: inclined plane, electronic stopwatch, cylinders of different masses.
Theoretical information
Let the cylinder of radius R and weight m rolls down an inclined plane forming an angle α with the horizon (Fig. 1). There are three forces acting on the cylinder: gravity P = mg, force of normal plane pressure on the cylinder N and the friction force of the cylinder on the plane F tr. lying in this plane.
The cylinder participates simultaneously in two types of motion: translational motion of the center of mass O and rotational motion about an axis passing through the center of mass.
Since the cylinder remains on the plane during movement, the acceleration of the center of mass in the direction of the normal to the inclined plane is zero, therefore
P cosα − N = 0. (1)
Equation of dynamics forward movement along an inclined plane is determined by the friction force F tr. and the component of gravity along the inclined plane mg∙sinα:
ma = mg sinα − F tr. , (2)
where a is the acceleration of the center of gravity of the cylinder along the inclined plane.
Equation of dynamics rotary motion relative to the axis passing through the center of mass has the form
Iε = F tr. R, (3)
where I– moment of inertia, ε – angular acceleration. moment of gravity and about this axis is zero.
Equations (2) and (3) are always valid, regardless of whether the cylinder moves along the plane with slip or without slip. But three unknown quantities cannot be determined from these equations: F tr. , a and ε, one more additional condition is necessary.
If the friction force is of sufficient magnitude, then the cylinder will roll along the inclined plane without slipping. Then the points on the circumference of the cylinder must travel the same path length as the center of mass of the cylinder. In this case, linear acceleration a and angular acceleration ε are related by the relation
a = Rε. (four)
From equation (4) ε = a/R. After substitution into (3), we obtain
. (5)
Replacing in (2) F tr. on (5), we get
. (6)
From the last relation we determine the linear acceleration
. (7)
From equations (5) and (7), the friction force can be calculated:
. (8)
The friction force depends on the angle of inclination α, gravity P = mg and from relationship I/mR 2. Without friction, there will be no rolling.
When rolling without sliding, the static friction force plays a role. The rolling friction force, like the static friction force, has a maximum value equal to μ N. Then the conditions for rolling without sliding will be satisfied if
F tr. ≤μ N. (9)
Taking into account (1) and (8), we obtain
, (10)
or finally
. (11)
In the general case, the moment of inertia of homogeneous symmetrical bodies of revolution about an axis passing through the center of mass can be written as
I = kmr 2 , (12)
where k= 0.5 for a solid cylinder (disk); k= 1 for a hollow thin-walled cylinder (hoop); k= 0.4 for a solid ball.
After substituting (12) into (11), we obtain the final criterion for a rigid body to roll down an inclined plane without slipping:
. (13)
Since when a solid body rolls on a solid surface, the rolling friction force is small, the total mechanical energy of the rolling body is constant. At the initial moment of time, when the body is at the top of the inclined plane at a height h, its total mechanical energy is equal to potential:
W n = mgh = mgs∙sinα, (14)
where s is the path traveled by the center of mass.
The kinetic energy of a rolling body is the sum of the kinetic energy of the translational motion of the center of mass with a speed υ and rotational motion with speed ω about the axis passing through the center of mass:
. (15)
When rolling without sliding, the linear and angular velocities are related by the relation
υ = Rω. (16)
Let us transform the expression for the kinetic energy (15) by substituting (16) and (12) into it:
Movement on an inclined plane is uniformly accelerated:
. (18)
Let us transform (18) taking into account (4):
. (19)
Solving together (17) and (19), we obtain the final expression for the kinetic energy of a body rolling on an inclined plane:
. (20)
Description of the installation and measurement method
It is possible to study the rolling of a body along an inclined plane using the “plane” node and the SE1 electronic stopwatch, which are part of the MUK-M2 modular educational complex.
At
The rig is an inclined plane 1, which can be set with a screw 2 at different angles α to the horizon (Fig. 2). Angle α is measured using scale 3. Cylinder 4 with mass m. The use of two rollers of different weights is provided. The rollers are fixed at the top of the inclined plane with the help of an electromagnet 5, which is controlled by
electronic stopwatch SE1. The distance traveled by the cylinder is measured by a ruler 6 fixed along the plane. The roll time of the cylinder is measured automatically by sensor 7, which turns off the stopwatch at the moment the roller touches the finish point.
