By special agreement with the editorial board and editors of the journal “Kvant”
When solving mechanical problems, the use of the concept of the center of mass of a system of material points can provide invaluable assistance. Some problems simply cannot be solved without resorting to this concept; solving others with its help can become much simpler and more clear.
Before discussing specific problems, let us recall the basic properties of the center of mass and illustrate them with examples.
The center of mass (center of inertia) of a system of material points is a point that characterizes the distribution of masses in the system, the coordinates of which are determined by the formulas
Here m i- masses of material points forming the system, x i, y i, z i- coordinates of these points. Readers familiar with the concept of a radius vector will prefer the vector notation:
(1)
Example 1. Let us find the position of the center of mass, the simplest system consisting of two points whose masses m 1 and m 2 and the distance between them l(Fig. 1).
Directing the axis X from the first point to the second, we find that the distance from the first point to the center of mass (i.e., the coordinate of the center of mass) is equal to and the distance from the center of mass to the second point is equal to i.e. the ratio of distances is inverse to the ratio of masses. This means that in this case the position of the center of mass coincides with the center of gravity.
Let us discuss some properties of the center of mass, which, it seems to us, will fill the somewhat formal definition of this concept given above with physical content.
1) The position of the center of mass will not change if some part of the system is replaced by one point with a mass equal to the mass of this subsystem and located at its center of mass.
Example 2. Let's consider a flat homogeneous triangle and find the position of its center of mass. Divide the triangle into thin strips parallel to one of the sides, and replace each strip with a point located in its middle. Since all such points lie on the median of the triangle, the center of mass must also lie on the median. Repeating the reasoning for each side, we find that the center of mass is at the intersection of the medians.
2) The speed of the center of mass can be found by taking the time derivative of both sides of equality (1):
(2)
Where - system impulse, m- total mass of the system. It can be seen that the speed of the center of mass of the closed system is constant. This means that if we associate a translationally moving reference frame with the center of mass, then it will be inertial.
Example 3. Let us place a uniform rod of length l vertically onto a smooth plane (Fig. 2) and release. During the fall, both the horizontal component of its momentum and the horizontal component of the velocity of the center of mass will remain equal to zero. Therefore, at the moment of the fall, the center of the rod will be in the place where the rod originally stood, and the ends of the rod will shift horizontally by .
3) The acceleration of the center of mass is equal to the derivative of its speed with respect to time:
(3)
where on the right side of the equality there are only external forces, since all internal forces cancel according to Newton’s third law. We find that the center of mass moves as an imaginary point with a mass equal to the mass of the system would move under the action of the resulting external force. This is probably the most physical property of the center of mass.
Example 4. If you throw a stick, causing it to rotate, then the center of mass of the stick (its middle) will move with constant acceleration along a parabola (Fig. 3).
4) Let the system of points be in a uniform gravitational field. Then the total moment of gravity relative to any axis passing through the center of mass is equal to zero. This means that the resultant of gravity passes through the center of mass, i.e. the center of mass is also the center of gravity.
5) The potential energy of a system of points in a uniform gravitational field is calculated by the formula
Where h ts - height of the system's center of mass.
Example 5. When digging a hole in a uniform pound deep h and scattering of soil over the surface, its potential energy increases by , where m- mass of excavated soil.
6) And one more useful property of the center of mass. The kinetic energy of a system of points can be represented as the sum of two terms: the kinetic energy of the general translational motion of the system, equal to , and the kinetic energy E relative to the motion relative to the reference system associated with the center of mass:
Example 6. The kinetic energy of a hoop rolling without slipping on a horizontal surface with a speed υ is equal to
since the relative motion in this case is pure rotation, for which the linear speed of the points of the hoop is equal to υ (the total speed of the lower point must be equal to zero).
Now let's start analyzing problems using the center of mass.
Problem 1. A homogeneous rod lies on a smooth horizontal surface. Two horizontal forces of equal magnitude but opposite in direction are applied to the rod: one force is applied to the middle of the rod, the other to its end (Fig. 4). Relative to what point will the rod begin to rotate?
At first glance, it may seem that the axis of rotation will be the point lying in the middle between the points of application of forces. However, equation (3) shows that since the sum of the external forces is zero, the acceleration of the center of mass is also zero. This means that the center of the rod will remain at rest, i.e. serve as an axis of rotation.
