The simplest type of motion of matter is mechanical motion, which is the movement in space of bodies or their parts relative to each other.
There are three types of mechanical motion of bodies - translational, rotational and oscillatory. At forward movement solid body all its points describe exactly the same (coinciding when superimposed) lines and have the same speed and the same acceleration (at a given time). Definition rotary motion bodies are given in § 21, vibrational in § 27.
If the shape and dimensions of the body do not have a significant effect on the nature of its movement, then such a body can be considered as a material point. A material point is a body whose shape and dimensions can be neglected in this problem. The last reservation is very significant: when considering one motion of a body, one can consider it a material point, while when considering another motion of the same body, this may turn out to be inadmissible. For example, when studying the motion of the Earth around the Sun, one can consider both the Earth and the Sun to be material points. When studying the motion of the Earth around its axis, one cannot take the Earth as a material point, since the nature of the rotational motion of the Earth is significantly influenced by its shape and size.
The displacement of a body can only be considered relative to some other body or group of bodies. Therefore, when studying the motion of a material point, it is necessary first of all to choose a reference system, i.e., a coordinate system associated with the body, relative to which the motion of the material point is considered. Such a reference system can be, for example, a rectangular XYZ coordinate system associated with some point O on the earth's surface (Fig. 7). Then the position of the material point A at any moment of time will be determined by the coordinates xyz. We will return to the question of frames of reference in § 14.
The line described by a moving material point is called a trajectory. The segment of the trajectory traversed by a point in a certain period of time represents the path traversed by the point
for this period of time (Fig. 7). The movement is called rectilinear if the trajectory is a straight line, and curvilinear if the trajectory is a curved line.
Let a material point, moving along a curvilinear trajectory, pass a short distance in a short period of time (Fig. 8). Let's draw a tangent to the trajectory at point A and a chord A B. The ratio of the path traveled by a material point to the time interval for which this path has been traveled is called the average speed of movement
In the general case of curvilinear (and rectilinear) motion, the value of the average speed can be different in different parts of the trajectory and depend on the value of the path under consideration or, which is the same, on the value of the time interval. We will infinitely decrease the time interval, i.e., we set Then point B will be tend to the point of the chord to the arc and both of them coincide in the limit with the tangent. Thus, the curvilinear motion along a small arc will turn into rectilinear motion along an infinitely small segment of the tangent to the trajectory near the point a average speed on a small path will go into instantaneous, or true, speed at point A. Therefore, the value of instantaneous speed
As can be seen from fig. 8, the instantaneous speed is directed tangentially to the trajectory.
So, the instantaneous speed of movement at any point of the trajectory is a vector directed tangentially to the trajectory, and in magnitude equal to the limit of the average speed when the time interval tends to zero:
From formulas (1) and (2) it follows that the speed is measured in The movement of a material point is called uniform if its speed does not change over time; otherwise, the motion is called non-uniform. The uneven movement is characterized by a physical quantity called acceleration.
Let the material point move in a short period of time from where it had a speed to B, where it has a speed (Fig. 9). The figure shows that the change (increment) in the speed of a point is a vector equal to the difference between the vectors of the final and initial speeds:
The ratio of the change in speed to the time interval during which this change occurred is called the average acceleration.
It follows from the rule of dividing a vector by a scalar that the average acceleration is directed in the same way as the velocity increment, i.e., at an angle to the trajectory towards its concavity (see Fig. 9).
In the general case, the value of the average acceleration can be different in different sections of the trajectory and depend on the value of the time interval over which the averaging is carried out. We will reduce the time interval. In the limit at, point B will tend to point and the average acceleration on the path A B will turn into instantaneous, or true, acceleration a at point Therefore
So, the instantaneous acceleration of motion at any point of the trajectory is a vector directed at an angle to the trajectory towards its concavity, and in magnitude equal to the limit of the average acceleration when the time interval tends to zero.
From formulas (3) and (4) it follows that the acceleration is measured in
It is customary to decompose the acceleration vector into two components, one of which is directed tangentially to the trajectory and is called tangent, or tangential, acceleration, the other is normal to the trajectory and is called normal, or centripetal, acceleration (Fig. 10). acceleration and
components are interconnected by obvious relationships:
Tangential acceleration changes only the magnitude of the speed, and centripetal acceleration - only its direction. Obviously, curvilinear motion always occurs with acceleration, since in this case the speed will necessarily change (at least in direction).
Using concepts higher mathematics, you can replace the limits of the relations in formulas (2) and (4) with derivatives and write:
Mean, respectively, infinitesimal changes (differentials) of displacement, speed, and time. Therefore, velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity with respect to time.
We got acquainted with the general case of non-uniform motion of a material point along a curvilinear trajectory of arbitrary shape. In the following paragraphs, we will consider special cases: rectilinear motion and circular motion.
Curvilinear movements- movements, the trajectories of which are not straight, but curved lines. Planets and river waters move along curvilinear trajectories.
Curvilinear motion is always motion with acceleration, even if the absolute value of the speed is constant. Curvilinear motion with constant acceleration always occurs in the plane in which the acceleration vectors and the initial velocities of the point are located. In the case of curvilinear motion with constant acceleration in the plane xOy projections v x And v y its speed on the axis Ox And Oy and coordinates x And y points at any time t determined by the formulas
A special case of curvilinear motion is circular motion. Circular motion, even uniform, is always accelerated motion: the velocity modulus is always directed tangentially to the trajectory, constantly changing direction, so circular motion always occurs with centripetal acceleration where r is the radius of the circle.
The acceleration vector when moving along a circle is directed towards the center of the circle and perpendicular to the velocity vector.
In curvilinear motion, acceleration can be represented as the sum of the normal and tangential components:
Normal (centripetal) acceleration is directed towards the center of curvature of the trajectory and characterizes the change in speed in the direction:
v- instantaneous speed, r is the radius of curvature of the trajectory at a given point.
Tangential (tangential) acceleration is directed tangentially to the trajectory and characterizes the change in speed modulo.
The total acceleration with which a material point moves is equal to:
In addition to centripetal acceleration, the most important characteristics of uniform motion in a circle are the period and frequency of revolution.
Period of circulation is the time it takes for the body to complete one revolution .
The period is denoted by the letter T(c) and is determined by the formula:
Where t- turnaround time P- the number of revolutions made during this time.
Frequency of circulation- this is a value numerically equal to the number of revolutions made per unit of time.
The frequency is denoted by the Greek letter (nu) and is found by the formula:
The frequency is measured in 1/s.
Period and frequency are mutually inverse quantities:
If a body moving in a circle with a speed v, makes one revolution, then the path traveled by this body can be found by multiplying the speed v for one turn:
l = vT. On the other hand, this path is equal to the circumference 2π r. That's why
vT= 2π r,
Where w(from -1) - angular velocity.
At a constant rotation frequency, the centripetal acceleration is directly proportional to the distance from the moving particle to the center of rotation.
Angular velocity (w) is a value equal to the ratio of the angle of rotation of the radius on which the rotating point is located to the time interval during which this rotation occurred:
.
Relationship between linear and angular speeds:
The motion of a body can be considered known only when it is known how each of its points moves. The simplest motion of rigid bodies is translational. Translational called the movement of a rigid body, in which any straight line drawn in this body moves parallel to itself.
Plis, V., On the dynamics of curvilinear motion, Kvant. - 2005. - No. 2. - S. 30-31, 34-35.
