Let us now return to our task - to find the acceleration with which the body moves in a circle with a constant speed in absolute value.

Acceleration, as is known, is determined by the formula

where is the speed of the body at some initial moment of time, and is its speed after a period of time . In our case, the speed modules and are equal to each other.

Suppose that the body moves along a circle with a radius and that at some point in time it is at point A (Fig. 67).

What is the acceleration at this point? The speed at this point is directed tangentially to the circle at point A. After a second, the body is at point B, and its speed is now

directed tangentially to the circle at point B. Modulo velocity and 10 are equal (the lengths of the arrows and are the same).

We want to find the acceleration at point A of the circle (instantaneous acceleration). Therefore, we must take points A and B close to each other, so close that the arc, as it were, contracts into a point.

Let us first find out how this acceleration is directed.

Let's draw radii from the center O of the circle to points A and B. The radius of the circle is perpendicular to the tangent at the point of contact, therefore, the radii and are perpendicular to the vectors and To find out the direction of the acceleration vector, you need to find a vector equal to the difference of the vectors and Its direction is the direction of the vector acceleration. We already know how to subtract vectors (see § 6). To find the difference, we arrange the vectors so that they emanate from one point (Fig. 68), and connect their ends by directing the arrow from the subtracted to the reduced (from the end of the vector to the end of the vector. The vector is the difference of the vectors. Therefore, the acceleration is directed along the vector. What can be said about this direction?

The triangle (see Fig. 68) is isosceles. The angle at the vertex A is equal to the angle between the radii and (Fig. 67), since they are formed by mutually perpendicular sides. Points A and B are close to each other, so the angle is very small (close to zero). Each of the angles at the base of the triangle is close to a right angle, since the sum of the angles of a triangle is equal to two right angles. This means that the vector

perpendicular to the velocity vector. Hence, the acceleration is perpendicular to the velocity. But the speed is tangent to the circle at point A, and the tangent is perpendicular to the radius. This means that the acceleration is directed along the radius towards the center of the circle. That is why it is called centripetal acceleration.

When a body moves uniformly along a circle, acceleration at any point is perpendicular to the speed of movement and is directed towards the center of the circle.

This interesting feature of acceleration when moving along a circle with a constant modulo speed is shown in Figure 69.

Let us now find the modulus of centripetal acceleration. To do this, you need to find what the absolute value of the quantity is. From Figure 68, it can be seen that the modulus of the difference of vectors is equal to the length of the segment. Since the angle is very small, the segment differs little from the arc of a circle (shown by a dotted line) centered at point A. The radius of this circle is numerically equal to But, as we know (see § 24), the length of such an arc is Therefore, the absolute value of the acceleration is . But the angular velocity So

The acceleration of a body moving along a circle is the product of its linear velocity and the angular velocity of the turn of the radius drawn towards the body.

It is more convenient to represent the formula for centripetal acceleration in such a form that it includes the value of the radius of the circle along which the body moves. Since the angular and linear velocities are related by the relation ( - radius of the circle), then, substituting this expression into the formula, we get:

But therefore, the formula for centripetal acceleration can also be written as follows:

In uniform circular motion, a body moves with

acceleration, which is directed along the radius to the center of the circle and whose modulus is determined by the expression

Therefore, the opposite is also true: if it is known that the speed of the body is equal and the acceleration of the body at all points is perpendicular to the vector of its speed and is equal in absolute value, then it can be argued that such a body moves in a circle, the radius of which is determined by the formula

This means that if we know the initial velocity of the body and the absolute value of its centripetal acceleration, we can draw a circle along which the body will move and find its position at any time (the initial position of the body must, of course, be known). Thus, the main problem of mechanics will be solved.

Recall that we are interested in acceleration during uniform motion along a circle because any motion along a curvilinear trajectory is motion along arcs of circles of different radii.

Now we can say that with uniform motion at any point of a curvilinear trajectory, the body moves with acceleration directed towards the center of the circle of which the given trajectory is a part near this point. The numerical value of the acceleration depends on the speed of the body at this point and on the radius of the corresponding circle. Figure 70 shows some complex trajectory and indicates the vectors of centripetal acceleration at various points of the trajectory.