Work order
1. After loosening screw 2 (Fig. 2), set the plane at some angle α to the horizon. Place roller 4 on an inclined plane.
2. Switch the toggle switch for controlling the electromagnets of the mechanical unit to the “flat” position.
3. Set stopwatch SE1 to mode 1.
4. Press the start button of the stopwatch. Measure the roll time.
5. Repeat the experiment five times. Record the measurement results in table. one.
6. Calculate the mechanical energy before and after rolling. Make a conclusion.
7. Repeat steps 1-6 for other angles of inclination of the plane.
Table 1
t i, c |
(t i − <t>) 2 |
way s, m |
Tilt angle |
roller, kg |
W p, J |
W k, J |
|
t(a, n) |
<t> |
å( t i – <t>) 2 |
Δ s, m |
Δ m, kg |
|||
8. Repeat steps 1-7 for the second roller. Record the results in table. 2, similar to table. one.
9. Draw conclusions on all the results of the work.
test questions
1. Name the types of forces in mechanics.
2. Explain the physical nature of friction forces.
3. What is called the coefficient of friction? Its dimension?
4. What factors affect the value of the coefficient of friction of rest, sliding, rolling?
5. Describe the general nature of the motion of a rigid body during rolling.
6. How is the moment of friction force directed when rolling on an inclined plane?
7. Write down the system of equations of dynamics when the cylinder (ball) rolls on an inclined plane.
8. Derive formula (13).
9. Derive formula (20).
10. Ball and cylinder with the same masses m and equal radii R simultaneously begin to roll down an inclined plane from a height h. Will they hit bottom at the same time? h = 0)?
11. Explain the reason for the braking of a rolling body.
Bibliographic list
1. Savelyev, I. V. A course of general physics in 3 volumes. Vol. 1 / I. V. Savelyev. - M.: Nauka, 1989. - § 41-43.
2. Khaikin, S. E. Physical foundations of mechanics / S. E. Khaikin. - M: Nauka, 1971. - § 97.
3. Trofimova T. I. Course of physics / T. I. Trofimova. - M: Higher. school, 1990. - § 16–19.
A rigid body, that is, a movement in which the points of the body describe trajectories lying in parallel planes. An example of such a movement is the rotation of the wheel of a car when it is moving in a straight line. You can take any point 0 body and mentally draw an axis of rotation through it perpendicular to the planes in which the trajectories of the points of the body lie. Then the axis of rotation will move forward, remaining all the time parallel to itself.
Video 7.2. Plane motion of a rigid body in a uniform gravitational field. Flight of a flat cardboard figure
Accordingly, the speed of the elementary mass of a rigid body is the sum of the speed of the translational movement of the point 0 and the linear speed of rotation around the associated (mentally drawn) axis:
where is the radius vector that determines the position of the elementary mass with respect to the point 0 .
The kinetic energy of the elementary mass is then:
vector product
has a module equal to , where is the distance of the mass from the axis of rotation. Therefore, the third term in brackets is . The second term, which is a mixed product of vectors, does not change when the factors are cyclically rearranged:
As a result, we obtain the following expression for the kinetic energy of a solid body element
To find the kinetic energy of the body, we sum over all elementary masses:
Sum of elementary masses
is the mass of a rigid body. Expression
where is the radius vector of the center of mass of the body relative to the point 0 .
There is a moment of inertia of the body about the axis of rotation. Therefore, for the kinetic energy of a solid body, we can write the formula:
Since the choice of the mental axis of rotation is entirely in our power, we will simplify the resulting expression by taking as a point 0 center of mass of the body. Then = 0 and kinetic energy of a body in plane motion is equal to
Here is the speed of the center of mass, a is the moment of inertia about the axis passing through the center of mass and the orthogonal plane where the trajectories of the points of the body lie. Thus, the kinetic energy of a rigid body during plane motion is composed of the energy of translational motion with a speed equal to the velocity of the center of mass and the energy of rotation around an axis passing through the center of mass of the body.