By special agreement with the editorial board and the editors of the journal "Kvant"
It is known from a school course in physics that uniform motion along a circle - this is the name given to the movement of a material point along a circle with a constant velocity - is a motion with acceleration.
This acceleration is due to a uniform change in the direction of the point's velocity over time. At any moment of time, the acceleration vector is directed towards the center of the circle, and its value is constant and equal to
where v - point linear speed, R- radius of the circle, ω - angular velocity of the radius-vector of the point, T - circulation period. In this case, the acceleration is called centripetal, or normal, or radial.
Obviously, curvilinear motion is possible not only in a circle and not necessarily uniform. Let's talk a little about the kinematics of arbitrary curvilinear motion. Moreover, last year in the program of entrance examinations in physics, for example, at Moscow State University. M.V. Lomonosov, included the question of the acceleration of a material point during arbitrary movement along a curvilinear trajectory.
Consider first the non-uniform motion of a material point along a circle. With such a movement, not only the direction of the velocity vector changes with time, but also its magnitude. In this case, the increment of the velocity vector in a short time from t before t +Δ t It is convenient to represent in the form of a sum: 
(Fig. 1). Here, is the tangential tangential component of the velocity increment, co-directed with the velocity vector and due to the increment in the magnitude of the velocity vector by 
, a is the normal component, due (as in the case of uniform motion along a circle) by the rotation of the velocity vector. Then it is natural to represent the acceleration as the sum of the tangent (tangential) and normal components:
For the projections of the acceleration vector on the tangential and normal directions, the relations are valid
Note that the tangent component aτ of acceleration characterizes the rate of change of the velocity value, and the normal component a n characterizes the rate of change in the direction of velocity. According to the Pythagorean theorem,
In the case of movement along an arbitrary curvilinear trajectory, all the above relations are also valid, while in the formula for normal acceleration An under magnitude R it is necessary to understand the radius of such a circle, with the elementary arc of which the section of the curvilinear trajectory coincides in a small neighborhood of the place where the moving material point is located. the value R is called the radius of curvature of the trajectory at a given point.
Now let's consider several specific problems for curvilinear motion, which have been offered in recent years at entrance exams and physics competitions in the leading universities of the country.
Task 1. A stone is thrown with a speed υ 0 at an angle α to the horizon. Find the radius R curvature of the trajectory in the vicinity of the starting point. Acceleration free fall g known.
To answer the question of the problem, we use the relation for normal acceleration:
In a small neighborhood of the starting point υ = υ 0 (Fig. 2). Normal acceleration An is the projection of free fall acceleration g to the normal to the trajectory: An= g cos a. This gives
Task 2. Determine the weight P body mass m at geographic latitude φ. The acceleration due to gravity is g. Consider the earth to be a uniform sphere with radius R.
Recall that the weight of a body is the force due to gravity with which the body acts on a support or suspension. Let us assume that the body lies on the surface of the rotating Earth. It is affected by the force of gravity, directed to the center of the Earth, and the reaction force of the support (Fig. 3). According to Newton's third law, . Therefore, to determine the weight of the body, we find the reaction force.
In an inertial frame of reference, the center of which is at the center of the Earth, the body moves uniformly along a circle with a radius r= R cos φ with a period of one day, i.e. T= 86400 s, and cyclic frequency
7.3 10 -5 s -1 .
The acceleration of the body is equal in magnitude
An= ω 2 r= ω 2 R cos phi
and directed toward the Earth's axis of rotation. From this it follows that the resultant of the forces of gravity and the reaction of the support must also be directed to the axis of rotation of the Earth. Then at 0< φ< π/2 сила реакции образует с перпендикуляром к оси вращения некоторый угол α ≠ φ. По второму закону Ньютона,
Let's move on to the projections of forces and acceleration on the radial direction:
and to the direction perpendicular to the plane in which the movement occurs:
Eliminating α from the last two relations, we find the weight of the body resting on the rotating Earth:
Task 3. The distance from Earth to a binary star in the constellation Centaurus is L= 2.62 10 5 a.u. The observed angular distance between stars periodically changes with a period T= 80 years old and reaches the greatest valueφ \u003d 0.85 10 -5 rad. Determine the total mass M stars. Universal gravitational constant G\u003d 6.67 10 -11 (N m 2 / kg 2), 1 au \u003d 1.5 10 11 m. Consider the orbits of stars circular.
Under the influence of gravitational forces
stars move uniformly with a period T along circles of radii r 1 and r 2 around the center of mass of the system with velocities υ 1 and υ 2, respectively (Fig. 4).
According to Newton's second law,
Adding these equalities (after reducing by m 1 and m 2, respectively), we get
Hence, taking into account the relations
come to an answer
= 3.5 10 27 kg.
Task 4. A vessel with water stands on a horizontal platform (Fig. 5). A thin rod is fixed in the vessel AB inclined to the horizon at an angle α. Uniform ball with radius R can slide without friction along the rod passing through its center. The density of the material of the ball ρ 0, the density of water ρ, ρ 0< ρ. При вращении системы с постоянной угловой скоростью вокруг вертикальной оси, проходящей через нижний конец A rod, the center of the ball is set at a distance L from this end. With what strength F the ball acts on the rod? What is the angular velocity ω of the platform rotation? At what minimum angular velocity ω min will the ball “sink”, i.e. will be at the bottom of the vessel?
Denote the volume of the ball V. Three forces will act on the ball: gravity ρ 0 V· g, normal reaction force N from the side of the rod (the ball acts on the rod with the same magnitude and opposite force) and the force of Archimedes F A. Let's find the Archimedean force.
Consider the motion of a fluid in the absence of a ball. Any elementary volume of water moves uniformly along a circle with a radius r in the horizontal plane. Consequently, the vertical component of the sum of pressure forces (Archimedes force) balances the force of gravity acting on the liquid in the volume under consideration, and the horizontal component imparts centripetal acceleration to this liquid An= ω 2 r. When a liquid is replaced by a ball, these components do not change, and the force acting on the water ball from the side of a thin rod is zero. Then the vertical component of the Archimedes force is equal in magnitude to the gravity of the water ball:
F Az = ρ· V· g,
and the component of the Archimedes force directed towards the axis of rotation imparts centripetal acceleration to the water ball An= ω 2 L cos α and is equal in magnitude to
F An = ρ· Vω 2 L cos α.
Under the action of applied forces, the ball moves uniformly along a circle with a radius L cos α in the horizontal plane (Fig. 6).
According to Newton's second law,
Turning to the projections of forces and accelerations on the vertical axis, we find
ρ· V· g – ρ 0 · V· g – N cosα = 0.
Projecting forces and accelerations in the horizontal plane onto the radial direction, we obtain
ρ 0 · Vω 2 L cos α = ρ Vω 2 L cos α - N sinα.
From the last two relations, we determine the magnitude of the force of the normal reaction of the rod, and hence the pressure force of the ball on the rod:
and angular speed:
As we can see, with an increase in the angular velocity ω, the distance L decreases. At the moment when the ball approaches the bottom, , while
Task 5. A uniform chain of length L R so that one of its ends is fixed at the top of the sphere. The upper end of the chain is released. With what magnitude of acceleration a t will move immediately after the release of each element of the chain? The mass of a unit length of the chain ρ. Acceleration of gravity g.