Problem 2. Thin uniform rod length l and mass m set in motion along a smooth horizontal surface so that it moves translationally and simultaneously rotates with angular velocity ω. Find the tension of the rod depending on the distance x to its center.
Let us move to the inertial reference system associated with the center of the rod. Let us consider the movement of a piece of a rod enclosed between the point of the rod under consideration (located at a distance x from the center) and its end (Fig. 5).
The only external force for this piece is the required tension force F n, the mass is equal to , and its center of mass moves in a circle of radius 
with acceleration. Writing down the equation of motion of the center of mass of the selected piece, we obtain
Problem 3. A binary star consists of two component stars with masses m 1 and m 2, the distance between which does not change and remains equal L. Find the rotation period of the binary star.
Let us consider the motion of the component stars in an inertial frame of reference associated with the center of mass of the binary star. In this reference frame, stars move with the same angular velocity along circles of different radii (Fig. 6).
Radius of rotation of a star with mass m 1 is equal (see Example 1), and its centripetal acceleration is created by the force of attraction towards another star:
We see that the rotation period of a double star is equal to
and is determined by the total mass of the binary star, regardless of how it is distributed among the component stars.
Problem 4. Two point masses m and 2 m tied with a weightless thread length l and move along a smooth horizontal plane. At some point in time the speed of mass 2 m is equal to zero, and the mass velocity m equal to υ and directed perpendicular to the thread (Fig. 7). Find the thread tension and the rotation period of the system.
Rice. 7
The center of mass of the system is at a distance from mass 2 m and moves with speed. In the reference system associated with the center of mass, a point of mass 2 m moves in a circle of radius with speed . This means that the rotation period is equal to (check that the same answer is obtained if we consider a point with mass m). We find the thread tension from the equation of motion of any of the two points:
Problem 5. Two identical blocks of mass m each connected by a light spring stiffness k(Fig. 8). The first bar is given a speed υ 0 in the direction from the second bar. Describe the motion of the system. How long will it take for the spring deformation to reach its maximum value for the first time?
The center of mass of the system will move at a constant speed. In the reference frame of the center of mass, the initial speed of each block is , and the stiffness of the half spring that connects it to the stationary center of mass is 2 k(the spring stiffness is inversely proportional to its length). The period of such oscillations is equal to
and the amplitude of vibration of each bar, which can be found from the law of conservation of energy, is
For the first time, the deformation will become maximum after a quarter of the period, i.e. after a while .
Problem 6. Ball mass m impinges with speed v on a stationary ball of mass 2 m. Find the velocities of both balls after the elastic central impact.
In the reference frame associated with the center of mass, the total momentum of the two balls is zero both before and after the collision. It is easy to guess which answer for final velocities satisfies both this condition and the law of conservation of energy: the velocities will remain the same in magnitude as before the impact, but will change their directions to the opposite. The speed of the center of mass of the system is equal to . In the center of mass system, the first ball moves with speed , and the second ball moves towards the first with speed . After the impact, the balls will fly away at the same speeds. It remains to return to the original frame of reference. Applying the law of addition of velocities, we find that the final speed of a ball with mass m equal and directed backward, and the speed of the previously at rest ball of mass 2 m equal and directed forward.
Note that in the center of mass system it is obvious that upon impact the relative speed of the balls does not change in magnitude, but changes in direction. And since the difference in speeds does not change when moving to another inertial reference system, we can assume that we have derived this important relation for the original reference system:
υ 1 – υ 2 = u 1 – u 2 ,
where the letter υ is used to denote initial velocities, and u- for the final ones. This equation can be solved together with the law of conservation of momentum instead of the law of conservation of energy (where velocities come in to the second power).
Problem 7. It is known that during an elastic off-center impact of two identical balls, one of which was at rest before the impact, the angle of expansion is 90°. Prove this statement.
In the center of mass system, an off-center impact can be described as follows. Before the impact, the balls approach with equal impulses; after the impact, they fly apart with impulses of the same magnitude, but in opposite directions, and the line of expansion rotates at a certain angle relative to the line of approach. To go back to the initial frame of reference, each final velocity must be added (vectorally!) with the velocity of the center of mass. In the case of identical balls, the speed of the center of mass is equal to , where υ is the speed of the incident ball, and in the reference frame of the center of mass, the balls approach and fly apart at the same speeds. The fact that after adding each final velocity to the velocity of the center of mass, mutually perpendicular vectors are obtained can be seen from Figure 9. Or you can simply check that the scalar product of vectors and vanishes due to the fact that the modules of the vectors are equal to each other.
