In order to illustrate the application of the laws of rigid body dynamics, let's solve the problem of a cylinder rolling down an inclined plane (Fig. 10.5).
Solid Mass Cylinder m and radius R rolls down an inclined plane without slipping. The angle of inclination of the plane is a, and the height H (H » R). The initial speed of the cylinder is zero. Let's determine the rolling time - T and the velocity of the center of mass of the cylinder at the base of the inclined plane.
When the cylinder is rolling, three forces act on it: the force of gravity, the elastic reaction force of the support, and the force of friction rest(after all, rolling without slipping!).
Let us represent this movement as the sum of two movements: translational with a speed V C , with which the axis of the cylinder moves, and rotation around the axis of the cylinder with an angular velocity w.
Rice. 10.5
This relationship between the speeds of translational and rotational motions follows from the condition "motion without slipping".
Differentiating equation (10.9) with respect to time, we obtain the ratio of the angular and linear accelerations of the cylinder:
I.e .
Using the theorem on the motion of the center of mass point, we describe the translational motion of the cylinder:
To describe the rotation, we use the basic equation of the dynamics of rotational motion:
M C= I C×e. (10.11)
Projecting equation (10.10) onto the directions of the axes x And y, we obtain two scalar equations:
x: mg Sina- F tr = ma C; (10.12)
y: N – mg cosa = 0. (10.13)
Let us now turn to equation (10.11). Of the three named forces, the moment relative to the axis of the cylinder is created only by the friction force:
The moment of inertia of a solid cylinder about its axis is (see lecture No. 9):
Considering all this, we can rewrite equation (10.11) as follows:
Solving equations (10.12) and (10.14) together, we obtain the following values of unknown quantities:
From equation (10.15) it follows that with an increase in the angle of inclination a, the static friction force should also increase F tr. But, as you know, its growth is limited by the limit value:
Since the static friction force (10.15) cannot exceed the limit value (10.17), the inequality must be satisfied:
⅓mg Sina ≤ m mg cosa.
It follows from this that rolling will occur without slipping until the angle a exceeds the value a before:
a before = arctg3m.
Here m is the coefficient of friction of the cylinder along the plane.
The linear acceleration of the cylinder (10.16) is a constant value, therefore, the translational motion of the cylinder is uniformly accelerated. With such a movement without an initial velocity, the cylinder will reach the base of the inclined plane in the time:
Here: l= - plane length;
a=, (see 10.16).
So the rolling time is:
Calculate the final speed forward movement cylinder axis:
Note that this problem can be solved more simply by using the law of conservation of mechanical energy.
True, there is a friction force in the system, but its work is zero, since the point of application of this force during the descent remains motionless: after all, the movement occurs without slipping. Since there is no work done by the friction force, the mechanical energy of the system does not change.
Consider the energy of the cylinder at the initial moment - at a height h and at the end of the descent. The total energy of the cylinder in these positions is the same:
Recall that and . Then the equation of the law of conservation of energy can be rewritten as follows:
From here we can easily find the final speed of the cylinder:
which brilliantly confirms our earlier result (10.19).
Lecture 11 "Elements of fluid mechanics"
Lecture plan
1. Fluid pressure. The laws of hydrostatics.
2. Stationary fluid flow. Flow continuity equation.
3. Basic law of dynamics for an ideal fluid. Bernoulli equation.
4. Application of the Bernoulli equation for solving problems of hydrodynamics.
4.1. Leakage of fluid from a vessel.
4.2. Manometric flowmeter.
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.
The task. 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. physical meaning 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
to laboratory work on the course of general physics
section: section: "Mechanics. Mechanical vibrations. Statistical physics and thermodynamics»
LAB #9
Coefficient definition
internal friction 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 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 uniform motion a small ball of radius in the liquid.
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 with different times relaxation. 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 viscous friction less 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?
1) Trofimova T.I. Physics course: tutorial for engineering and technical specialties of universities - M .: Academia, 2006.
