Theory
If a body is thrown at an angle to the horizon, then in flight it is affected by gravity and air resistance. If the resistance force is neglected, then the only force left is the force of gravity. Therefore, due to Newton's 2nd law, the body moves with an acceleration equal to the free fall acceleration; acceleration projections on the coordinate axes are a x = 0, and at= -g.
Any complex movement of a material point can be represented as an imposition of independent movements along the coordinate axes, and in the direction of different axes, the type of movement may differ. In our case, the motion of a flying body can be represented as a superposition of two independent motions: uniform motion along the horizontal axis (X-axis) and uniformly accelerated motion along the vertical axis (Y-axis) (Fig. 1).
The velocity projections of the body therefore change with time as follows:
,
where is the initial speed, α is the throwing angle.
The body coordinates therefore change like this:
With our choice of the origin of coordinates, the initial coordinates (Fig. 1) Then
The second value of the time at which the height is equal to zero is equal to zero, which corresponds to the moment of throwing, i.e. this value also has a physical meaning.
The flight range is obtained from the first formula (1). Flight range is the value of the coordinate X at the end of the flight, i.e. at a point in time equal to t0. Substituting the value (2) into the first formula (1), we obtain:
| . | (3) |
From this formula it can be seen that longest range flight is achieved at a throw angle of 45 degrees.
The highest lifting height of the thrown body can be obtained from the second formula (1). To do this, you need to substitute in this formula the value of time equal to half the flight time (2), because it is at the midpoint of the trajectory that the flight altitude is maximum. Carrying out calculations, we get
Consider the motion of a body thrown horizontally and moving under the action of gravity alone (neglecting air resistance). For example, imagine that a ball lying on a table is given a push, and it rolls to the edge of the table and begins to fall freely, having an initial velocity directed horizontally (Fig. 174).
Let's project the movement of the ball on the vertical axis and on the horizontal axis. The movement of the projection of the ball onto the axis is a movement without acceleration with a speed of ; the motion of the projection of the ball on the axis is a free fall with acceleration beyond the initial velocity under the action of gravity. We know the laws of both motions. The velocity component remains constant and equal to . The component grows in proportion to time: . The resulting speed is easily found using the parallelogram rule, as shown in Fig. 175. It will lean downward and its slope will increase with time.
Rice. 174. Movement of a ball rolling off a table
Rice. 175. A ball thrown horizontally with a speed has a speed at the moment
Find the trajectory of a body thrown horizontally. The coordinates of the body at the moment of time matter
To find the trajectory equation, we express from (112.1) the time through and substitute this expression in (112.2). As a result, we get
The graph of this function is shown in Fig. 176. The ordinates of the trajectory points turn out to be proportional to the squares of the abscissas. We know that such curves are called parabolas. A parabola depicted a graph of the path of uniformly accelerated motion (§ 22). Thus, a freely falling body whose initial velocity is horizontal moves along a parabola.
The path traveled in the vertical direction does not depend on the initial speed. But the path traveled in the horizontal direction is proportional to the initial speed. Therefore, with a large horizontal initial velocity, the parabola along which the body falls is more elongated in the horizontal direction. If a jet of water is fired from a horizontally located tube (Fig. 177), then individual particles of water will, like the ball, move along a parabola. The more open the tap through which water enters the tube, the greater the initial velocity of the water and the farther from the tap the jet gets to the bottom of the cuvette. By placing a screen with parabolas pre-drawn on it behind the jet, one can verify that the water jet really has the shape of a parabola.
112.1. What will be the speed of a body thrown horizontally at a speed of 15 m/s after 2 seconds of flight? At what moment will the velocity be directed at an angle of 45° to the horizontal? Ignore air resistance.
112.2. A ball rolled down from a table of height 1m fell at a distance of 2m from the edge of the table. What was the horizontal speed of the ball? Ignore air resistance.
Here is the initial speed of the body, is the speed of the body at the moment of time t, s- horizontal flight distance, h is the height above the ground from which a body is thrown horizontally with a speed .
1.1.33. Kinematic equations of velocity projection:
1.1.34. Kinematic coordinate equations:
1.1.35. body speed at the time t:
In the moment falling to the ground y=h, x = s(Fig. 1.9).