Exercises
1. Rod of mass m and length l hinged at one end. The rod was deflected at a certain angle from the vertical position and released. At the moment of passing the vertical position, the speed of the lower point is equal to υ. Find the tension at the midpoint of the rod at this point in time.
2. Rod of mass m and length l rotate in a horizontal plane with angular velocity ω around one of its ends. Find the relationship between the tension of the rod and the distance x to the axis of rotation, if a small weight of mass is attached to the other end M.
3. Find the period of oscillation for the system described in problem 5 of the article, but for bars of different masses m 1 and m 2 .
4. Derive the known general formulas for the elastic central impact of two balls, using the transition to the center of mass reference frame.
5. Ball of mass m 1 collides with a ball at rest of lesser mass m 2. Find the maximum possible angle of deflection of the incoming ball during an elastic off-center impact.
1. 
2. 
3. 
Center of mass Equation of motion of the center of mass. The law itself: Bodies act on each other with forces of the same nature directed along the same straight line, equal in magnitude and opposite in direction: The center of mass is a geometric point characterizing the movement of a body or a system of particles as a whole. Definition The position of the center of mass of the center of inertia in classical mechanics is defined as follows: where the radius vector of the center of mass is the radius vector of the ith point of the system and the mass of the ith point.
7.Newton's third law. Center of mass Equation of motion of the center of mass.
Newton's third lawstates: the action force is equal in magnitude and opposite in direction to the reaction force.
The law itself:
Bodies act on each other with forces of the same nature, directed along the same straight line, equal in magnitude and opposite in direction:
Center of mass this is a geometric point characterizing movement body or system of particles as a whole.
Definition
The position of the center of mass (center of inertia) in classical mechanics is determined as follows:
where radius vector of the center of mass, radius vector i th point of the system,
mass of the i-th point.
.
This is the equation of motion of the center of mass of a system of material points with a mass equal to the mass of the entire system, to which the sum of all external forces is applied (the main vector of external forces) or the theorem on the motion of the center of mass.
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The basic law of dynamics can be written in a different form, knowing the concept of the center of mass of the system:
It's there equation of motion of the center of mass of the system, one of the most important equations of mechanics. It states that the center of mass of any system of particles moves as if the entire mass of the system were concentrated at that point and all external forces were applied to it.
The acceleration of the center of mass of the system is completely independent of the points of application of external forces.
If , then , then and is the case of a closed system in an inertial reference frame. Thus, if the center of mass of a system moves uniformly and in a straight line, this means that its momentum is conserved during the movement.
Example: a homogeneous cylinder of mass and radius rolls down an inclined plane making an angle with the horizontal without slipping. Find the equation of motion?
| |
The joint solution gives the values of the parameters
The equation of motion of the center of mass coincides with the basic equation of the dynamics of a material point and is its generalization to a system of particles: the acceleration of the system as a whole is proportional to the resultant of all external forces and inversely proportional to the mass of the system.
A reference system rigidly connected to the center of mass, which moves translationally relative to the ISO, is called the center of mass system. Its peculiarity is that the total momentum of the particle system in it is always equal to zero, as .