2) Aleksandrov I.V. and others. Modern physics [Electronic resource]: a textbook for students of all forms of education studying in technical and technological areas and specialties - Ufa: USATU, 2008.
3) Grinkrug M.S., Vakulyuk A.A. Laboratory workshop in physics [Electronic resource] - St. Petersburg: Lan, 2012.
4) Kalashnikov N. P. Fundamentals of Physics: a textbook for universities: in 2 volumes / N. P. Kalashnikov, M. A. Smondyrev - M .: Bustard, 2007.
The sliding speed of the point of contact of the body with the surface is obviously equal to the difference between the linear speed of the points on the surface of a round body and the speed of the translational motion of the body:
By the time
the sliding speed becomes equal to zero and the pure rolling mode sets in.
In the pure rolling mode, the equality
The length of the rolling stabilization section is
The amount of heat released can be determined using the law of conservation of energy or by calculating the work of the moment of friction force:
Note that the speed of steady translational motion u c and the amount of heat released Q do not depend on the value of the coefficient of sliding friction m .
Task 44. A round body rolls along an inclined plane without slipping. The plane is inclined to the horizontal at an angle a. Neglecting rolling friction and air resistance, determine the acceleration of the rolling body. At what values of the coefficient of friction m is it possible to roll without slipping?
Solution. Consider the energy solution. Since the body rolls without slipping, and air resistance and rolling friction can be neglected, its mechanical energy is conserved when the body moves. At first, the body is at rest, and its mechanical energy is equal to the potential energy mgh, and after rolling, the mechanical energy is equal to the sum of the kinetic energy of translational motion and the kinetic energy of rotation:
where J c \u003d CmR 2- the moment of inertia of a round body about the axis passing through the center of inertia, m- body weight (С - form parameter), u c is the speed of the body's center of inertia, w is the angular velocity of rotation of the body ( R is the radius of a round body).
Since the body is rolling without slipping, the speed of the center of inertia u c and angular velocity of rotation of the body w related by the ratio u c =wR(see problem 43).
From the equation (1) for the speed of the body's center of inertia u c after rolling we find
where h- the height from which the body rolls.
The body's center of inertia moves with acceleration
because S=h/sina, u with (0)=0.
The question of the value of the coefficient of friction m energetically unresolved.
Consider a dynamic solution to the problem. The body is affected by gravity, reaction force and static friction force (see figure). Under the action of these forces, the body rotates and moves forward according to the equations of dynamics:
Excluding from the system of equations (4) And (5) friction force, taking into account that u c =wR And J c \u003d CmR 2, we obtain a formula for calculating the acceleration of the body's center of inertia (3) .
Consider the question of estimating the value of the friction coefficient m .
Express from the system of equations (4) And (5) friction force
The static friction force is limited to the maximum value
F tr max =mN=mmgsina.
From the condition F tr £F tr max we obtain a relation that limits the value of the friction coefficient
Rolling without slip for set value friction coefficient m possible for tilt angles a, satisfying the condition
Task 45. Round body with radius r rolls without slipping on an inclined plane, which smoothly passes into a cylindrical surface with a radius R. From what minimum height must a body be rolled so that it can overcome an obstacle in the form of a “dead loop”? Ignore air resistance and rolling friction.
Solution. The speed of the center of inertia of a round body at a point BUT
(see problem 44).
Movement along the inner surface of the cylinder is described by a system of dynamic equations:
where Jc=Cmr2 is the moment of inertia of a round body about its own axis of rotation ( m- body mass, FROM– form parameter).
To the equations (1)-(3) it is necessary to add a relation that relates the speed of the translational motion of the body and the angular speed of rotation in the absence of slip:
From the equations (1) And (3) taking into account (4) we find the equation for the speed of the translational motion of the body
We integrate the last equation (see Problem 32), taking into account that u=uA at j=0.
Let us analyze the physical situation in critical point B. The body must reach point B and not break away from it.