1.1.36. Maximum range horizontal flight:
1.1.37. Height above ground from which the body is thrown
horizontally:
Motion of a body thrown at an angle α to the horizon
with initial speed
1.1.38. The trajectory is a parabola(Fig. 1.10). Curvilinear movement along a parabola is due to the result of adding two rectilinear movements: uniform movement along the horizontal axis and equally variable movement along the vertical axis.
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| Rice. 1.10 |
( is the initial speed of the body, are the projections of the velocity on the coordinate axes at the moment of time t, is the flight time of the body, hmax- the maximum height of the body, smax is the maximum horizontal flight distance of the body).
1.1.39. Kinematic projection equations:
; 
1.1.40. Kinematic coordinate equations:
; 
1.1.41. The height of the body lift to the top point of the trajectory:
At the moment of time , (Figure 1.11).
1.1.42. Max Height body lift: 
1.1.43. Body flight time: 
At the point in time , (Fig. 1.11).
1.1.44. Maximum horizontal flight range of the body:
1.2. Basic equations of classical dynamics
Dynamics(from Greek. dynamic- force) - a branch of mechanics devoted to the study of the movement of material bodies under the action of forces applied to them. Classical dynamics are based on Newton's laws . All equations and theorems necessary for solving problems of dynamics are obtained from them.
1.2.1. Inertial Reporting System - it is a frame of reference in which the body is at rest or moving uniformly and in a straight line.
1.2.2. Strength is the result of the interaction of the body with environment. One of the simplest definitions of force: the influence of a single body (or field) that causes acceleration. Currently, four types of forces or interactions are distinguished:
· gravitational(manifested in the form of forces gravity);
· electromagnetic(existence of atoms, molecules and macrobodies);
· strong(responsible for the connection of particles in nuclei);
· weak(responsible for the decay of particles).
1.2.3. The principle of superposition of forces: if several forces act on a material point, then the resulting force can be found by the rule of vector addition:
.
The mass of a body is a measure of the inertia of a body. Any body resists when trying to set it in motion or change the module or direction of its speed. This property is called inertia.
1.2.5. Pulse(momentum) is the product of the mass t body by its speed v:
1.2.6. Newton's first law: Any material point (body) maintains a state of rest or uniform rectilinear motion until the impact from other bodies causes her (him) to change this state.
1.2.7. Newton's second law(basic equation of the dynamics of a material point): the rate of change of the momentum of the body is equal to the force acting on it (Fig. 1.11):
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| Rice. 1.11 | Rice. 1.12 |
The same equation in projections onto the tangent and normal to the point trajectory:
and 
.
1.2.8. Newton's third law: the forces with which two bodies act on each other are equal in magnitude and opposite in direction (Fig. 1.12):
1.2.9. Law of conservation of momentum for a closed system: the momentum of a closed system does not change in time (Fig. 1.13):
,
where P is the number of material points (or bodies) included in the system.
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| Rice. 1.13 |
The law of conservation of momentum is not a consequence of Newton's laws, but is fundamental law of nature, which knows no exceptions, and is a consequence of the homogeneity of space.
1.2.10. The basic equation of the dynamics of the translational motion of a system of bodies:
where is the acceleration of the center of inertia of the system; is the total mass of the system from P material points.
1.2.11. Center of mass of the system material points (Fig. 1.14, 1.15):
.
The law of motion of the center of mass: the center of mass of the system moves like a material point, the mass of which is equal to the mass of the entire system and which is affected by a force equal to the vector sum of all forces acting on the system.
1.2.12. Impulse of the body system:
where is the speed of the center of inertia of the system.
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| Rice. 1.14 | Rice. 1.15 |
1.2.13. Theorem on the motion of the center of mass: if the system is in an external stationary uniform force field, then no actions inside the system can change the motion of the center of mass of the system:
.
1.3. Forces in mechanics
1.3.1. Body weight relationship with gravity and support reaction:
Free fall acceleration (Fig. 1.16).
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| Rice. 1.16 |
Weightlessness is a state in which the weight of a body is zero. In a gravitational field, weightlessness occurs when a body moves only under the action of gravity. If a a = g, then p=0.
1.3.2. Relationship between weight, gravity and acceleration:
1.3.3. sliding friction force(Fig. 1.17):
where is the coefficient of sliding friction; N is the force of normal pressure.