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Kinematics of translational motion
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All topics in this section:
Mechanical movement
Matter, as is known, exists in two forms: in the form of substance and field. The first type includes atoms and molecules from which all bodies are built. The second type includes all types of fields: gravity
Space and time
All bodies exist and move in space and time. These concepts are fundamental to all natural sciences. Any body has dimensions, i.e. its spatial extent
Reference system
To unambiguously determine the position of a body at an arbitrary moment in time, it is necessary to select a reference system - a coordinate system equipped with a clock and rigidly connected to an absolutely rigid body, according to
Kinematic equations of motion
When t.M moves, its coordinates change with time, therefore, to specify the law of motion, it is necessary to indicate the type of function
Movement, elementary movement
Let point M move from A to B along a curved path AB. At the initial moment its radius vector is equal to
Acceleration. Normal and tangential acceleration
The movement of a point is also characterized by acceleration—the rate of change in speed. If the speed of a point for an arbitrary time
Forward movement
The simplest type of mechanical motion of a rigid body is translational motion, in which a straight line connecting any two points of the body moves with the body, remaining parallel | its
Law of Inertia
Classical mechanics is based on Newton’s three laws, formulated by him in his essay “Mathematical Principles of Natural Philosophy,” published in 1687. These laws were the result of a genius
Inertial reference frame
It is known that mechanical motion is relative and its nature depends on the choice of reference system. Newton's first law does not hold true in all frames of reference. For example, bodies lying on a smooth surface
Weight. Newton's second law
The main task of dynamics is to determine the characteristics of the motion of bodies under the influence of forces applied to them. It is known from experience that under the influence of force
The basic law of the dynamics of a material point
The equation describes the change in the motion of a body of finite dimensions under the influence of force in the absence of deformation and if it
Newton's third law
Observations and experiments indicate that the mechanical action of one body on another is always an interaction. If body 2 acts on body 1, then body 1 necessarily counteracts those
Galilean transformations
They make it possible to determine kinematic quantities during the transition from one inertial reference system to another. Let's take
Galileo's principle of relativity
Acceleration of any point in all reference systems moving relative to each other rectilinearly and uniformly in the same way:
Conservation quantities
Any body or system of bodies is a collection of material points or particles. The state of such a system at some point in time in mechanics is determined by specifying coordinates and velocities in
Center of mass
In any system of particles you can find a point called the center of mass
Conservative forces
If at each point in space a force acts on a particle placed there, the particle is said to be in a field of forces, for example, in the field of gravity, gravitational, Coulomb and other forces. Field
Central forces
Every force field is caused by the action of a specific body or system of bodies. The force acting on the particle in this field is about
Potential energy of a particle in a force field
The fact that the work of a conservative force (for a stationary field) depends only on the initial and final positions of the particle in the field allows us to introduce the important physical concept of potential
Relationship between potential energy and force for a conservative field
The interaction of a particle with surrounding bodies can be described in two ways: using the concept of force or using the concept of potential energy. The first method is more general, because it also applies to forces
Kinetic energy of a particle in a force field
Let a particle of mass move in force
Total mechanical energy of a particle
It is known that the increment in the kinetic energy of a particle when moving in a force field is equal to the elementary work of all forces acting on the particle:
Law of conservation of particle mechanical energy
It follows from the expression that in a stationary field of conservative forces the total mechanical energy of a particle can change
Kinematics
You can rotate your body through a certain angle
Momentum of a particle. Moment of power
In addition to energy and momentum, there is another physical quantity with which the law of conservation is associated - this is angular momentum. The angular momentum of the particle
Moment of impulse and moment of force about the axis
Let us take an arbitrary fixed axis in the reference system of interest to us
Law of conservation of angular momentum of a system
Let us consider a system consisting of two interacting particles, which are also acted upon by external forces and
Thus, the angular momentum of a closed system of particles remains constant and does not change with time
This is true for any point in the inertial reference system: . Moments of impulse of individual parts of the system m
Moment of inertia of a rigid body
Consider a solid body that can
Equation of dynamics of rigid body rotation
The equation for the dynamics of rotation of a rigid body can be obtained by writing the equation of moments for a rigid body rotating around an arbitrary axis
Kinetic energy of a rotating body
Let us consider an absolutely rigid body rotating around a fixed axis passing through it. Let's break it down into particles with small volumes and masses
Work of rotation of a rigid body
If a body is rotated by force
Centrifugal force of inertia
Let's consider a disk that rotates together with a ball on a spring put on a spoke, Fig. 5.3. The ball is located
Coriolis force