From the basic equation of dynamics (2) for point B
It can be seen that the reaction force NB determined by the speed of translational motion at that point. The body does not come off at a point IN, if N B >0. The minimum speed of the body at a point IN, at which it does not break away from a given point, we estimate by setting NB=0:
From the formula (5) for the minimum height of the descent, we obtain
The same result can be obtained from energy considerations (check it out).
Task 46. The edge profile of the horizontal table is rounded into a semicircle with a radius R. A round body of radius r rolls on a table without slipping at a speed u 0. Neglecting rolling friction and air resistance, determine the place of body separation from the table surface and the speed of the body at the moment of separation.
Solution. When the body moves along the horizontal surface of the table, the speed of translational motion and the angular speed of rotation w=u 0 /r do not change.
The movement of the body along the rounding is described by the equations of dynamics:
where Jc=Cmr2, u c =wr(see problems 44, 45).
Solving the system of equations (1) – (3) taking into account the initial condition u=u 0 at j=0, find
Translational body speed u c increases with increasing polar angle j, and the reaction force N decreases. At the point of break N=0. From here we obtain the relation for determining the polar angle corresponding to the separation point:
The speed of the translational motion of the body at the moment of separation is equal to
The resulting expressions for j otp And u otp contain many special cases (check it out).
It is of interest to check the assumption about rolling without slipping. Such rolling is possible if the static friction force does not exceed the maximum static friction force at any point:
Do this analysis yourself.
Task 47. A person whose mass m 1 \u003d 65 kg, goes from the edge of the rotating platform to its middle. Considering the platform as a homogeneous circle, and the person as a material point, evaluate the changes in the kinetic energy of the system. The mass and radius of the platform are respectively equal m 2 \u003d 210 kg, R=2.1m. The initial angular velocity of rotation of the system is w 0 \u003d 2.3 rad / s
Solution. Question: “Will the kinetic energy of the system change?”
For the approximations indicated in the condition of the problem, the platform-man system can be considered isolated. Therefore, when a person moves to the center of the platform, the angular momentum of the system relative to the axis of rotation of the platform will be preserved:
where J 0 \u003d J (1 + 2m 1 / m 2), J=0.5m2R2 is the moment of inertia of the platform about its own axis of rotation, w- angular speed of rotation of the system after the transition of the person to the center of the platform.
The kinetic energy of the system is not conserved in this case. To conserve mechanical energy, it is not enough to require the system to be isolated. The system of interacting bodies must also be conservative.
Is our system conservative? A person has the ability to move relative to the platform only due to the presence of friction force. The static friction force allows the human muscle energy to be converted into rotational kinetic energy. Our system is non-conservative. The kinetic energy of the system increases due to human bioenergy. When moving to the center of the platform, a person “unwinds” the platform due to the force of rest energy.
The increment of kinetic energy can be found by calculating the work associated with the transition of a person to the center of the platform, or as a difference in the kinetic energies of the system:
It is taken into account here that L=L 0 = J 0 w 0 .
Note that a person can move on the platform if the friction coefficient satisfies the condition m³w 0 2 R/g.
Problem 48. At the edge of the rotating platform is a washer with mass m 1 \u003d 0.21 kg. An inextensible thread is tied to the washer at one end, the other end of which is passed through a small hole in the center of the platform. With the help of a thread, the puck is moved to the center of the platform. The coefficient of friction between the washer and the platform is equal to m=0.4. Estimate the work spent on moving the washer, neglecting its size, friction in the axis of the platform and air resistance. The radius and mass of the platform are respectively equal R=0.57m, m 2 \u003d 5.6 kg.
Solution. Although the system under consideration is not isolated, nevertheless, the law of conservation of angular momentum can be applied to it, since the moment of the thread tension about the axis of rotation is zero. Therefore, one can write
where J 0 \u003d J (1 + 2 m 1 / m 2), J \u003d 0.5m 2 R 2.