Task. The aircraft, leaving the peak, moves along an arc, which in its lower part is an arc of a circle with a radius of 500 m (Fig. 71). Calculate the acceleration of the aircraft at its nadir if its speed is 800 km/h and compare this value with the acceleration due to gravity.

4. A grinding wheel with a radius of 10 cm makes 1 revolution in 0.2 seconds during rotation. Find the speed of the points furthest from the axis of rotation.

5. A car is moving along a rounding of the road with a radius of 100 m at a speed of 54 km/h. What is the centripetal acceleration of the car?

6. The period of revolution of the first ship-satellite "Vostok" around the Earth was 90 minutes. The average height of the spacecraft above the Earth can be considered equal to 320 km. The radius of the Earth is 6,400 km. Calculate the speed of the ship.

7. What is the speed of the car if its wheels with a radius of 30 cm make 10 revolutions in 1 second?

8. Two pulleys, the radii of which are connected by an endless belt. The rotation period of a pulley with a smaller radius is 0.5 sec. What is the speed at which the belt points move? What is the period of rotation of the second pulley?

9. The moon moves around the Earth at a distance of 385,000 km from it, making one revolution in 27.3 days. Calculate the centripetal acceleration of the Moon.

In the study of motion in physics, the concept of a trajectory plays an important role. It is she who largely determines the type of movement of objects and, as a result, the type of formulas that describe this movement. One of the common trajectories of movement is a circle. In this article, we will consider what centripetal acceleration is when moving in a circle.

The concept of full acceleration

Before characterizing centripetal acceleration when moving along a circle, let's consider the concept of total acceleration. Under it, a physical quantity is assumed, which simultaneously describes the change in the value of the absolute and the velocity vector. In mathematical form, this definition looks like this:

Acceleration is the total derivative of speed with respect to time.

As is known, the velocity v¯ of the body at each point of the trajectory is tangential. This fact allows us to represent it as a product of the module v and the unit tangent vector u¯, i.e.:

Then the total acceleration can be calculated as follows:

a¯ = d(v*u¯)/dt = dv/dt*u¯ + v*du¯/dt

The value a¯ is the vector sum of two terms. The first term is directed tangentially (as the speed of the body) and is called tangential acceleration. It determines the rate of change of the speed modulus. The second term is the normal acceleration. We will consider it in more detail later in the article.

The above expression for the normal acceleration component an¯ can be written explicitly:

an¯ = v*du¯/dt = v*du¯/dl*dl/dt = v2/r*re¯

Here dl is the path traveled by the body along the trajectory in time dt, re¯ is the unit vector directed to the center of curvature of the trajectory, r is the radius of this curvature. The resulting formula leads to several important features of the an¯ component of the total acceleration:

• The value of an¯ increases as the square of the velocity and decreases inversely with the radius, which distinguishes it from the tangential component. The latter is not equal to zero only in the case of a change in the velocity modulus.
• Normal acceleration is always directed towards the center of curvature, which is why it is called centripetal.

Thus, the main condition for the existence of a nonzero quantity an¯ is the curvature of the trajectory. If such curvature does not exist (rectilinear displacement), then an¯ = 0, since r->∞.

Centripetal acceleration in circular motion

A circle is a geometric line, all points of which are at the same distance from some point. The latter is called the center of the circle, and the distance mentioned is its radius. If the speed of the body during rotation does not change in absolute value, then they speak of uniformly variable motion in a circle. Centripetal acceleration in this case is easy to calculate using one of the two formulas below:

Where ω is the angular velocity, measured in radians per second (rad/s). The second equality is obtained thanks to the formula for the relationship between the angular and linear velocities:

Centripetal and centrifugal forces

With a uniform motion of a body along a circle, centripetal acceleration occurs due to the action of the corresponding centripetal force. Its vector is always directed towards the center of the circle.

The nature of this force can be very diverse. For example, when a person spins a stone tied to a rope, then on its trajectory it is held by the tension force of the rope. Another example of the action of the centripetal force is the gravitational interaction between the Sun and the planets. It is it that makes all the planets and asteroids move in circular orbits. The centripetal force is not able to change the kinetic energy of the body, since it is directed perpendicular to its speed.