When a body moves relative to a rotating CO, in addition, another force appears - the Coriolis force or Coriolis force
Small fluctuations
Consider a mechanical system whose position can be determined using a single quantity, such as x. In this case, the system is said to have one degree of freedom. The value of x can be
Harmonic vibrations
The equation of Newton's 2nd Law in the absence of friction forces for a quasi-elastic force of the form has the form:
Math pendulum
This is a material point suspended on an inextensible thread of length, oscillating in a vertical plane
Physical pendulum
This is a solid body that vibrates around a fixed axis connected to the body. The axis is perpendicular to the figure and
Damped oscillations
In a real oscillatory system there are resistance forces, the action of which leads to a decrease in the potential energy of the system, and the oscillations will be damped. In the simplest case
Self-oscillations
With damped oscillations, the energy of the system gradually decreases and the oscillations stop. In order to make them undamped, it is necessary to replenish the energy of the system from the outside at certain moments
Forced vibrations
If the oscillatory system, in addition to resistance forces, is subject to the action of an external periodic force that changes according to the harmonic law
Resonance
The curve of the dependence of the amplitude of forced oscillations on leads to the fact that at some specific for a given system
Wave propagation in an elastic medium
If a source of oscillation is placed in any place in an elastic medium (solid, liquid, gaseous), then due to the interaction between particles the oscillation will propagate in the medium from particle to hour
Equation of plane and spherical waves
The wave equation expresses the dependence of the displacement of an oscillating particle on its coordinates,
Wave equation
The wave equation is a solution to a differential equation called the wave equation. To establish it, we find the second partial derivatives with respect to time and coordinates from the equation
The center of mass of the system is the point with the radius vector
For a continuous distribution of mass with density 
. If the gravitational forces applied to each particle of the system are directed one way, then the center of mass coincides with the center of gravity. But if 
not parallel, then the center of mass and the center of gravity do not coincide.
Taking the time derivative of 
, we get:
those. the total momentum of the system is equal to the product of its mass and the speed of the center of mass.
Substituting this expression into the law of change in total momentum, we find:
The center of mass of the system moves like a particle in which the entire mass of the system is concentrated and to which the resulting mass is applied external strength
At progressive In motion, all points of a rigid body move in the same way as the center of mass (along the same trajectories), therefore, to describe translational motion, it is enough to write down and solve the equation of motion of the center of mass.
Because 
, then the center of mass closed system must maintain a state of rest or uniform linear motion, i.e. 
=const. But at the same time, the entire system can rotate, fly apart, explode, etc. as a result of action internal forces.
Jet propulsion. Meshchersky equation
Reactive called the movement of a body in which it occurs accession or discarding masses. During the process of movement, a change in the mass of the body occurs: during the time dt, a body of mass m attaches (absorbs) or rejects (emits) mass dm with a speed 
relative to the body; in the first case dm>0, in the second dm<0.
Let's consider this movement using the example of a rocket. Let us move to the inertial reference frame K", which at a given moment of time t is moving at the same speed 
, the same as a rocket - this is called an ISO accompanying– in this frame of reference the rocket is currently t rests(rocket speed in this system 
=0). If the sum of external forces acting on the rocket is not equal to zero, then the equation of motion of the rocket in the K system, but since all ISOs are equivalent, then in the K system the equation will have the same form:
This - Meshchersky equation, describing the movement any body with variable mass).
In the equation, mass m is a variable quantity, and it cannot be included under the derivative sign. The second term on the right side of the equation is called reactive force
For a rocket, the reactive force plays the role of a traction force, but in the case of adding mass dm/dt>0, the reactive force will also be a braking force (for example, when a rocket is moving in a cloud of cosmic dust).
Energy of a particle system
The energy of a particle system consists of kinetic and potential. The kinetic energy of a system is the sum of the kinetic energies of all particles in the system
and is, according to definition, the quantity additive(like impulse).
The situation is different with the potential energy of the system. Firstly, interaction forces act between the particles of the system 
. ThereforeA ij =-dU ij, where U ij is the potential energy of interaction between the i-th and j-th particles. Summing U ij over all particles of the system, we find the so-called own potential energy systems:
It is essential that the system's own potential energy depends only on its configuration. Moreover, this quantity is not additive.
Secondly, each particle of the system, generally speaking, is also affected by external forces. If these forces are conservative, then their work will be equal to the decrease in external potential energy A=-dU ext, where
where U i is the potential energy of the i-th particle in an external field. It depends on the positions of all particles in the external field and is additive.
Thus, the total mechanical energy of a particle system located in an external potential field is defined as
E syst =K syst +U int +U ext
Lesson "Center of Mass"
Schedule: 2 lessons
Target: Introduce students to the concept of “center of mass” and its properties.
Equipment: figures made of cardboard or plywood, tumbler, penknife, pencils.