1.3.5. Basic ratios for a body on an inclined plane(Fig. 1.19). :
· friction force: 
;
· resultant force: ;
· rolling force: 
;
· acceleration:
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| Rice. 1.19 |
1.3.6. Hooke's law for a spring: spring extension X proportional to the elastic force or external force:
where k- spring stiffness.
1.3.7. Potential energy elastic spring :
1.3.8. The work done by the spring:
1.3.9. Voltage- a measure of internal forces arising in a deformable body under the influence of external influences (Fig. 1.20):
where is the cross-sectional area of the rod, d is its diameter, is the initial length of the rod, is the increment of the rod length.
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| Rice. 1.20 | Rice. 1.21 |
1.3.10. Strain diagram - plot of normal stress σ = F/S on relative elongation ε = Δ l/l when stretching the body (Fig. 1.21).
1.3.11. Young's modulus is the value characterizing the elastic properties of the rod material:
1.3.12. Bar length increment proportional to voltage:
1.3.13. Relative longitudinal tension (compression):
1.3.14. Relative transverse tension (compression):
where is the initial transverse dimension of the rod.
1.3.15. Poisson's ratio- the ratio of the relative transverse tension of the rod to the relative longitudinal tension:
1.3.16. Hooke's law for a rod: relative increment of the length of the rod is directly proportional to the stress and inversely proportional to the Young's modulus:
1.3.17. Bulk potential energy density:
1.3.18. Relative shift ( pic1.22, 1.23 ):
where is the absolute shift.
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| Rice. 1.22 | Fig.1.23 |
1.3.19. Shear modulusG- a value that depends on the properties of the material and is equal to such a tangential stress at which (if such huge elastic forces were possible).
1.3.20. Tangential elastic stress:
1.3.21. Hooke's law for shear:
1.3.22. Specific potential energy bodies in shear:
1.4. Non-inertial frames of reference
Non-inertial frame of reference is an arbitrary frame of reference that is not inertial. Examples of non-inertial systems: a system moving in a straight line with constant acceleration, as well as a rotating system.
The forces of inertia are due not to the interaction of bodies, but to the properties of the non-inertial frames of reference themselves. Newton's laws do not apply to inertial forces. The forces of inertia are not invariant with respect to the transition from one frame of reference to another.
In a non-inertial system, you can also use Newton's laws if you introduce inertial forces. They are fictitious. They are introduced specifically to use Newton's equations.
1.4.1. Newton's equation for non-inertial frame of reference
where is the acceleration of a body of mass t relative to the non-inertial system; – force of inertia is a fictitious force due to the properties of the frame of reference.
1.4.2. Centripetal force- inertia force of the second kind, applied to a rotating body and directed along the radius to the center of rotation (Fig. 1.24):
,
where is the centripetal acceleration.
1.4.3. Centrifugal force- the force of inertia of the first kind, applied to the connection and directed along the radius from the center of rotation (Fig. 1.24, 1.25):
,
where is the centrifugal acceleration.
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| Rice. 1.24 | Rice. 1.25 |
1.4.4. Gravity acceleration dependence g from the latitude of the area is shown in fig. 1.25.
Gravity is the result of the addition of two forces: and; thus, g(and hence mg) depends on latitude:
,
where ω is the angular velocity of the Earth's rotation.
1.4.5. Coriolis force- one of the forces of inertia that exists in a non-inertial frame of reference due to rotation and the laws of inertia, which manifests itself when moving in a direction at an angle to the axis of rotation (Fig. 1.26, 1.27).
where is the angular velocity of rotation.
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| Rice. 1.26 | Rice. 1.27 |
1.4.6. Newton's equation for non-inertial frames of reference, taking into account all forces, takes the form
where is the force of inertia due to the translational motion of a non-inertial frame of reference; and 
– two inertial forces due to the rotational motion of the reference frame; is the acceleration of the body relative to the non-inertial frame of reference.
1.5. Energy. Job. Power.
Conservation laws
1.5.1. Energy- universal measure various forms motion and interaction of all kinds of matter.
1.5.2. Kinetic energy is the function of the state of the system, determined only by the speed of its movement:
Kinetic energy of the body - scalar physical quantity equal to half the product of the mass m body per square of its speed.