In this case, the kinetic energy of the system increases by the value (see problem 47)
The work of moving the washer is equal to the sum of the increment of the kinetic energy of the system and the work of the friction force:
A=DK+mmgR=23.5J.
The same work can be calculated directly from the work formula:
where F=mm 1 g+ m 1 w 2 x is the force applied to the thread ( is the angular velocity of rotation of the system).
Problem 49. A person walks along the edge of a round platform and returns to the starting point. Considering the person as a point, and the platform as a uniform disk, estimate the angle through which the platform will turn. The mass of the person and the platform, respectively, are equal m 1 \u003d 75 kg, m 2 \u003d 150 kg. Neglect friction in the platform axis and air resistance.
Solution. When a person moves along the edge of the platform, the platform itself with the person will rotate relative to the Earth in the opposite direction of the person's movement relative to the platform. For simplicity, let's assume that a person moves along the edge of the platform uniformly with an angular velocity w/ relative to the platform. In this case, the platform will rotate relative to the Earth with an angular velocity w pl, and a person - with an angular velocity, equal to the sum
To determine the angle of rotation of the platform, we use the law of conservation of angular momentum:
Jw+J pl w pl =0,
where J=m 1 R 2, J pl \u003d 0.5m 2 R 2 are the moments of inertia of the person and the platform.
Whence for the angular velocity of rotation of the platform we obtain
Multiplying the last relation by the time of movement, for the angle of rotation of the platform we find
The sign "-" indicates that the platform turns in the opposite direction of the movement of a person along the edge of the platform.
The angle of rotation does not depend on the nature of the movement of a person along the edge of the platform.
Task 50. Homogeneous rod with mass m=250g and length l=1.2m hung from one end. A small body of mass m 0 \u003d 120g moving horizontally at a speed u 0 \u003d 4.2 m / s, collides in such a way that the rod deviates to the maximum possible angle after the collision. Determine the point of collision of the body with the rod (the distance from the point of suspension to the point of collision) and the angle of deflection of the rod, assuming that the collision is absolutely elastic, neglecting air resistance and friction in the axis.
Solution. Let's use the law of conservation of angular momentum and mechanical energy
where J=(1/3)ml 2 is the moment of inertia of the rod relative to the suspension point, P 0 \u003d m 0 u 0, P=m 0 u are the impulses of the body before and after the collision, L is the angular momentum of the rod after the collision.
From the equations (1) And (2) for the unknown P And L find
As you can see, the values P and L depend on the coordinate of the collision site. Function L(x) has a maximum. From the extremum condition, we obtain
The maximum angular momentum that the rod receives upon collision is equal to
In this case, the momentum of the body after the collision is zero (make sure of this).
The deflection of a rod in the uniform gravity field of the Earth after a collision can be estimated by solving a dynamic problem or using the law of conservation of mechanical energy.
To calculate the deflection angle of the rod, the following relation is obtained:
After calculations, we get x m =1.0 m, j=730 .
Tasks for independent solution
(Translational and rotational motion of a rigid body)
1 . Flywheel, the mass of which m=5.2kg distributed along the rim, freely rotates around a horizontal axis passing through its center, with a frequency 720rpm. When braking, the flywheel stops after 20s. Determine the braking torque if the radius of the flywheel is 36cm (2.5Nm).
2 . For a uniform cylinder of mass 5.1kg an inextensible thread is wound, to the end of which a weight of mass is attached 0.25kg. At the point in time t=0 the system was in motion. Determine the kinetic energy of the entire system at the time 3.3s (2J).
3 . An inextensible thread is wound around a fixed block, to the end of which is attached a load of mass 1.7kg. Determine with what acceleration the load will fall if the mass of the block is equal to 2.2kg. The block is considered to be a homogeneous disk. Neglect air resistance and friction in the axis of the block (6.1m/s 2).