The motion of a rigid body is determined by the external forces acting on the body and the moments of these forces
The index in the notation for the moment of external force means the projection of the moment onto the axis of rotation.
In the following examples, we are dealing with flat motion.
Video 7.3. Dependence of the behavior of cylinders on an inclined plane on the nature of the mass distribution over their volume
Example 1. A round homogeneous body (hoop, cylinder, ball) with radius and mass rolls down without slipping on an inclined plane at an angle to the horizon from a height (Fig. 7.12). The initial velocity of the body is zero. Find the speed of the center of mass of each body at the end of the descent.
Rice. 7.12. Rolling a body down an inclined plane
This problem can be considered in two ways.
1st way. According to the condition, the body rolls without slipping. This condition is used twice. The friction force between the body and the plane acts at the point of contact and, in the absence of sliding, does not exceed its maximum value:
where is the coefficient of sliding friction.
It is convenient to direct the coordinate axes as follows: X- along movement, axis at- perpendicular to the inclined plane. The body moves under the action of three forces: gravity, friction force and normal pressure force, so the equation for the translational motion of the body's center of inertia has the form:
Along axis at the body is not moving. Projecting the equation of motion of the center of mass onto the axis at, we obtain the relation for the force of normal pressure:
Projection of the equation of motion on the axis X gives:
Since the linear speed of the points of contact of the cylinder with the inclined plane is equal to zero (again, we use the condition of no slippage), the speed (acceleration) of translational motion is related to the angular velocity (angular acceleration) of the body by the usual relations:
In addition to translational motion, the body also rotates. Rotation is conveniently described with respect to the z axis passing through the center of mass of the cylinder.
This choice is due to the fact that the lines of action of the force of gravity and the force of normal pressure of the plane pass through the axis of rotation and, therefore, the moments of these forces are equal to zero. Thus, the cylinder rotates only under the action of the friction force, and the equation of rotational motion has the form:
Thus, a system of 4 equations is obtained that describes translational and rotational motion with an additional inequality expressing the law of friction. Solving the system of equations, we find:
The greater the moment of inertia about the axis passing through the center of mass, the lower the acceleration of the body. We have already received an answer to one of the questions of the problem: the ball will move faster than the cylinder, and the cylinder will move faster than the hoop. Substituting the solution for the friction force into the inequality expressing the law of friction, we find the condition under which there will be no slippage:
The meaning of this condition is simple: the slope should not be too steep.
So, center of gravity body moves along a plane with constant acceleration a, so that the dependence of the distance traveled and the speed on time has the form:
From this follows the relationship between speed and distance traveled:
By the end of the descent, the body passes the path
so that its speed reaches the value
Substituting here the moments of inertia of the hoop (), cylinder () and ball (), we find, respectively:
2nd way. We use the conservation law full energy. At the end of the descent, the body acquires kinetic energy
This kinetic energy is acquired at the expense of potential energy. From here it follows that the expression for the speed of the body at the end of the descent was found above. This method is much shorter, but it does not allow you to find out the details of the process: the forces acting on the body, etc.
In the example above, we assumed that we were dealing with a case where there was no slippage. This made it possible to assert simple connection() between the angular and linear velocities of the body and its radius. The static friction force was found in this case as a result of solving the equations of motion. In the case when the body moves with slippage, there is no known relationship between the linear and angular velocities. But we know the friction force in advance: since the point of contact of the body with the surface slides over the surface, the friction force is the sliding friction force, the modulus of which is related to the force of normal pressure by the Amonton-Coulomb law.
The forces of friction, as already mentioned, are directed so that prevent relative slippage bodies in contact. This possible slippage is often confused with the ongoing forward movement. It must be clearly understood that cases are not uncommon when the friction force does not slow down, but accelerates the body, that is, it is directed along its movement. The most famous example is starting a car. The wheels start spinning and skid backwards on the ground. Accordingly, the friction force is directed forward, and it is she who makes the car move. Let's take a look at an example to better understand these cases.
Example 2. A circus artist throws a hoop of mass and radius into the arena, which begins to roll in a horizontal direction with a speed (Fig. 7.13). In this case, the hoop is given a reverse rotation with an angular velocity . Let us find at what angular velocity the hoop will roll back towards the artist after stopping, as well as the final speed of the translational motion of the hoop.