1.5.3. Theorem on the change in kinetic energy. The work of the resultant forces applied to the body is equal to the change in the kinetic energy of the body, or, in other words, the change in the kinetic energy of the body is equal to the work A of all forces acting on the body.
1.5.4. Relationship between kinetic energy and momentum:
1.5.5. Force work is a quantitative characteristic of the process of energy exchange between interacting bodies. Work in mechanics 
.
1.5.6. Work of a constant force:
If a body is moving in a straight line and a constant force is acting on it F, which makes a certain angle α with the direction of movement (Fig. 1.28), then the work of this force is determined by the formula:
,
where F is the modulus of force, ∆r is the modulus of displacement of the force application point, is the angle between the direction of force and displacement.
If a< /2, то работа силы положительна. Если >/2, then the work done by the force is negative. At = /2 (the force is directed perpendicular to the displacement), then the work of the force is zero.
| Rice. 1.28 | Rice. 1.29 |
Work of constant force F when moving along the axis x at a distance (Fig. 1.29) is equal to the force projection on this axis multiplied by displacement:
.
On fig. 1.27 shows the case when A < 0, т.к. >/2 - obtuse angle.
1.5.7. elementary work d A strength F on elementary displacement d r called a scalar physical quantity equal to dot product moving forces:
1.5.8. Variable force work on the trajectory section 1 - 2 (Fig. 1.30):
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| Rice. 1.30 |
1.5.9. Instant Power is equal to the work done per unit of time:
.
1.5.10. Average power for a period of time:
1.5.11. Potential energy body at a given point is a scalar physical quantity, equal to the work done by the potential force when moving the body from this point to another taken as the zero of the potential energy reference.
Potential energy is determined up to some arbitrary constant. This is not reflected in the physical laws, since they include either the difference in potential energies in two positions of the body or the derivative of the potential energy with respect to coordinates.
Therefore, the potential energy in a certain position is considered equal to zero, and the energy of the body is measured relative to this position (zero reference level).
1.5.12. The principle of minimum potential energy. Any closed system tends to move to a state in which its potential energy is minimal.
1.5.13. The work of conservative forces is equal to the change in potential energy
.
1.5.14. Vector circulation theorem: if the circulation of any force vector is zero, then this force is conservative.
The work of conservative forces along a closed loop L is zero(Fig. 1.31):
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| Rice. 1.31 |
1.5.15. Potential energy of gravitational interaction between the masses m and M(Fig. 1.32):
1.5.16. Potential energy of a compressed spring(Fig. 1.33):
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| Rice. 1.32 | Rice. 1.33 |
1.5.17. Total mechanical energy of the system is equal to the sum of kinetic and potential energies:
E = E to + E P.
1.5.18. Potential energy of the body on high h over the ground
E n = mgh.
1.5.19. Relationship between potential energy and force:
Or 
or
1.5.20. Law of conservation of mechanical energy(for a closed system): the total mechanical energy of a conservative system of material points remains constant:
1.5.21. Law of conservation of momentum for a closed system of bodies:
1.5.22. Law of conservation of mechanical energy and momentum with absolutely elastic central impact (Fig. 1.34):
where m 1 and m 2 - masses of bodies; and are the speeds of the bodies before the impact.
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| Rice. 1.34 | Rice. 1.35 |
1.5.23. Body speeds after a perfectly elastic impact (Fig. 1.35):
.
1.5.24. Body speed after a completely inelastic central impact (Fig. 1.36):
1.5.25. Law of conservation of momentum when the rocket is moving (Fig. 1.37):
where and are the mass and speed of the rocket; and the mass and velocity of the ejected gases.
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| Rice. 1.36 | Rice. 1.37 | |
1.5.26. Meshchersky equation for the rocket.
Using several examples (which I initially solved, as usual, on otvet.mail.ru), we will consider a class of problems of elementary ballistics: the flight of a body launched at an angle to the horizon with a certain initial speed, without taking into account air resistance and curvature earth's surface(that is, the direction of the free fall acceleration vector g is assumed to be unchanged).
Task 1. The flight range of the body is equal to the height of its flight above the Earth's surface. At what angle is the body thrown? (in some sources, for some reason, the wrong answer is given - 63 degrees).