Lesson Plan
Lesson stages time methods and techniques
I Introduction to students 10 frontal survey, students’ work at the blackboard.
to the lesson problem
II. Learning something new 15-20 Teacher’s story, problem solving,
material: 10 experimental task
III Practicing new 10 student messages
material: 10-15 problem solving,
15 frontal poll
IV. Conclusions. Homework 5-10 Oral summary of the material by the teacher.
task Writing on the board
During the classes.
I Repetition 1. Frontal survey: shoulder of force, moment of force, equilibrium condition, types of equilibrium
Epigraph: The center of gravity of each body is a certain point located inside it - such that if you mentally hang the body from it, then it remains at rest and maintains its original position.
II. Explanationnew material
Let a body or system of bodies be given. Let us mentally divide the body into arbitrarily small parts with masses m1, m2, m3... Each of these parts can be considered as a material point. The position in space of the i-th material point with mass mi is determined by the radius vector ri(Fig. 1.1). The mass of a body is the sum of the masses of its individual parts: m = ∑ mi.
The center of mass of a body (system of bodies) is such a point C, the radius vector of which is determined by the formula
r= 1/m∙∑mi ri
It can be shown that the position of the center of mass relative to the body does not depend on the choice of the origin O, i.e. The definition of the center of mass given above is unambiguous and correct.
The center of mass of homogeneous symmetrical bodies is located in their geometric center or on the axis of symmetry; the center of mass of a flat body in the form of an arbitrary triangle is located at the intersection of its medians.
The solution of the problem
PROBLEM 1. Homogeneous balls with masses m1 = 3 kg, m2 = 2 kg, m3 = 6 kg, and m4 = 3 kg are attached to a light rod (Fig. 1.2). Distance between the centers of any nearby balls
a = 10 cm. Find the position of the center of gravity and the center of mass of the structure.
SOLUTION. The position of the center of gravity of the structure relative to the balls does not depend on the orientation of the rod in space. To solve the problem, it is convenient to place the rod horizontally, as shown in Figure 2. Let the center of gravity be on the rod at a distance L from the center of the left ball, i.e. from t. A. At the center of gravity, the resultant of all gravitational forces is applied and its moment relative to axis A is equal to the sum of the moments of gravity of the balls. We have r = (m1 + m2 + m3 + m4) g ,
R L = m2gα + m 3 g 2 a + m 4 g 3 a.
Hence L=α (m1 +2m3 + 3m4)/ (m1 + m2 + m3 + m4) ≈ 16.4 cm
ANSWER. The center of gravity coincides with the center of mass and is located at point C at a distance L = 16.4 cm from the center of the left ball.
It turns out that the center of mass of a body (or system of bodies) has a number of remarkable properties. In dynamics it is shown that the momentum of an arbitrarily moving body is equal to the product of the mass of the body and the speed of its center of mass and that the center of mass moves as if all external forces acting on the body were applied at the center of mass, and the mass of the entire body was concentrated in him.
The center of gravity of a body located in the gravitational field of the Earth is called the point of application of the resultant of all gravity forces acting on all parts of the body. This resultant is called the force of gravity acting on the body. The force of gravity applied at the center of gravity of the body has the same effect on the body as the forces of gravity acting on individual parts of the body.
An interesting case is when the size of the body is much smaller than the size of the Earth. Then we can assume that parallel gravity forces act on all parts of the body, i.e. the body is in a uniform gravitational field. Parallel and identically directed forces always have a resultant force, which can be proven. But at a certain position of the body in space, it is possible to indicate only the line of action of the resultant of all parallel forces of gravity; the point of its application will remain undetermined for now, because for a solid body, any force can be transferred along the line of its action. What about the application point?
It can be shown that for any position of the body in a uniform field of gravity, the line of action of the resultant of all gravitational forces acting on individual parts of the body passes through the same point, motionless relative to the body. At this point the equal force is applied, and the point itself will be the center of gravity of the body.
The position of the center of gravity relative to the body depends only on the shape of the body and the distribution of mass in the body and does not depend on the position of the body in a uniform field of gravity. The center of gravity is not necessarily located in the body itself. For example, a hoop in a uniform field of gravity has its center of gravity at its geometric center.
In a uniform field of gravity, the center of gravity of a body coincides with its center of mass.
In the overwhelming majority of cases, one term can be painlessly replaced by another.