Each person could pay attention to the fact that while turning the car, for example, to the left, passengers are pressed to the right edge of the vehicle interior. This process is the result of the action of the centrifugal force of rotational motion. In fact, this force is not real, since it is due to the inertial properties of the body and its desire to move along a straight path.

The centrifugal and centripetal forces are equal in magnitude and opposite in direction. If this were not the case, then the circular trajectory of the body would be violated. If we take into account Newton's second law, then it can be argued that during rotational motion, centrifugal acceleration is equal to centripetal.

Aslamazov L.G. Circular motion // Kvant. - 1972. - No. 9. - S. 51-57.

By special agreement with the editorial board and the editors of the journal "Kvant"

To describe motion in a circle, along with linear velocity, the concept of angular velocity is introduced. If a point moving along a circle in time Δ t describes an arc, the angular measure of which is Δφ, then the angular velocity.

The angular velocity ω is related to the linear velocity υ by the relation υ = ω r, where r- the radius of the circle along which the point moves (Fig. 1). The concept of angular velocity is especially convenient for describing the rotation of a rigid body around an axis. Although the linear velocities of points located at different distances from the axis will not be the same, their angular velocities will be equal, and we can talk about the angular velocity of rotation of the body as a whole.

Task 1. Disk Radius r rolls without slipping on a horizontal plane. The speed of the center of the disk is constant and equal to υ p. With what angular velocity does the disk rotate in this case?

Each point of the disk participates in two movements - in translational motion with a speed υ p together with the center of the disk and in rotational motion around the center with a certain angular velocity ω.

To find ω, we use the absence of slippage, that is, the fact that at each moment of time the speed of a disk point in contact with the plane is zero. This means that for the point BUT(Fig. 2) the speed of translational motion υ p is equal in magnitude and opposite in direction to the linear speed of rotational motion υ vr = ω· r. From here we immediately get .

Task 2. Find speed points AT, With and D the same disk (Fig. 3).

Consider first the point AT. The linear speed of its rotational movement is directed vertically upwards and is equal to , that is, equal in magnitude to the speed of translational motion, which, however, is directed horizontally. Adding these two speeds vectorially, we find that the resulting speed υ B is equal in magnitude and forms an angle of 45º with the horizon. At the point With rotational and translational speeds are directed in the same direction. Resulting speed υ C equal to 2υ n and directed horizontally. Similarly, the speed of a point is found D(See Fig. 3).

Even in the case when the speed of a point moving along a circle does not change in magnitude, the point has some acceleration, as the direction of the velocity vector changes. This acceleration is called centripetal. It is directed towards the center of the circle and is equal to ( R is the radius of the circle, ω and υ are the angular and linear velocities of the point).

If the speed of a point moving along a circle changes not only in direction, but also in magnitude, then along with centripetal acceleration, there is also the so-called tangential acceleration. It is directed tangentially to the circle and is equal to the ratio (Δυ is the change in the velocity over time Δ t).

Task 3. Find Accelerations of Points BUT, AT, With and D disk radius r rolling without slipping on a horizontal plane. The speed of the center of the disk is constant and equal to υ p (Fig. 3).

In the coordinate system associated with the center of the disk, the disk rotates with an angular velocity ω, and the plane moves forward with a speed υ p. There is no slippage between the disk and the plane, therefore, . The speed of translational motion υ p does not change, therefore the angular velocity of rotation of the disk is constant and the points of the disk have only centripetal acceleration directed towards the center of the disk. Since the coordinate system moves without acceleration (with a constant speed υ n), then in a fixed coordinate system, the accelerations of the disk points will be the same.

Let us now turn to problems on the dynamics of rotational motion. Let us first consider the simplest case, when the motion along a circle occurs at a constant speed. Since the acceleration of the body is directed towards the center, then the vector sum of all forces applied to the body must also be directed towards the center, and according to Newton's second law.