Rice. 7.13. Reverse hoop movement
When the hoop rotates backwards, the point of contact with the arena moves forward due to both the rotation and translational motion of the hoop. Therefore, slippage inevitably exists and, therefore, the friction force reaches its maximum value. It slows down both forward movement and rotation of the hoop. It may happen that the forward movement of the hoop will be stopped at the moment when it still retains the reverse rotation. Further, the friction force will begin to accelerate the hoop towards the artist. This acceleration will stop when the tendency to slip disappears, after which the hoop will roll back uniformly with a certain steady speed. However, it may also happen that the reverse rotation is stopped earlier, and then the hoop will continue to move forward, changing the direction of rotation to forward. To distinguish between these two cases, qualitative reasoning is not enough, and we turn to formulas.
Let's direct the axis OH to the right (in the direction of the red arrow in Fig. 7.13), the axis of rotation OZ send on us(see the following example, where it is more convenient to direct this axis away from us, that is, behind the drawing), that is, in the direction of "reverse" rotation, the axis OY direct as usual, up. Let us represent the planar motion of the hoop as a superposition of its translational motion together with the center of mass (geometric center, since the hoop is assumed to be homogeneous). We project the linear and angular velocities onto the corresponding axes. Then, as long as the friction force is the sliding friction force and it is directed to the left, the equations of motion have the form
Equation (7.3.1) describes the motion of the center of mass of the hoop, and equation (7.3.2) describes its rotation around the axis passing through the center of mass in the reference frame in which it rests (center of mass system). In (7.3.2), it is taken into account that the moment of inertia of a homogeneous hoop about its axis of symmetry is equal to . After elementary integration, we get
The translational motion will stop, that is, it will become equal to zero, at the moment of time
The rotation will stop, that is, it will become equal to zero, at the moment of time
Their attitude
can be any due to the independence of the initial speeds of translational and rotational motions.
For further analysis, we introduce into consideration the speed of the lower point of the hoop - that point that touches the surface of the arena. We note already here that the condition for the disappearance of slippage is the vanishing of the velocity of this particular point, because the velocity of the corresponding point on the surface of the arena (the one that the hoop touches) is obviously equal to zero in our frame of reference, where the arena is stationary. The absence of slippage is the immobility of these two points relative to each other. With the chosen direction of the axes oz and OX, we have
If , then the translational motion of the hoop will stop first. At the moment of time, the speeds (7.3.3) and (7.3.8) will have the values
The lower point of the hoop, due to the continued rotation, will continue to slide relative to the arena to the right (to the right in Figure 7.13), the sliding friction force will retain its magnitude and direction to the left. Accordingly, the center of the hoop will begin to accelerate to the left, that is, it will become less than zero and begin to increase in absolute value, the counterclockwise rotation (in Figure 7.13) will continue to slow down. In other words, at , the hoop at time (7.3.5) begins to return to the artist who threw it.
As follows from (7.3.8), at the moment of time
the speed of the lower point of the hoop from (7.3.8) vanishes, the slippage stops, the sliding friction force is abruptly replaced by the static friction force equal to zero (we neglect the rolling friction force) and the hoop begins to roll towards the artist with a constant speed of the center of mass
rotating counterclockwise at a constant angular velocity
If , then the first, at the moment of time (7.3.6), the rotation of the hoop will stop. At the moment of time, the speed (7.3.8) of the lower point of the hoop will be equal to the speed of its center and be positive:
Sliding remains, the sliding friction force retains its magnitude and direction to the left, but the hoop under the action of this sliding friction force begins to rotate clockwise (recall: left, right, clockwise or counterclockwise - in Figure 10). As a result, the speed of the center of mass (the center of the hoop) will decrease, the rotation speed will increase, at the time
the slippage of the hoop will stop and the hoop will start moving away from the performer evenly with the center speed (7.3.10) and the angular speed of rotation (7.3.11). Recall that in this case , so that a
Thus, the answer to the question: "Will the hoop return or roll away?" is determined by the initial conditions, and more specifically by the value of the parameter , which has a simple physical meaning: this is the ratio of the modulus
the speed of any point of the hoop due to its translational movement together with the center of mass to the modulus of the velocity of the same point due to the rotation of the hoop around an axis passing through its center of mass at the initial moment of time.