Consider an elementary section of a chain of length Δ L = RΔφ (Fig. 7). Its mass is Δ m = ρ·Δ L. The forces acting on the selected area are shown in the figure. According to Newton's second law,
Passing to the projections of forces and accelerations on the tangential direction, we obtain
Let us rewrite the resulting relation in the form
Let us sum the increments of the tension force along the entire length of the chain:
Now we take into account that at the free ends of the chain, the tension forces vanish, i.e. that acceleration aτ is the same for all elementary fragments, , and we get
Task 6. The driving wheels of the locomotive are connected by a rack and pinion gear, one link of which is a flat horizontal rod hinged to the spokes of adjacent wheels at a distance R/2 from the axis, where R- wheel radius. When examining the locomotive, the mechanic put a box on this rod and absent-mindedly forgot it there. The locomotive moves away and picks up speed very slowly. Estimate the speed υ 1 of the locomotive at which the box begins to slip relative to the bar. Coefficient of sliding friction of the box on the rod μ = 0.4, wheel radius R= 0.8 m, free fall acceleration g\u003d 10 m / s 2.
Let's move to the reference system associated with the steam locomotive (Fig. 8). Since acceleration is very slow, this system can be considered inertial.
Before slipping, the box moves along a circle with a radius r=R/2. According to Newton's second law,
The acceleration vector of the box is directed towards the center of the circle and is equal in magnitude to a =ω 2 r, where ω is the angular speed of rotation of the wheels of the locomotive. Let us denote the angle that the acceleration vector forms at a given moment of time with the horizon by the letter β. Turning to the projections of forces and accelerations on the horizontal and vertical axes, taking into account the fact that F tr ≤ μ N, we get
Eliminating the support reaction force from here, we arrive at the inequality
The largest value of the expression
where the angle α is such that and , is achieved at β = α and is equal to . The movement of the load will occur without slipping as long as the angular speed of rotation of the wheels of the locomotive satisfies the inequality
From here, for the desired speed of the steam locomotive υ 1, we obtain
= 2.4 m/s.
Task 7. A smooth gutter consists of a horizontal part AB and circular arcs BD radius R= 5 m (Fig. 9). The washer slides along the horizontal part with a speed υ 0 = 10 m/s. Determine the amount of acceleration of the puck at the point WITH and the angle β that the puck acceleration vector makes with the thread at this moment. Radius OS forms an angle α = 60° with the vertical. Acceleration of gravity g\u003d 10 m / s 2.
To find the acceleration of the puck at a point WITH find the tangential aτ and normal a n the magnitude of the acceleration components at this point.
On a body moving in a vertical plane along an arc BD, gravity forces act at any point mg and support reactions N. According to Newton's second law,
Let's move on to the projections of forces and acceleration on the tangential direction:
ma τ = – mg sinα, whence a τ = – g sin α ≈ -8.7 m / s 2.
To determine the normal component of acceleration, we find the value υ of the speed of the puck at the point WITH(because the ). Let's turn to energy considerations. The potential energy of the washer on the horizontal part of the chute will be considered equal to zero. Then, according to the law of conservation of total mechanical energy,
\u003d 10 m / s 2.
The amount of acceleration of the puck at the point WITH find by the Pythagorean theorem:
≈ 13.2 m / s 2.
At the point WITH the acceleration vector forms an angle β with the thread, such that
≈ 0.87, whence β ≈ 41°.
Task 8. Along a smooth wire helix with a radius R step by step h, whose axis is vertical, a bead of mass m. In what time T the bead will descend vertically on H? With what size F the force of the bead acts on the wire at this moment? Acceleration of gravity g.
The force of gravity acts on the bead and normal reaction , where it is directed horizontally (perpendicular to the plane of Figure 10), and lies in the same plane with the vectors and .
To answer the questions of the problem, we find the tangent and normal components of the acceleration. According to Newton's second law,
Turning to the projections of forces and acceleration on the tangential direction, we find a τ = g sinα. Here α is the angle of inclination of the velocity vector to the horizon such that
According to the law of conservation of energy,
The tangential component of the acceleration is constant, the initial speed is zero, therefore, the modulus of the velocity vector grows linearly with time. Hence, for the desired time, we obtain
To determine the normal component of the acceleration, we pass to a moving frame of reference moving forward relative to the laboratory vertically downwards with a speed υ ·
sinα. In this system, the bead moves rapidly along a circle with a radius R at a speed v ·
cos α, while the normal component of the acceleration of the bead is equal in magnitude to 
. Since the acceleration of the moving system is co-directed with , the normal component of the bead's acceleration will not change upon transition to the laboratory frame of reference (this follows from the rule for adding accelerations).
From Newton's second law, we find the components of the force with which the wire acts on the bead:
Where 
.
According to Newton's third law, the bead acts on the wire with a force 
, the value (modulus) of which is equal to
Exercises
1. Spherical balloon with radius R= 5 m is held by a vertical rope, its center is at a height H= 6 m above the horizontal surface. From this surface, a stone is thrown so that it overshoots the ball, almost touching it at the top point. With what minimum speed υ 0 should a stone be thrown and at what distance s from the center of the ball will be in this case the point of throwing?
Note: the free fall acceleration near the Earth's surface in this and subsequent problems is equal to g = 10 m/s 2 .
2. It is known that a satellite in an orbit whose height h = 3.610 4 km, revolves around the Earth in one day and can "hang" over the same point on the equator. Let's assume that the issue of launching a satellite at the same height, which will "hang" over St. Petersburg, is being discussed. What is the magnitude and direction of the traction force F must develop the satellite's engine to keep it in a given orbit? Satellite mass m= 10 3 kg, St. Petersburg is located at latitude φ = 60°, Earth radius R= 6.4 10 3 km.
3. Two bodies with masses move on a smooth table m 1 and m 2, connected by a light inextensible thread of length L.At some moment the first body stops, and the speed of the second is equal to υ and is perpendicular to the thread. Find strength T thread tension.
4. Homogeneous chain of mass m and length L placed on a smooth spherical surface with radius R= 4L so that one of its ends is fixed at the top of the sphere. The upper end of the chain is released. Find the largest value T max chain tension force immediately after its release. indication: for the angles considered in the problem, consider sin α ≈ α, cos α ≈ 1 – α 2 /2.