But: the center of mass of a body exists regardless of the presence of a gravitational field, and we can talk about the center of gravity only in the presence of gravity.
It is convenient to find the location of the center of gravity of the body, and therefore the center of mass, taking into account the symmetry of the body and using the concept of moment of force.
If the arm of the force is zero, then the moment of the force is zero and such a force does not cause rotational motion of the body.
Consequently, if the line of action of the force passes through the center of mass, then it moves translationally.
Thus, you can determine the center of mass of any flat figure. To do this, you need to secure it at one point, giving it the opportunity to rotate freely. It will be installed so that the force of gravity, turning it, passes through the center of mass. At the point where the figure is secured, hang a thread with a load (nut), draw a line along the suspension (i.e., the line of gravity). Let's repeat the steps, securing the figure at another point. The intersection of the lines of action of gravity forces is the center of mass of the body
Experimental task: determine the center of gravity of a flat figure (based on the figures prepared earlier by students from cardboard or plywood).
Instructions: fix the figure on a tripod. We hang a plumb line from one of the corners of the figure. We draw the line of action of gravity. Rotate the figure and repeat the action. The center of mass lies at the point of intersection of the lines of action of gravity.
Students who quickly complete the task can be given an additional task: attach a weight (metal bolt) to the figure and determine the new position of the center of mass. Draw a conclusion.
The study of the remarkable properties of “centers”, which are more than two thousand years old, turned out to be useful not only for mechanics - for example, in the design of vehicles and military equipment, calculating the stability of structures or for deriving the equations of motion of jet vehicles. It is unlikely that Archimedes could even imagine that the concept of the center of mass would be very convenient for research in nuclear physics or in the physics of elementary particles.
Student messages:
In his work “On the Equilibrium of Flat Bodies,” Archimedes used the concept of the center of gravity without actually defining it. Apparently, it was first introduced by an unknown predecessor of Archimedes or by himself, but in an earlier work that has not reached us.
Seventeen long centuries had to pass before science added new results to Archimedes’ research on centers of gravity. This happened when Leonardo da Vinci managed to find the center of gravity of the tetrahedron. He, thinking about the stability of Italian leaning towers, including the Pisa tower, came to the “theorem about the support polygon.”
The conditions of equilibrium of floating bodies, discovered by Archimedes, subsequently had to be rediscovered. This was done at the end of the 16th century by the Dutch scientist Simon Stevin, who used, along with the concept of the center of gravity, the concept of “center of pressure” - the point of application of the pressure force of the water surrounding the body.
Torricelli's principle (and the formulas for calculating the center of mass are also named after him), it turns out, was anticipated by his teacher Galileo. In turn, this principle formed the basis of Huygens’s classic work on pendulum clocks, and was also used in Pascal’s famous hydrostatic studies.
The method that allowed Euler to study the motion of a rigid body under the action of any forces was to decompose this motion into the displacement of the center of mass of the body and rotation around the axes passing through it.
To keep objects in a constant position when their support moves, the so-called cardan suspension has been used for several centuries - a device in which the center of gravity of a body is located below the axes around which it can rotate. An example is a ship's kerosene lamp.
Although the gravity on the Moon is six times less than on Earth, it would be possible to increase the high jump record there “only” by four times. Calculations based on changes in the height of the center of gravity of the athlete’s body lead to this conclusion.
In addition to the daily rotation around its axis and the annual revolution around the Sun, the Earth takes part in another circular motion. Together with the Moon, it “spins” around a common center of mass, located approximately 4,700 kilometers from the center of the Earth.
Some artificial Earth satellites are equipped with a folding rod several or even tens of meters long, weighted at the end (the so-called gravitational stabilizer). The fact is that an elongated satellite, when moving in orbit, tends to rotate around its center of mass so that its longitudinal axis is vertical. Then it, like the Moon, will always be facing the Earth with one side.
Observations of the movement of some visible stars indicate that they are part of binary systems in which “celestial partners” rotate around a common center of mass. One of the invisible companions in such a system could be a neutron star or, possibly, a black hole.
Teacher's explanation
Center of mass theorem: the center of mass of a body can change its position only under the influence of external forces.
Corollary of the theorem on the center of mass: the center of mass of a closed system of bodies remains motionless during any interactions of the bodies of the system.