DETERMINATION OF THE MOMENT OF INERTIA OF A BODY ROLLING FROM AN INCLINED PLANE
GOAL : acquire the skill of calculating the moment of inertia of bodies consisting of simple elements, determine the moment of inertia of the body relative to the instantaneous axis of rotation by the calculation and experimental method
EQUIPMENT : installation, body set, stopwatch
THEORETICAL PART
INSTALLATION DESCRIPTION
In this work, bodies are used, the axis of which is a cylindrical rod of radius r. One of the fig. 1) are placed on parallel guides 2, forming angles α1 and α2 with the horizon.
If the body is released, then it, rolling down, will reach the bottom point and, moving further by inertia, will rise up along the guides. The motion of a body, in which the trajectories of all points lie in parallel planes, is called flat. Plane motion can be represented in two ways: either as a set of translational motion of the body with the speed of the center of mass and rotation around an axis passing through the center of mass; or as just a rotational movement around an instantaneous axis of rotation (MOB), whose position is continuously changing. In our case, this instantaneous Z-axis passes through the points of contact of the guides with the moving rod.
DESCRIPTION OF THE MEASUREMENT METHOD
When rolling down the body, falling from a height goes the way l0, and rising by inertia to a height passes the path l. At the bottom point, the speed of the translational motion of the center of mass , and the angular velocity of the body
where t- the time of movement from the top point to the bottom, r is the radius of the rod (axis).
The rolling body is affected by the moment of resistance forces Mtr. Its work on the path l0 is equal to A = Mtrφ where the angular path φ = l0/r.
The law of conservation of energy on a segment of the path l0 has the form
, (2)
where J is the moment of inertia of the rolling body relative to MOB, m - the mass of the body, including the mass of the rod.
When the body moves down from a height h0 and rolls it to a height h, the work of the resistance forces on the way ( l + l0) is equal to the loss of potential energy
https://pandia.ru/text/80/147/images/image008_41.gif" width="146 height=48" height="48"> (4)
Here, the value (α1 and α2) is a constant for the given installation.
The moment of inertia of the body relative to MOB is determined by the Steiner theorem J = J0 + ma2, (5)
8. What functions are called integrals of motion?
9. List the additive integrals of motion.
10. How do you understand the following physical categories: “homogeneity of time”, “homogeneity of space”, “isotropy of space” and how do they relate to additive integrals of motion?
TEST QUESTIONS
1. What is the method for determining the moment of inertia of a body?
2. Specify possible systematic measurement errors.
3. Indicate the values of the kinetic and potential energy during the rolling of the body: at the beginning and at the end of the movement, at the bottom point and at an arbitrary point.
4. Describe the nature of the movement of the body along the guides. What force creates a moment about the axis of rotation?
5. How is the angular velocity ω measured in this paper?
6. What quantities are measured to determine the speed ω, the moment of friction forces, the work of friction forces?
7. What equations underlie the dynamic methods for determining the moment of inertia?
8. Specify possible sources of random errors in measurements.
9. A homogeneous cylinder of mass m and radius R rolls without slipping on a horizontal plane. The center of the cylinder moves with a speed υ0. Find an expression for determining the kinetic energy of a cylinder.
10. Calculate the angular momentum of the Earth, due to its movement around the axis. Compare this moment with the angular momentum due to the motion of the Earth around the Sun. Consider the Earth to be a uniform sphere, and the Earth's orbit to be a circle.
Sterlitamak
Study of bodies rolling down an inclined plane
Objective : to acquire some skills of independent research of physical phenomena and processing of the obtained results.
Equipment and accessories : inclined plane (tribometer), scale ruler, set of bodies, scales, stopwatch.
Exercise. Investigate the rolling of cylinders and a ball on an inclined plane.
Note: if a cylinder or ball rolls down an inclined plane at a slight angle to the horizontal, then the rolling occurs without slipping. If the angle of inclination of the plane exceeds a certain limit value, then the rolling will occur with slippage.