5. In problem 6 from the text of the article, find the speed υ 2 at which the box will start bouncing.
transcript
1 DYNAMICS OF A MATERIAL POINT CURVILINEAR MOVEMENT
2 federal agency of Education Russian Federation Ural State Technical University UPI named after the first President of Russia B.N. Yeltsin DYNAMICS OF A MATERIAL POINT CURVILINEAR MOTION Published by decision of the editorial and publishing council of USTU UPI from Yekaterinburg USTU UPI 009
3 UDC (075.8) Compiled by: G.S. Novikova Scientific Editor Associate Professor, Ph.D. Phys.-Math. Sciences Druzhinina T.V. Dynamics of a material point. Curvilinear motion: a collection of tasks for independent work at the rate " Theoretical mechanics»/ comp. G.S. Novikov. Yekaterinburg: USTU UPI, p. The collection is intended for the issuance of homework assignments, computational and control works for students of all specialties and all forms of education. Rice. 30 Prepared by the Department of Theoretical Mechanics Ural State Technical University UPI, 009
4 INTRODUCTION The collection contains 30 tasks on the topic “Dynamics of a material point. Curvilinear motion. It is assumed that it will be used by students when performing individual calculation tasks provided for by the standard program of the course "Theoretical Mechanics". In problems, given forces are assumed to be linear functions of the coordinates of a point, its absolute or relative velocity. Therefore, the differential equations will be linear and have an analytical solution. When solving, it is possible to use computer technology both for the numerical integration of the equations of motion, and for plotting motion and trajectory graphs in the analytical solution of systems of equations. Instructions for performing tasks When working on a task, it is necessary to build a calculated mechanical model, replacing the given body with a material point, show the acting forces in the figure for an arbitrary position M (x, y), and write down the equation of motion in vector form. Express the acting elastic and resistance forces in terms of the radius vector r (x, y) and the absolute speed of the point ν r (x, y). Then compose differential equations of motion in projections on the selected coordinate axes. Integrating the equations analytically or numerically, we obtain solutions x (t), y(t). In most problems, the solution is damped oscillations. Find the period T and the decrement D of these oscillations. The construction of motion graphs x (t), y(t) is carried out point by point on a section of one period (if the periods for solutions are different, then take the largest one) with a step, for example, T / 4. For numerical integration, take a step h = T / 40. To continue building for the entire period of the transition mode for steady motion, you can use T and D. The time of the transition mode can be estimated approximately by the formula 3 τ = 3 / n, where n = μ / m. At "us-
5 "laying" tasks, it is recommended that the resistance forces be considered proportional to the square of the speed 0 R = μν ν, where ν = ν / ν 0 is the unit vector, ν and \ν is the vector and the velocity modulus. In options 4, 5, 10, 14, 3, 5, 7, take the resistance force as 1 x μ y R = μ V i V j. An example of solving the problem Determine the movement of a heavy material point, the mass of which is equal to m, attracted to a fixed center O by a force directly proportional to the distance to this center. Movement takes place in emptiness; the force of attraction per unit distance is μ m ; at time t = 0: M O = x = a x& = 0; y=0; y& 0, 0 0 ; = and the y-axis is directed vertically down (see figure). According to Newton's second law m a = P + F, where F = μ m OM. In projections on the coordinate axes, we get m & x = μ m OM sin α ; where x = OM sinα, y = OM cosα. m & y = mg μ m OM cosα, Then m& x = μ mx, m& y = mg μ my. Finally, the differential equations of motion will have the form 4
6 && x = μ x, && y = g μ y. We look for the solution of the first linear homogeneous differential equation of the second order & x& + μ x = 0 depending on the type of roots of the characteristic equation, for which we substitute x = e in the equation and obtain the characteristic equation λt λ + μ = 0, whence λ = ± i .. 1, μ Since the roots of the characteristic equation are imaginary and different, the solution to the equation is x = c1 coskt + c sinkt. To determine the integration constants c 1 and c, we define the speed x & = c1k sin kt + ck coskt. The solution of the second inhomogeneous differential equation with a constant right-hand side & y μ y = g = will be the sum of common solution homogeneous equation& y& + μ y 0 and g of the partial solution of the inhomogeneous & y + μ y =, that is, y = A & y 0, then μ A = g, A = g. μ Full solution y = y 1 + y: y = c 1 coskt + c g sinkt + μ., = Velocity y & = c1k sinkt + ck coskt. According to the initial conditions: y =, y& 0 from these equations we obtain c g = 1 = ; c = μ 0.5
7 Then the law of motion of the point in the projection on the y-axis will be g y = (1 coskt). μ Finally, the law of motion of a material point in projections on the coordinate axes will be x = acoskt, g y = (1 coskt). μ Eliminating the time t from these equations, we obtain the trajectory of the point: a straight line segment g x g y = 1 ; a x a; 0y. μ a μ 6
8 Problem 1. A cable car of mass m is lifted by a given force Q. The cable is elastic, its elastic force is considered proportional to the transverse strain of the velocity AM. The resistance of the medium is proportional. The straight line OO 1 defines the points where the transverse strain of the cable is zero. The movement of the trolley started from point O, the initial speed is shown in the figure. Find the equations of motion of the trolley. Build motion graphs and trajectory. Given: µ = 1.4 10³ N s/m; α = 30 ; Q = 7 10³ N; = 1.8 m/s; m = 1.3 10³ kg; c = 1 10³ N/m. Task. A balloon with a mass m is towed at a constant speed V A. The difference between the Archimedean force and its weight is directed vertically upwards and is equal to 0.1mg. The cable is elastic, the elastic force is considered proportional to the distance AM, AM. The resistance force of the medium is proportional to the speed. At the initial moment of time, the speed of the balloon is vertical, point A was at the origin of coordinates. Take AM = 0. Find the equations of motion of the balloon. Build motion graphs and trajectory. Given: m = 0.8 10³ kg; = 0.9 m/s. O V A = 5 m/s; c \u003d 1.1 10³ N / m; µ = 0.8 10³ N s/m; 7
9 Problem 3. An elastic thread fixed at point A passes through a fixed smooth ring O; a ball M is attached to its free end, the mass of which is m. The length of the undrawn thread l = AO. Thread stiffness coefficient c. By pulling the thread vertically twice, the initial horizontal velocity was given to the ball. When moving, the drag force of the medium acts on the ball, which is proportional to the speed. Find the equations of motion of the ball. Build motion graphs and trajectory. Given: m = 0, kg; c = 0 N/m; µ = 0.8 N s/m; = 0 m/s; l = 1m. Problem 4. A platform of mass m on an air cushion is accelerated by a constant force Q. The elastic forces are realized by the forces of the air cushion system. Consider the equivalent elastic force proportional to the vertical deflection AM. The straight line OA corresponds to the level where F = 0. The viscous drag forces in the horizontal and vertical directions are proportional to the respective velocity components, the proportionality factors are equal to µ 1 and µ. The initial speed of the platform is shown in the figure. Find the equations of motion of the platform. Build motion graphs and trajectory. Given: m = kg; c = 1, N/m; Q = 4, N; µ 1 = 0, N s/m; µ = 1, N s/m; = 0.7 m/s. 8