4 . An inextensible thread is wound on a fixed block, to the ends of which weights are attached with masses 1.6kg And 1.2kg. Determine the kinetic energy of the system through 1.8s after the start of the movement. Block weight 3.2kg. The block is considered to be a homogeneous disk. The thread does not slip on the block. Neglect air resistance and axle friction (15J).
5 . On a fixed block whose mass is equal to 25kg, wrapped rope. A monkey is hanging on a rope, which is trying to climb up it. With what acceleration does the rope move if the monkey remains at the same height from the floor all the time? Monkey mass 5.0kg. Friction in the axis of the block and the mass of the rope can be neglected (4.0m/s 2).
6 . The system of bodies (see figure) moves with acceleration 1.4m/s 2, masses of cargo m 2 \u003d 2.3 kg, m b \u003d 1.6 kg, coefficient of friction m=0.2. The thread is inextensible and does not slip on the block. Neglecting air resistance and friction in the axis of the block, determine the mass m 1. Consider a block as a homogeneous disk (1.0kg).
7 . The coupled system consists of three bodies (see figure): a fixed block of mass m 2 \u003d 1.8 kg, moving block mass m 3 \u003d 2.0 kg and a load of mass m 1 \u003d 1.5 kg. Determine the acceleration with which the load falls if the thread is inextensible and does not slip over the blocks (1.6m/s 2).
8 . Hockey puck spun to angular velocity 31rad/s and laid flat on ice. Determine the braking time of the puck if the mass and radius of the puck are respectively equal 0.21kg And 3.2cm. The coefficient of friction between the puck and ice is 0.13 (0.57s).
9 . A ball rotating around its own axis with a frequency 10r/s, placed on a horizontal surface. Determine the angular velocity of the ball and the fraction of its initial kinetic energy that is converted into heat ( 18 rad/s, 71%).
10 . Hollow thin-walled cylinder rotating at angular velocity 15rad/s placed on a horizontal surface. How long does it take the cylinder to cover the distance 5.7m if its radius is 12cm, and the coefficient of friction between the cylinder and the horizontal surface is equal to 0,25 ( 6,6c).
11 . The horizontal surface smoothly turns into a flat hill with an angle of inclination a=250 to the horizon. Homogeneous cylinder rotating at angular velocity 45rad/s, placed on a horizontal surface away from the foot of the slide. Determine the height to which the cylinder will roll if the coefficient of friction between the cylinder and the surface is everywhere equal to 0,2 . The radius of the cylinder is 13cm(29cm).
12 . A uniform ball descends an inclined plane from a height 1.5m. The angle of inclination of the plane to the horizon is equal to 33 0 . The coefficient of friction between the ball and the plane everywhere, including the horizontal surface, is equal to 0,15 . Determine the steady speed of the ball rolling on a horizontal surface if rolling friction and air resistance can be neglected ( 4.5m/s).
13 . A homogeneous cylinder moves along an inclined plane from a certain height without initial velocity. The plane is inclined to the horizontal at an angle 26 0 . The coefficient of friction between the body and the plane is equal to 0,1 . Determine the ratio of the kinetic energy at the end of the descent to the initial value of the potential energy of the body. Ignore rolling friction and air resistance. (0,9) .
14 . From what minimum height must a ball with radius be rolled? r=1.1cm, so that he can overcome the barrier in the form of a "dead loop" with a radius R=13cm? The ball rolls without slipping. Neglect air resistance and rolling friction 33cm).
15. A hollow thin-walled cylinder rolls on a horizontal surface, which smoothly turns into a cylindrical one, without slipping. At what minimum translational speed will the cylinder roll over the cylindrical surface without falling out if the radius of the cylindrical surface is 41cm, and the radius of the hollow cylinder 2.0cm. Ignore rolling friction and air resistance. (3.4m/s).
16 . The inclined plane smoothly passes into a cylindrical surface with a radius R=1.2m. A ball rolls without slipping down an inclined plane from a height 2.5m no initial speed. Determine the height of the point of separation of the ball from the surface of the cylinder. The radius of the ball is 0.15m. Ignore rolling friction and air resistance. (1.9m).