When performing the task, it is necessary to determine the limiting angle at which the rolling of the bodies will begin to occur with slippage. Based on the results of the study, draw up a report in which to reflect the research methodology, provide a table of observation results and give an explanation why, at an angle exceeding a certain value, the rolling of bodies occurs with slippage.
In addition, the problem includes determining the moment of inertia of the cylinders and the ball according to the results of observations of their rolling down an inclined plane.
Suppose a cylinder rolls down an inclined plane without slipping. External forces act on the cylinder: gravity, friction, and the reaction force from the side of the plane. We consider the motion as translational with a speed equal to the speed of the center of mass, and rotational about the axis passing through the center of mass.
Equation for the motion of the center of mass of the ball (cylinder)
or in scalar form in projections:
to the OX axis: .
on the y axis:
The equation of moments about the axis
With no slippage
Let us find the acceleration that the cylinder acquires under the action of the indicated forces. It can be found by using the expression for the kinetic energy of a rolling body
, | (1) |
where is the mass of the ball (cylinder), is the speed of the translational movement of the center of mass, is the moment of inertia of the ball, relative to the axis of rotation, is the angular velocity of rotation, relative to the axis of rotation.
The change in the kinetic energy of the body is equal to the work of external forces acting on the body. The elementary work of the force of friction and reaction, the plane is equal to zero, because their lines of action pass through the instantaneous axis of rotation (). Therefore, the change in the kinetic energy of the body occurs only due to the work of gravity
where is the final velocity of the center of mass at the end of the inclined plane, is the initial velocity, it is equal to zero, therefore
, | (6) |
where is the time of the body rolling down the inclined plane, is the radius of the ball (cylinder), is the mass of the ball (cylinder), is the angle of the plane to the horizon, is the length of the inclined plane.
By measuring the above quantities, the moment of inertia of the rolling cylinder can be calculated. It can be solid, hollow, with grooves on its generatrix, etc. Formula (9): is valid both for cylinders and for a ball.
The experiment with each of the bodies should be carried out at least three times. Record the results of observations and calculations in Table 1.
Table 1
No. p / p | Rolling body shape | Weight, kg | Radius, m | Inclined plane length (m) | Rolling time, s | Moment of inertia, kg m 2 | |||
Determine for each case the error in determining .
Determine the value of the moment of inertia for each body theoretically. Compare the value of the moment of inertia of the bodies determined theoretically and from the experiment, and in case of their discrepancy, explain the reason.
test questions
1. Define the moment of forces. Write in vector form. How is the moment of force directed relative to the force? What is the radius vector of force action? Draw and show.
2. What is the direction of angular acceleration, angular velocity?
3. Define the moment of inertia of a material point and an absolutely rigid body. The physical meaning of inertia.
4. Display the moment of inertia of the ball and cylinder.
5. Prove Steiner's theorem.
6. Formulate the law of conservation of energy during rotational motion.
7. Derive the formula for the day of calculation of kinetic energy, taking into account the rotation of the body.
8. Derive the law of conservation of the angular momentum of the system of bodies.
9. Define the center of mass of the thermal system.
10. Formulate the condition under which the body rolls without slipping and derive the formulas used in the calculation.
11. Formulate the laws of dynamics for rotational motion and derive them for a material point and for an absolutely rigid body.
12. Explain how the measurement error was calculated in the work.
MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION
BRANCH FGBOU VPO "UFA STATE AVIATION TECHNICAL UNIVERSITY"
IN THE CITY OF STERLITAMAK
Guidelines
to laboratory work on the course of general physics
section: section: "Mechanics. Mechanical vibrations. Statistical physics and thermodynamics»
LAB #9
Coefficient definition
internal friction of liquids
Sterlitamak
Objective: determine the coefficient of internal friction of an unknown fluid using the Stokes method.
Devices and equipment: glass cylinder with test liquid, stopwatch, balls different diameter, micrometer.
Brief theory
Any body moving in a viscous fluid is subject to a drag force. In the general case, the magnitude of this force depends on many factors: on the internal friction of the fluid, on the shape of the body, on the nature of the flow, etc.