10 Problem 5. A load M of mass m is towed at a given constant speed V A. The cable is elastic, its elastic force is considered proportional to the longitudinal deformation F1 = c1 AM. The shock absorbers generate an elastic force proportional to the vertical deflection from the undeformed state BM. The resistance forces of the medium in the horizontal and vertical directions are proportional to the corresponding velocity components. The coefficients of proportionality are equal to μ 1 and μ, the initial velocity is vertical. Find the equations of motion. Build motion graphs and trajectory. Given: m = kg; V A = 4, m/s; s 1 = 3, N/m; c = 1.105 N/m; µ 1 \u003d 1, N s / m; µ = N s/m; V m (O) \u003d 1.6 m / s; B 0 M 0 = 1.5 m; OB 0 = 0; OA 0 = 0.4 m. Problem 6. A load M of mass m is attached to the end of a horizontally stretched elastic thread AM, fixed at point A and passing through a fixed smooth ring O. At the initial moment, the thread is stretched by the value OM 0 and the load is released without initial speed. The elastic force is proportional to the elongation. The coefficient of proportionality is s. The length of the undeformed thread l = AO. The resistance force of the medium is proportional to the speed. Find the equations of motion of the load. Build motion graphs and trajectory. Given: m = 0.6 kg; c = 15 N/m; µ =.4 N s/m; l = 1 m; OM 0 = 0.8 m. 9
11 Problem 7. A load of mass m is suspended on an elastic cable, the elastic force of which is proportional to the longitudinal deformation = c OM. A constant force Q acts on it, directed at an angle α to the horizon. The force of viscous resistance to movement is proportional to the speed F. Find the equations of motion of the load, if at the initial moment its speed is horizontal, the cable was vertical, OM 0 is the initial deformation of the cable. Build motion graphs and trajectory. Given: m = 1.5 10 kg; c = 1, N/m; µ =.6 10 N s/m; α = 30 ; Q =.8 10 N; =, m/s; OM 0 = 0.8 m. Problem 8. A pontoon of mass m, located in a fluid flow, is held by an elastic cable. The elastic force is proportional to the longitudinal strain F1 = c1 AM. The flow rate U is shown in the figure. The Archimedean force is proportional to the magnitude of the immersion BM, The viscous drag force is proportional to the relative velocity rel. Find the equations of motion for the pontoon if its initial velocity is vertical. Build motion graphs and trajectory. Given: m = kg; c 1 = N/m; c = 4, N/m; µ = 4, N s/m; U =.6 m/s; = 0.3 m/s; AM 0 = 1 m; BM 0 = 0.10
12 Problem 9. A cable car of mass m is freely lowered along a cable. The cable is elastic, the elastic force is assumed to be proportional to the transverse deformation AM. The resistance of the medium is proportional to the speed. The straight line OO 1 defines the points where the transverse strain of the cable is zero. The movement of the trolley started from point O, the initial speed is shown in the figure. Find the equations of motion of the trolley. Build motion graphs and trajectory. Given: m = 5 10 kg; c = 6, N/m; µ = 4.3 10 N s/m; α = 10 ; = 1.8 m/s. Problem 10. An airship of mass m is in an air flow whose speed is U. The cable that holds the airship at the mooring mast is elastic, the elastic force is proportional to the longitudinal deformation OM. The difference between the Archimedean force and weight is directed vertically upwards, and is equal to 0.mg. Viscous drag forces in the vertical and horizontal directions are proportional to the respective relative velocity components. The coefficients of proportionality are equal to μ 1 and μ. At the initial moment, the speed of the airship. Find the equations of motion for the airship. Build motion graphs and trajectory. Given: m = kg; c = 1, N/m; µ 1 = 5, N s/m; µ = 1, N s/m; U = 5 m/s; = 1.7 m/s; OM 0= 0.5 m; OM 0U.11
13 Problem 11. A boat of mass m is accelerated by a horizontal constant force. At the same time, having the initial speed of immersion in water, it oscillates under the action of the Archimedean force proportional to the depth of the submerged part of the boat AM. The force of water resistance acting on the boat is proportional to the speed. Find the equations of motion of the boat. Build motion graphs and trajectory. Given: m = 1, kg; c = 4, N/m; µ = 1, N s/m; Q = 3, N; = 1.3 m/s; point A is the projection of the center of mass of the boat on the surface of the water. Problem 1. An underwater vehicle of mass m is towed at a given speed V. The towing rope is elastic, the elastic force is A AM longitudinal deformation. The difference between the Archimedean force and the weight of the device is 0.3mg and is directed vertically downwards. The resistance force of the environment. Find the equations of motion of the apparatus if its F = c AM, where the initial velocity is vertical. Build motion graphs and trajectory. Given: m = 5, kg; VA = m/s; c = N/m; µ = 5, N s/m; = 0.6 m/s; at t = 0 the vehicle is under the tug at a depth of 0.5 m. 1
14 Problem 13. A load of mass m hanging on a cable with side shock absorbers performs free vibrations under the action of the elastic force of the cable F1 = c1 OM (OM longitudinal deformation) and the elastic forces of the shock absorbers, the resultant of which can be considered horizontal and proportional to the horizontal deviation from the undeformed state of the springs: F x = c x. The resistance force of the medium is proportional to the speed. Find the equations of motion of the load, if its initial velocity is horizontal, the cable OM 0 is vertical. Build motion graphs and trajectory. Given: m =, kg; c 1 = N/m; c = N/m; BM 0 = 0.0 m; µ = 8, N s/m; = 0.9 m/s; OM 0 = 0, m. Problem 14. A buoy of mass m is accelerated by the wind, whose speed U is constant. The ice surface on which the ice glider slides is assumed to be elastic. The elastic force is proportional to the transverse strain AM. The forces of viscous friction in the vertical and horizontal directions are proportional to the components of the relative speed of the buoy in these directions, the coefficients of proportionality are equal to μ 1 and μ. The straight line OO 1 indicates the positions of the buoy, where F = 0. The initial speed of the buoy is directed vertically down. Find the equations of motion of the ice floe. Build motion graphs and trajectory. Given: m = 3.5 10 kg; c = 7, N/m; µ 1 \u003d N s / m; µ =,1 10 N s/m; = 1.4 m/s; U = 5 m/s. 13
15 Problem 15. A load of mass m hanging on an elastic cable is in a fluid flow moving at a constant speed U. The elastic force of the cable is proportional to the longitudinal deformation OM. The difference between the weight of the load and the Archimedean force is directed vertically downward and is equal to Q = 0.8mg. The force of viscous friction is proportional to the relative velocity of the load R μv = rel. At the initial moment, the load was in an equilibrium position and received an initial velocity directed at an angle α to the horizon. Find the equations of motion of the load. Build motion graphs and trajectory. Given: m = kg; U = 8 m/s; c = 1, N/m; µ = 1, N s/m; α = 30 ; = 1, m/s. Problem 16. A barge of mass m is towed at a given horizontal speed V A in a fluid flow with a speed U. The buoyancy force from the water side is proportional to the immersion depth, the coefficient of proportionality is c 1. The elastic force of the cable is proportional to its longitudinal deformation AM. The force of water resistance is proportional to the relative velocity rel. The initial speed is shown in the figure. Take the initial position of point A as the origin of coordinates, consider AM 0 = 0. Find the equations of motion of the barge. Build motion graphs and trajectory. Given: m = kg; VA = 4 m/s; U = 3 m/s; s 1 =, N/m; c = 6, 10 5 N/m; µ = N s/m; α = 30 ; = 0.7 m/s. A 14
16 Problem 17. A body of mass m, thrown with an initial velocity at an angle α to the horizon, moves under the influence of gravity and air resistance force proportional to the velocity. Find the equations of motion of the body, the maximum height of the lift, the horizontal distance when this height is reached, the flight range. Construct graphs of movement and body trajectory. Given: m = 5 kg; = 0 m/s; α = 60 ; µ = 0.3 N s/m. Problem 18. A load of mass m, suspended on an elastic cable, is lifted by a crane at a constant speed V A. The elastic force of the cable is proportional to the longitudinal deformation AM. The force of air resistance is proportional to the speed of the load. The initial speed is horizontal, the cable was vertical, A0M 0 initial deformation. Find the equations of motion of the load. Build motion graphs and trajectory. Given: m = kg; VA = m/s; c = 6, 10 4 N/m; µ = 4, N s/m; = 1.3 m/s; A 0M = 0.5 m. 0 15