17 . A disk rolls without slipping on an inclined plane with an angle of inclination to the horizontal 27 0 , smoothly turning into a cylindrical surface with a radius of curvature 25cm. Determine the minimum height from which the disk must be rolled in order for it to break away from the surface at the transition line of the inclined plane to the cylindrical surface. The disk radius is 5cm (0.2m).
18 . A ball rolls without slipping from the top of a sphere of radius 0.50m with initial speed 1.0m/s. Determine the polar angle corresponding to the place where the ball leaves the spherical surface if air resistance and rolling friction can be neglected. ball radius 10cm (49 0).
19 . The ball rolls without sliding along an inclined plane, which smoothly passes into a cylindrical surface with a radius R=1.5m. ball radius r=11cm. The ball rolls down from a height h=2.9m no initial speed. Determine the coordinate of the point of separation of the ball from the surface of the cylinder (polar angle)( 130 0).
20 . A horizontally flying body hits a homogeneous rod suspended at one of its ends and sticks to it. Determine the angle by which the rod deviates from the vertical position. The length and mass of the rod are respectively equal 0.51cm, 980 g. Body weight 12g. The distance from the suspension point to the line of motion of the body is 34cm. The speed of the body before the collision 30m/s (15 0).
21 . Ball mass 2.1kg suspended on a light rod. A horizontally flying bullet of mass 9.0g and gets stuck in the middle of the ball. Determine the speed of the bullet if the system deviated from the equilibrium position by an angle 40 0 .The length of the rod and the radius of the ball are respectively equal 6.5cm, 35cm. Neglect air resistance and friction in the suspension axis (520 m/s).
22 . The period of rotation of the Sun around its own axis is 27 earthly days. The sun is a hydrogen star. After the hydrogen is completely burnt out, the Sun will experience a gravitational collapse. Estimate the radius of the Sun before it shatters into pieces. mass of the sun 2.0×10 30 kg, radius of the Sun 7.0×10 8 (14km).
Examples of problem solving
(oscillatory movement)
Problem 51. The maximum oscillation frequency of a physical pendulum with a mass m=2.3kg is equal to n max =1.3Hz. Determine the moment of inertia of the pendulum about the axis passing through its center of inertia.
Solution. The pendulum performs a rotational oscillatory motion relative to the swing axis under the action of the moment of gravity
where x \u003d OS, J 0 \u003d J c + mx 2, Jc- the moment of inertia of the pendulum about the axis passing through the center of inertia FROM, m is the mass of the pendulum.
In this approximation, we neglect air resistance and friction in the swing axis of the pendulum.
At small deflection angles, the pendulum performs a harmonic oscillatory motion with an angular frequency
Dependence of the angular frequency on the position of the swing axis w(x) has a maximum at
The maximum angular frequency is
Where do we find
A physical pendulum is used to measure free fall acceleration.
Problem 52. On the inner surface of a cylinder with a radius R a round body rolls without slipping. Determine the period of small oscillations of the body around the equilibrium position. The radius of a round body is r. Ignore air resistance and rolling friction.
Solution. Consider a dynamic solution to the problem.
The translational and rotational motions of a body under the action of gravity, reaction and friction (see figure) are described by the basic equations of the dynamics of a rigid body
F tr -mgsin , (1)
F tr × r=J c , (2)
where J c = cmr 2 is the moment of inertia of a round body about its own axis of rotation.
From the equations (1) And (2) , given that u c = wr(no slippage), to accelerate the translational motion of the body, we obtain the following equation:
Whence at small angles for displacement S=(R-r) find
So the offset S(t) is described by a harmonic function with an angular frequency
and oscillation period
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)
The equation for the dynamics of translational motion 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.
The equation for the dynamics of rotational motion about an axis passing through the center of mass has the form
Iε = F tr. R, (3)
where I is the moment of inertia, ε is the 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. (4)
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 training 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.
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 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. flat 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.