The force of internal friction arising from macroscopic movements in a fluid is directly proportional to the velocity gradient. The coefficient of proportionality is called the coefficient of internal friction, or simply the viscosity of the fluid. Viscosity (or dynamic viscosity) is numerically equal to the force of internal friction acting on the unit area of the interface of parallel moving fluid layers, when the speed of their movement decreases by one when moving in the direction perpendicular to the boundary, per unit length, i.e. ~ at .
. | (1) |
Law (1) was obtained by Newton from the analysis of experimental data and became the basis for studying the motion of a viscous liquid and gas.
Consider, for example, the uniform motion of a small ball of radius in a fluid.
Let's denote the speed of the ball relative to the liquid as .
The distribution of velocities in neighboring layers of liquid entrained by the ball should have the form shown in Figure 1. In the immediate vicinity of the surface of the ball, this velocity is equal to , and as it moves away, it decreases and practically becomes equal to zero at some distance from the surface. It is obvious that the larger the radius of the ball, the more the mass of liquid is involved in its movement, and should be proportional to
The surface of the ball and the total friction force experienced by the moving ball is
Formula (5) is called the Stokes law.
The Stokes formula is applicable only in the case of bodies of sufficiently small sizes and low speeds of their movement. At high speeds around moving bodies, complex vortex motions of the fluid arise, and the drag force increases in proportion to the square of the speed, and not to its first power.
The role of friction is characterized by a dimensionless quantity called the Reynolds number:
, |
where are the linear dimensions characteristic of the fluid flow under consideration. In the case of fluid flow through a pipe - the radius of the pipe, - the average speed. The ratio is called the kinematic viscosity coefficient.
In order to explain the role of the Reynolds number, consider the volume element of the liquid with edge length . The kinetic energy of this volume is:
The frictional force acting on a fluid volume element is proportional to its surface, viscosity coefficient, and velocity gradient. Assuming that the velocity drops to zero at a distance equal in order of magnitude (in the case of a flow through a pipe, in the radial direction), we obtain that the velocity gradient is equal to . So the force of friction
The role of friction in fluid flow is small if the work is small compared to the kinetic energy of the fluid volume, that is, if the inequality
,
But - Re is the Reynolds number.
Thus, the role of friction forces in fluid flow is small at high Reynolds numbers.
Consider the free fall of a ball in a viscous fluid. There are 3 forces acting on the ball: gravity, Archimedean force, resistance force, which depends on speed. Let's find the equation of motion of a ball in a liquid. According to Newton's second law
where is the volume of the ball; - its density; is the density of the liquid; - acceleration of gravity.
Solving this equation, we find
. | (9) |
As can be seen from (7), the speed of the ball is exponentially approaching the steady speed . The establishment of the speed is determined by the quantity , which has the dimension of time and is called the relaxation time. If the fall time is several times longer than the relaxation time, the process of velocity establishment can be considered completed.
By measuring experimentally the steady velocity of the ball falling and the values , it is possible to determine the coefficient of internal friction of the liquid by the formula
, | (10) |
following from (8).
Note: balls with different radii move in a liquid with equal velocities and different relaxation times. If in the entire range of encountered velocities and relaxation times, the values calculated by formula (10) turn out to be the same, then formula (5) correctly conveys the dependence of forces on the radius of the ball. Dependence or independence from serves as a sensitive indicator of the correctness of the theory and the reliability of the experiment. It makes sense to process the results of the experiment only if the value does not show a systematic dependence on . If such a dependence is observed, then most often this is due to the influence of the walls of the vessel.
3. What characterizes the Reynolds number?
4. Laminar and turbulent flow and their connection with the Reynolds number.
5. What are the limits of applicability of Stokes' law?
6. What methods for determining the friction force exist?
7. How to explain the mechanism of the phenomenon of viscous friction?
8. What physical quantities does friction depend on?
9. What energy transformations occur when bodies move, taking into account the friction force?
10. What is the magnitude of the static friction force, sliding?
11. Tell us about sliding friction, rest friction, viscous friction and rolling friction.
12. Why is sliding friction greater than rolling friction?
13. Why is viscous friction less than sliding friction?
14. How does friction manifest itself in nature? When does it play a positive or negative role? How to get rid of friction?
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