17 Problem 19. A climber of mass m descends an elastic rope, which, in an unloaded state, coincides with the straight line OO 1, which makes an angle α with the horizon. The elastic force of the rope is considered proportional to the transverse deformation AM. The air resistance force is proportional to the speed. The initial speed is shown in the figure. Find the climber's equations of motion. Build motion graphs and trajectory. Given: m = 80 kg; α = 15; c = 6.5 10 = 1.5 m/s. N/m; AM 0 = 0; µ = 75 N s/m; Problem 0. A load of mass m, suspended on an elastic cable, is moved by a crane with a constant horizontal speed proportional to its longitudinal deformation V. The elastic force of the cable is A AM. The movement occurs in a medium moving at a constant speed U. The drag force of the medium is proportional to the relative speed of the load = rel. At the initial moment of time, the speed of the load R μv was horizontal, the cable was vertical, A 0M 0 =1 m. Take the initial position of point A as the origin of coordinates. Find the equations of motion of the load. Build motion graphs and trajectory. Given: m = kg; V A \u003d, 5 m / s; c = 5, N/m; U = 3.3 m/s; µ = 6, N s/m; = 1.4 m/s. 16
18 Problem 1. A buoy of mass m is held in a liquid by an elastic cable. The elastic force is proportional to the longitudinal strain OM. The buoy is acted upon by a constant modulo force Q, directed at an angle α to the horizon. The difference between the Archimedean force and the weight of the buoy is 0.5mg and is directed vertically upwards (positive buoyancy). When the buoy moves, it is affected by the fluid resistance force proportional to the speed. Find the equations of motion of the buoy, if at the initial moment its speed is vertical and directed upwards, the cable was vertical and OM 0 = 0.1 m. Build graphs of motion and trajectory. Given: m = 1.10 kg; c = 6, 10 3 N/m; = 0.7 m/s; Q = 4, 10; α = 40 ; µ = 3.8 10 N s/m. Task. A man of mass m jumps into a boat of mass m 1, tied to the shore with an elastic cable, while the boat receives an initial speed directed at an angle α to the horizon. The initial strain of the cable is zero. The coefficient of rigidity of the cable is c 1. The Archimedean force acting on the boat during its oscillations is proportional to the depth of immersion. Proportionality coefficient s. The force of viscous resistance depends on the speed according to a linear law. Find the equations of motion of a boat with a man. Build motion graphs and trajectory. Given: m 1 = 60 kg; m = 80 kg; = 5 m/s; α = 15; c 1 = 500 N/m; c = N/m; µ = 1.8 10 N s/m. 17
19 Problem 3. A ship of mass m drifts freely in a stream whose speed is constant and equal to U. Consider the Archimedean force acting on the ship as proportional to the immersion depth with a proportionality coefficient c. The forces of viscous resistance to movement in the horizontal and vertical directions are proportional to the corresponding components of the relative velocity, the coefficients of proportionality are equal to μ 1 and μ. At the initial moment, the ship had a speed. Find the ship motion equations. Build motion graphs and trajectory. Given: m = kg; U =.5 m/s; c = 6, N/m; µ 1 = 0, N s/m; µ = 1.105 N s/m; =.3 m/s. Problem 4. A load of mass m slides along an elastic conveyor belt. in an unloaded state, the tape occupies the position OO 1, which makes an angle α with the horizon. At some point in time, the load falls on the tape (at point O) with a speed perpendicular to the tape. The force of friction of the load on the belt is considered to be proportional to its speed. The force of transverse elasticity of the tape is proportional to its deflection AM. A constant force Q also acts on the load, parallel to OO 1 and slowing down the movement. Find the equations of motion of the load. Build motion graphs and trajectory. Given: m = 60 kg; α = 15; = 1.5 m/s; µ = 80 N s/m; c = 7, 10 N/m; Q = 45 N. 18
20 Problem 5. An airship of mass m is towed at a given speed The towing rope is elastic, the elastic force is assumed to be proportional to the longitudinal deformation V A. AM, The difference between the Archimedean force and the weight of the airship is 0.15 mg and is directed vertically upwards. The forces of air resistance in the horizontal and vertical directions are considered proportional to the corresponding components of the airship's speed. The coefficients of proportionality are μ and 1 μ. At the beginning of towing, the airship received the initial speed and AM 0. Take the initial position of the point A 0 = as the origin of coordinates. Find the equations of motion for the airship. Build motion graphs and trajectory. Given: m = kg; VA = 3 m/s; c = N/m; µ 1 \u003d 1, N s / m; µ = 8, 10 4 N s/m; = 0.9 m/s. Problem 6. At the bottom of the tank there is a load of mass m, tied with an elastic cord, the stiffness coefficient of which is c. At some point in time, the load was picked up and pulled out with a constant force Q at an angle α to the horizon. Negative buoyancy (the difference between the weight and the Archimedean force) is directed downward and is equal to N = 0.5G, where G is the weight of the load. The viscous friction of water is proportional to the speed of the load and is determined by the formula At the moment of engagement, the load touched the block O, the cord was not deformed, and the load received the initial horizontal speed. Find the equations of motion of the load.. Build graphs of motion and trajectory. Given: m = 50 kg; c = 00 N/m; µ = 100 N s/m; Q = 100 N; α = 30 ; = 8 m/s. 19
21 Problem 7. A ship of mass m is towed at a constant horizontal speed V A. The towing rope is elastic, the elastic force is assumed to be proportional to the longitudinal deformation F = c1 AM. At the initial moment, the ship touched the tug, the cable had no deformation, and the initial speed was directed vertically downwards. Consider the Archimedean force as proportional to the depth of the vessel's immersion, the coefficient of proportionality is equal to c. The forces of water resistance in the horizontal and vertical directions are proportional to the respective velocity components, μ 1 and μ coefficients of proportionality. Find the ship motion equations. Build motion graphs and trajectory. Given: m = kg; VA = 4.5 m/s; c 1 = 0, N/m; c = 1, N/m; µ 1 = 0, N s/m; µ = 1, N s/m; =.3 m/s. Problem 8. A boat of mass m moves against the current with the engines turned off, with an initial speed directed at an angle α to the horizon. The flow velocity U is constant. The Archimedean force is proportional to the immersion height, the proportionality factor is equal to c. From the side of the water, the boat experiences resistance proportional to the relative speed rel. Find the equations of motion of the boat. Build motion graphs and trajectory. Given: m = 50 kg; α = 10 ; = 3 m/s; µ = 1.7 10 N s/m; c =, N/m; U = 5 m/s. 0
22 Task 9. A load of mass m, suspended on an elastic cable, is moved by a crane with a constant speed V A directed at an angle α to the horizon. The elastic force of the cable is proportional to the longitudinal deformation F = c AM. The air resistance force is proportional to the speed. At the initial moment of time, the speed of the load is horizontal, the cable was vertical, A 0M 0 is the initial deformation of the cable. Take the origin of coordinates at the initial position of point A. Find the equations of motion of the load. Build motion graphs and trajectory. Given: m = 500 kg; VA = 3 m/s; α = 30 ; c = 8, N/m; = 1.8 m/s; µ = 9 10 N s/m; A 0 M 0 = 0, m. Problem 30. A pontoon of mass m is kept in the flow, the speed of which is U, by an elastic cable. The elastic force is proportional to the longitudinal strain F1 = c1 OM. The Archimedean force is proportional to the depth of the pontoon, the coefficient of proportionality c. The viscous resistance force acting on the pontoon from the liquid side is proportional to the relative velocity rel. At the initial moment of time, the pontoon touched the block (OM 0 = 0) and had a velocity directed along the vertical. Find the equations of motion of the pontoon. Build motion graphs and trajectory. Given: m = kg; U = m/s; c 1 = 8, µ = 3, N s/m; =.1 m/s. N/m; c = 9, 10 4 N/m; 1