Let's denote the flight time as 2*t (then during t the body rises, and during the next interval t it descends). Let the horizontal component of the velocity be V1 and the vertical component V2. Then the flight range S = V1*2*t. Flight altitude H \u003d g * t * t / 2 \u003d V2 * t / 2. Equate
S=H
V1*2*t = V2*t/2
V2/V1 = 4
The ratio of vertical and horizontal speeds is the tangent of the required angle α, whence α = arctan(4) = 76 degrees.
Task 2. A body is thrown from the Earth's surface with a speed V0 at an angle α to the horizon. Find the radius of curvature of the body trajectory: a) at the beginning of the movement; b) at the top of the trajectory.
In both cases, the source of the curvilinear motion is gravity, that is, the free fall acceleration g, directed vertically downwards. All that is required here is to find the projection g, perpendicular to the current velocity V, and equate it to the centripetal acceleration V^2/R, where R is the desired radius of curvature.
As can be seen from the figure, to start the movement, we can write
gn = g*cos(a) = V0^2/R
whence the desired radius R = V0^2/(g*cos(a))
For the upper point of the trajectory (see figure) we have
g = (V0*cos(a))^2/R
whence R = (V0*cos(a))^2/g
Task 3. (variation on a theme) The projectile moved horizontally at a height h and exploded into two identical fragments, one of which fell to the ground in time t1 after the explosion. How long after the first piece falls will the second one fall?
Whatever vertical velocity V the first fragment acquires, the second fragment acquires the same vertical velocity in absolute value, but directed in opposite side(this follows from the same mass of fragments and conservation of momentum). In addition, V is directed downward, because otherwise the second fragment will arrive on the ground BEFORE the first one.
h = V*t1+g*t1^2/2
V = (h-g*t1^2/2)/t1
The second one will fly up, lose vertical speed after the time V/g, and then after the same time will fly down to the initial height h, and the time t2 of its delay relative to the first fragment (not the flight time from the moment of explosion) will be
t2 = 2*(V/g) = 2h/(g*t1)-t1
updated on 2018-06-03
Quote:
A stone is thrown at a speed of 10 m/s at an angle of 60° to the horizontal. Determine the tangential and normal acceleration of the body after 1.0 s after the start of movement, the radius of curvature of the trajectory at this point in time, the duration and range of the flight. What angle does the total acceleration vector form with the velocity vector at t = 1.0 s
After 1 second, the vertical velocity will be 8.66 - 9.8*1 = -1.14 m/s (downwards). This means that the angle of velocity to the horizon will be arctan(1.14/5) = 12.8° (down). Since the total acceleration here is unique and unchanged (this is the acceleration of free fall g pointing vertically downwards), then the angle between the velocity of the body and g at this point in time will be 90-12.8 = 77.2°.
Tangential acceleration is a projection g to the direction of the velocity vector, which means it is g*sin(12.8) = 2.2 m/s2. Normal acceleration is a projection perpendicular to the velocity vector g, it is equal to g*cos(12.8) = 9.56 m/s2. And since the latter is related to the speed and radius of curvature by the expression V^2/R, we have 9.56 = (5*5 + 1.14*1.14)/R, whence the required radius R = 2.75 m.
If the velocity is not directed vertically, then the motion of the body will be curvilinear.
Consider the motion of a body thrown horizontally from a height h with a speed (Fig. 1). Air resistance will be neglected. To describe the movement, it is necessary to choose two coordinate axes - Ox and Oy. The origin of coordinates is compatible with the initial position of the body. Figure 1 shows that .
Then the motion of the body will be described by the equations:
An analysis of these formulas shows that in the horizontal direction the speed of the body remains unchanged, i.e. the body moves uniformly. In the vertical direction, the body moves uniformly with acceleration, i.e., in the same way as a freely falling body without an initial velocity. Let's find the trajectory equation. To do this, from equation (1) we find the time and, substituting its value in formula (2), we obtain
This is the equation of a parabola. Therefore, a body thrown horizontally moves along a parabola. The speed of the body at any moment of time is directed tangentially to the parabola (see Fig. 1). The modulus of speed can be calculated using the Pythagorean theorem:
Knowing the height h from which the body is thrown, you can find the time after which the body will fall to the ground. At this moment, the y-coordinate is equal to the height: . From equation (2) we find