Solving the problem (at the board)
PROBLEM 2. The boat is standing motionless in still water. The person in the boat moves from the bow to the stern. At what distance h will the boat move if the mass of a person is m = 60 kg, the mass of the boat is M = 120 kg, and the length of the boat is L = 3 m? Neglect water resistance.
SOLUTION. Let us use the condition of the problem that the initial velocity of the center of mass is zero (the boat and the man were initially at rest) and there is no water resistance (no external forces in the horizontal direction act on the “man-boat” system). Consequently, the coordinate of the center of mass of the system in the horizontal direction has not changed. Figure 3 shows the initial and final positions of the boat and the person. Initial coordinate x0 of the center of mass x0 = (mL+ML/2)/(m+M)
Final coordinate x of the center of mass x = (mh+M(h+L/2))/(m+M)
Equating x0 = x, we find h= mL/(m+M) =1m
Additionally: collection of problems by Stepanova G.N. No. 393
Teacher's explanation
Recalling the equilibrium conditions, we found that
For bodies with a support area, stable equilibrium is observed when the line of action of gravity passes through the base.
Consequence: the larger the support area and the lower the center of gravity, the more stable the equilibrium position.
Demonstration
Place the children's toy tumbler (Vanka - Vstanka) on a rough board and lift the right edge of the board. In what direction will the “head” of the toy deviate while maintaining its balance?
Explanation: The center of gravity C of the tumbler is located below the geometric center O of the spherical surface of the “torso”. In the equilibrium position, point C and point of contact A of a toy with an inclined plane should be on the same vertical; therefore, the tumbler’s “head” will deviate to the left
How to explain the preservation of equilibrium in the case shown in the figure?
Explanation: The center of gravity of the pencil-knife system lies below the fulcrum
IIIConsolidation. Frontal survey
Questions and tasks
1. When a body moves from the equator to the pole, the force of gravity acting on it changes. Does this affect the position of the body's center of gravity?
Answer: no, because the relative changes in the force of gravity of all elements of the body are the same.
2. Is it possible to find the center of gravity of a “dumbbell” consisting of two massive balls connected by a weightless rod, provided that the length of the “dumbbell” is comparable to the diameter of the Earth?
Answer: no. The condition for the existence of a center of gravity is the uniformity of the gravitational field. In a non-uniform gravitational field, rotations of the “dumbbell” around its center of mass lead to the fact that the lines of action L1 and L2, the resultant forces of gravity applied to the balls, do not have a common point
3. Why does the front part of a car drop when you brake sharply?
Answer: when braking, a frictional force acts on the wheels on the road side, creating a torque around the center of mass of the car.
4. Where is the center of gravity of the donut?
Answer: in the hole!
5. Water is poured into a cylindrical glass. How will the position of the center of gravity of the glass - water system change?
Answer: The center of gravity of the system will first decrease and then increase.
6. What length of end must be cut from a homogeneous rod so that its center of gravity shifts by ∆ℓ?
Answer: length 2∆ℓ.
7. A homogeneous rod was bent in the middle at a right angle. Where was his center of gravity now?
Answer: at point O - the middle of the segment O1O2 connecting the midpoints of sections AB and BC of the rod
9. The stationary space station is a cylinder. The astronaut begins a circular walk around the station along its surface. What will happen to the station?
Answer: With the station will begin to rotate in the opposite direction, and its center will describe a circle around the same center of mass as the astronaut.
11. Why is it difficult to walk on stilts?
Answer: the center of gravity of a person on stilts increases significantly, and the area of his support on the ground decreases.
12. When is it easier for a tightrope walker to maintain balance - during normal movement along a rope or when carrying a strongly curved beam loaded with buckets of water?
Answer: In the second case, since the center of mass of the rope walker with buckets lies lower, i.e. closer to the support - the rope.
IVHomework:(performed by those who wish - the tasks are difficult, those who solve them receive a “5”).
*1. Find the center of gravity of the system of balls located at the vertices of the equilateral weightless triangle shown in the figure
Answer: the center of gravity lies in the middle of the bisector of the angle at the vertex of which there is a ball of mass 2m
*2. The depth of the hole in the board into which the ball is inserted is half the radius of the ball. At what angle of the board's inclination to the horizon will the ball jump out of the hole?