23 Dynamics of a material point. Curvilinear motion Editor O.S. Smirnova Computer layout I.I. Ivanov Signed for printing Format 60х84 1/16 Writing paper Flat printing Convention. oven l. Uch.-ed. l. Circulation 100 copies. Order Editorial and publishing department USTU UPI 6006, Yekaterinburg, st. Mira, 19 Risography NICH USTU UPI 6006, Yekaterinburg, st. Mira, 19
Federal Agency for Railway Transport Ural State University of means of communication Department "Mechatronics" G. V. Vasilyeva V. S. Tarasyan
Ministry of Education of the Russian Federation educational institution higher vocational education"SAMARA STATE TECHNICAL UNIVERSITY" Department "MECHANICS" DYNAMICS
FEDERAL AGENCY FOR EDUCATION VOLGOGRAD STATE TECHNICAL UNIVERSITY VOLGA POLYTECHNICAL INSTITUTE (BRANCH) G.B. Potapova, K.V. Khudyakov FREE OSCILLATIONS OF A MATERIAL POINT
Conditions and solutions to problems of the II Mordovia State University Olympiad in Theoretical Mechanics (2013-2014 academic year) 1. The load is pulled up along a rough surface inclined at an angle
Federal Air Transport Agency
Ministry of Education and Science of the Russian Federation Federal State Budgetary Educational Institution of Higher Professional Education Altai State Technical University
TASK D-I Topic: The second main problem of point dynamics and the kinetostatics method (the principle of Hermann-Euler-D'Alembert). PLAN FOR SOLVING PROBLEMS 1. For task 1: a) arrange the forces acting on a material point
Theoretical Mechanics Tests 1: Which or which of the following statements is not true? I. The reference system includes the reference body and the associated coordinate system and the selected method
Ministry of Education and Science of the Russian Federation Ural federal university named after the first President of Russia B.N. Yeltsin DETERMINATION OF FREE-FALL ACCELERATION USING A REVERSIBLE PENDULUM
Excerpts from the book Gorbaty IN "Mechanics" 3 Work Power Kinetic energy
Explanation of phenomena 1. The figure shows a schematic view of the graph of changes in the kinetic energy of the body over time. Choose two correct statements describing the movement according to the given
Explanation of phenomena 1. The figure shows a schematic view of the graph of changes in the kinetic energy of the body over time. Choose two correct statements describing the movement according to the given
3 Conservation laws in mechanics Basic laws and formulas Newton's second law ma = F can be represented as: m υ = F t, the change in the momentum of the body (p = m υ = mυ mυ) is equal to the momentum n of the resultant
Physics. Grade 9 Training “Inertia. Newton's laws. Forces in mechanics» 1 Inertia. Newton's laws. Forces in mechanics Option 1 1 A metal bar is suspended from a spring and completely immersed in a vessel with water, being
Physics tasks A5 1. A body is dragged up a rough inclined plane. Which of the forces shown in the figure does positive work? 1) 1 2) 2 3) 3 4) 4 2. The figure shows a graph of dependence
Lecture 1. Sergey Evgenievich Muravyov Candidate of Physical and Mathematical Sciences, Associate Professor of the Department of Theoretical Nuclear Physics, National Research Nuclear University MEPhI We are starting! 1. The winners and prize-winners of the Olympiads must score 75 points in the Unified State Examination!.
Methodical materials on the topic "Mechanical phenomena" - grade 9 Part 1 1. The car starts moving in a straight line from rest with an acceleration of 0.2 m / s 2. How long will it take to acquire a speed of 20 m / s?
Ministry of Education of the Russian Federation State Educational Institution of Higher Professional Education "SAMARA STATE TECHNICAL UNIVERSITY" Department "MECHANICS" K O
"FOUNDATIONS OF DYNAMICS" Newton's laws: First: There are frames of reference called inertial, relative to which a progressively moving body retains its state of rest or rectilinear uniform
Lesson 11 Final 2. Mechanics. Task 1 The figure shows a graph of the path S of a cyclist versus time t. Determine the time interval after the start of movement, when the cyclist moved with
Differential equation of motion of a point Problem D2.1. 1 The braking distance of a car on a horizontal road at a speed v 0 is S. What is the braking distance of this car at the same speed
00-0 account year., cl. Physics. Basic laws of mechanics.. Dynamics In dynamics, mechanical motion is studied in connection with the causes that cause one or another of its character. In inertial frames of reference these
Examples of tasks from the database of tasks for the remote qualifying round of the Rosatom Olympiad, Grade 11 The database of tasks for the remote qualifying round of the Rosatom Olympiad (which is held only for schoolchildren
Establishing a correspondence, part 2 1. A rudder, located on a rough horizontal surface, begins to move uniformly accelerated under the action of a force In the frame of reference associated with the horizontal surface,
KINEMATICS of type B references Page 1 of 5 1. The body began to move along the OX axis from the point x = 0 with an initial speed v0x = 10 m/s and with a constant acceleration a x = 1 m/s 2. How will physical quantities change,
OLYMPIAD FUTURE RESEARCHERS FUTURE OF SCIENCE 2018-2019 Physics, round I, option 1 7 class 1. (30 points) Two cars left at the same time: one from point A to point B, the other from B to A. The speed of one
Ural Federal University named after the first President of Russia BN Yeltsin
Distance training bituru PHYSICS Article 8 Mechanical oscillatory systems Theoretical material In this article we will consider methods for solving problems on the oscillatory motion of bodies by oscillatory motion
Dynamics 1. A block of mass moves translationally along a horizontal plane under the action of a constant force directed at an angle to the horizon. The modulus of this force The coefficient of friction between the bar and the plane
TOPIC Lecture 3 Work, power, energy. The law of conservation and change of mechanical energy. Matronchik Aleksey Yurievich Candidate of Physical and Mathematical Sciences, Associate Professor of the Department of General Physics, National Research Nuclear University MEPhI, expert
OLYMPIAD FUTURE RESEARCHERS FUTURE OF SCIENCE 017-018 Physics, round I, option 1 SOLUTIONS Attention: the assessment quantum is 5 (only 5, 10, 15, etc. points can be put)! General recommendation: When checking,
Lesson 3. Basic principles of dynamics. Forces: gravity, reactions, elasticity Option 3 ... Several forces act on a body with a mass of 0 kg, the resultant of which is constant and equal to 5 N. Relative to the inertial
C1.1. Two identical bars connected by a light spring rest on a smooth horizontal table surface. At the moment t = 0, the right block starts moving so that in time x it picks up the final speed
Distance training Abituru PHYSICS Article Newton's laws Theoretical material In this article we will consider the tasks of applying Newton's laws
Test 1 on the topic: “Kinematics. Dynamics. Conservation laws» Grade 10 Questions for the test 1 1. What is called mechanical motion? 2. What is the reference body? 3..What are the ways to set the position
Bank of tasks in physics Grade 1 MECHANICS Uniform and uniformly accelerated rectilinear motion 1 The figure shows a graph of the dependence of the coordinate of the body on time during its rectilinear motion along the x axis.
Federal Agency for Education State Educational Institution of Higher Professional Education Ukhta State Technical University Physics MECHANICAL OSCILLATIONS AND WAVES
INTRODUCTION The condition of each task of the calculation and graphic work is accompanied by ten figures and two tables of numerical values of the given values. The choice of options is made according to the student's code.
Test 1 on the topics “Kinematics. Dynamics". Questions for the test: 1. What does kinematics study? 2. Basic concepts of kinematics: mechanical movement, material point, frame of reference, trajectory, traversed
Learning tasks on the topic "DYNAMICS" 1 (A) The bus moves in a straight line at a constant speed. Choose the correct statement. 1) Only gravity acts on the bus.) The resultant of all applied
Student's problem book izprtalru 6 Dynamics rectilinear motion The basic equation of the dynamics of a material point (Newton's second law) for a body of constant mass in inertial reference frames has the form
Ministry of Education and Science of the Russian Federation Ural Federal University named after the first President of Russia B.N. Yeltsin
Examples of problem solving Example 1 A weightless inextensible thread is thrown through a block rotating around a horizontal axis (Fig. 1a), to the ends of which weights 1 and
Solving problems on the movement of bodies using blocks Problem An inextensible thread is thrown through the block, to which two bodies are attached with masses and (moreover) Determine the accelerations with which they will move
OLYMPIAD FUTURE RESEARCHERS FUTURE OF SCIENCE 017-018 Physics, round I, SOLUTION variant Attention: the assessment quantum is 5 (only 5, 10, 15, etc. points can be put)! General recommendation: When checking, even
1.2.1. Inertial reference systems. Newton's first law. Galileo's principle of relativity 28(C1).1. A bus passenger at a bus stop tied a light balloon filled with
WORK, POWER, ENERGY, PRESSURE 008 1. A steel part (ρс = 7800kg/m) with a volume of 4 dm is at a height of m. Its potential energy is A) 9600 J B) 960 J C) 96000 J D) 96 J E) 9 .6 J. Determine
OLYMPIAD FUTURE RESEARCHERS FUTURE OF SCIENCE 017-018 Physics, round I, option 1 SOLUTIONS Grade 7 1. (40 points) Two cars simultaneously drive towards each other from different points and drive at speeds
ITT- 10.3.2 Option 2 CONSERVATION LAWS 1. What is the name of physical quantity equal to the product of the mass of the body and the vector of its instantaneous velocity? 2. What is the name of a physical quantity equal to half the product
Options for homework HARMONIC OSCILLATIONS AND WAVES Option 1. 1. Figure a shows a graph of oscillatory motion. Oscillation equation x = Asin(ωt + α o). Determine the initial phase. x O t
Value, its definition Designation Unit of measurement "MECHANICS" Formula Values in the formula TYPES OF MOVEMENT I. Uniform rectilinear motion is a motion in which the body for any equal intervals
Minimum in physics for students of the 10th grade for the 1st half of the year. Physics teacher - Turova Maria Vasilievna e-mail: [email protected] References: 1. Textbook of physics grade 10. Authors: G.Ya.Myakishev, B.B.
Lecture 4 Topic: Dynamics of a material point. Newton's laws. Dynamics of a material point. Newton's laws. Inertial reference systems. Galileo's principle of relativity. Forces in mechanics. Elastic force (law
Questions for the test in the course "Theoretical Mechanics", section "Dynamics" 1. Basic axioms of classical mechanics.. Differential Equations movement of a material point. 3. Moments of inertia of a system of points
Thematic diagnostic work in preparation for the exam in PHYSICS on the topic "Mechanics" December 18, 2014 Grade 10 Option PHI00103 (90 minutes) District. City (town). School Class Surname. Name.
Ministry of Education of the Republic of Belarus Educational Establishment "Mogilev State University of Food" Department of Physics
Demonstration variant_10 class (profile) Task 1 1. A truck passes by a stop along a straight street at a speed of 10 m/s. After 5 seconds, a motorcyclist drives off from the stop after the truck, moving
Nurusheva Marina Borisovna Senior Lecturer of the Department of Physics 3 National Research Nuclear University MEPhI Mechanical oscillations
IV Yakovlev Materials on physics MathUs.ru Newton's laws Problem 1. A rocket starts from the surface of the Earth and moves vertically upwards, accelerating with an acceleration of 5g. Find the weight of an astronaut of mass m, who is
OLYMPIAD FUTURE RESEARCHERS FUTURE OF SCIENCE 2018-2019 Physics, round I, option 2 7 class 1 (40 points) Two cars left at the same time: one from point A to point B, the other from B to A The speed of one car
006-007 account. year., 9 cells. Physics. Dynamics. 5. Forces Recording Newton's second law in the form of a formula () cannot be interpreted as the equality of the two forces F and ma. This entry is only an expression of the resultant
Conservation laws The momentum of a body (material point) is a physical vector quantity equal to the product of the body's mass and its speed. p = m υ [p] = kg m/s p υ Impulse of force is a vector physical quantity,
Consider the motion of a body along an arbitrary curvilinear trajectory. We have already noted above that when a body moves along a curvilinear trajectory, its velocity vector at any point is directed tangentially to the trajectory. The figure shows why this is so. The average speed is . This means that the direction of the average velocity vector always coincides with the direction of movement Δr. But if we bring the end point closer to the initial one, making the time interval Δt less and less, then, as can be seen from the figure, the direction of the vector Δr will approach the direction of the tangent to the trajectory at the starting point and merge with it in the limit. But in this limit, the average speed will turn into instantaneous speed.
Unlike velocity, acceleration when a body moves along a curvilinear trajectory is almost never directed tangentially to the trajectory. Since , the direction of the acceleration vector always coincides with the direction of the velocity change vector. As can be seen from the figure, the velocity change vector, and, hence, the acceleration is directed inside the curvature of the trajectory. In general, the angle between the velocity and acceleration vectors can vary from 0 to 180°.
Very often, the acceleration of a body when moving along a curvilinear trajectory is decomposed into two mutually perpendicular components: the direction of the tangent to the trajectory and the direction perpendicular to the tangent. The component of the total acceleration vector in the direction of the tangent to the trajectory is called tangential or tangential acceleration ( and τ). The component of the total acceleration vector in the direction perpendicular to the tangent is called centripetal or normal acceleration ( a c).
If α is the angle between the directions of acceleration and velocity, then we can write:
Besides:
The division of acceleration into two components is due to the fact that each component of the total acceleration characterizes the change in speed in one of two parameters. Tangential acceleration characterizes the change in speed in magnitude. The tangential acceleration coincides in direction with the velocity vector if the velocity increases in magnitude and is directed opposite to the velocity if it decreases. When moving at a constant speed, the tangential acceleration is zero. The module of tangential acceleration is equal to:
Centripetal acceleration characterizes the change in speed in direction. When moving along a rectilinear trajectory, the centripetal acceleration is zero.
An important particular case of movement along a curved path is movement along a circle. The fact is that any smooth curved line can be replaced by a set of conjugate arcs of circles of different radii. Let there be some curved line. At each point of the curve, you can draw a set of circles tangent to it at that point. But among all these circles, there is one that best describes the curvature of the curve at a given point. The radius of this circle is called the radius of curvature of the line at that point. Thus, the movement of a body along an arbitrary curvilinear trajectory can be represented as a sequential movement along circles of different radii.
Let the body move along a curved path. Consider two very close points of the trajectory A and B. Since the points are very close to each other, we can assume that they lie on an arc of a circle with a radius equal to the radius of curvature of the trajectory in this part of the trajectory - R. Suppose that the speed of the body is constant in magnitude . In this case, the tangential acceleration is equal to zero and the total acceleration of the body is equal to centripetal. Triangle built on vectors v A, v B And Δv isosceles and similar to triangle AOB. So you can write:
Let Δt be the time during which the body moved from point A to point B. Since points A and B are located very close to each other (in the figure, for clarity, they are located far from each other), the chord AB practically coincides with the arc AB. Therefore, you can write: . And so we get:
Since the tangential acceleration is zero, it is centripetal acceleration. Thus, we obtain a formula for centripetal acceleration when a body moves along a curvilinear trajectory:
Here v is the instantaneous velocity of the body, and R is the radius of curvature of the trajectory at a given point.
Loading...