It should be remembered that the right side of this equation includes only real forces acting on a given body from other bodies. No centripetal force does not occur when moving in a circle. This term is used simply to denote the resultant of forces applied to a body moving in a circle. Concerning centrifugal force, then it arises only when describing motion along a circle in a non-inertial (rotating) coordinate system. We will not use here the concept of centripetal and centrifugal force at all.

Task 4. Determine the smallest radius of curvature of the road that the car can pass at a speed of υ = 70 km/h and the coefficient of tire friction on the road k =0,3.

R = m g, road reaction force N and friction force F tr between the tires of the car and the road. Forces R and N directed vertically and equal in size: P = N. The friction force that prevents the car from slipping (“skidding”) is directed towards the center of the turn and imparts centripetal acceleration: . The maximum value of the friction force F tr max = k· N = k· m g, therefore, the minimum value of the radius of the circle, along which it is still possible to move at a speed υ, is determined from the equation . From here (m).

Road reaction force N when the car moves in a circle, it does not pass through the center of gravity of the car. This is due to the fact that its moment relative to the center of gravity must compensate for the frictional moment tending to overturn the car. The magnitude of the friction force is greater, the greater the speed of the car. At a certain speed, the moment of the friction force will exceed the moment of the reaction force and the car will overturn.

Task 5. At what speed is a car moving along an arc of a circle of radius R= 130 m, can tip over? The vehicle's center of gravity is at a height h= 1 m above road, vehicle track width l= 1.5 m (Fig. 4).

At the time of the car overturning, as the reaction force of the road N, and the force of friction F mp are attached to the "outer" wheel. When a car moves in a circle with a speed υ, a friction force acts on it. This force creates a moment about the vehicle's center of gravity. The maximum moment of the reaction force of the road N = m g relative to the center of gravity is (at the moment of overturning, the reaction force passes through the outer wheel). Equating these moments, we find the equation for the maximum speed at which the car will not tip over yet:

From where ≈ 30 m/s ≈ 110 km/h.

In order for a car to move at such a speed, a coefficient of friction is needed (see the previous problem).

A similar situation occurs when turning a motorcycle or bicycle. The frictional force that creates the centripetal acceleration has a moment about the center of gravity that tends to overturn the motorcycle. Therefore, to compensate for this moment by the moment of the reaction force of the road, the motorcyclist leans towards the turn (Fig. 5).

Task 6. A motorcyclist travels along a horizontal road at a speed of υ = 70 km/h, making a turn with a radius R\u003d 100 m. At what angle α to the horizon should he tilt in order not to fall?

The force of friction between the motorcycle and the road, as it imparts centripetal acceleration to the motorcyclist. Road reaction force N = m g. The condition of equality of the moments of the friction force and the reaction force relative to the center of gravity gives the equation: F tp l sinα = N· l cos α, where l- distance OA from the center of gravity to the trail of the motorcycle (see fig. 5).

Substituting here the values F tp and N, find something or . Note that the resultant of forces N and F tp at this angle of inclination of the motorcycle passes through the center of gravity, which ensures that the total moment of forces is equal to zero N and F tp .

In order to increase the speed of movement along the rounding of the road, the section of the road at the turn is made inclined. At the same time, in addition to the friction force, the reaction force of the road also participates in the creation of centripetal acceleration.

Task 7. With what maximum speed υ can a car move along an inclined track with an inclination angle α with a curvature radius R and coefficient of tire friction on the road k?

The force of gravity acts on the car m g, reaction force N, directed perpendicular to the track plane, and the friction force F tp directed along the track (Fig. 6).

Since we are not interested in this case, the moments of forces acting on the car, we have drawn all the forces applied to the center of gravity of the car. The vector sum of all forces must be directed towards the center of the circle along which the car is moving, and impart centripetal acceleration to it. Therefore, the sum of the projections of forces on the direction to the center (horizontal direction) is , that is

The sum of the projections of all forces on the vertical direction is zero:

N cos α - m g – F t p sinα = 0.

Substituting into these equations the maximum possible value of the friction force F tp = k N and excluding force N, find the maximum speed , with which it is still possible to move along such a track. This expression is always greater than the value corresponding to a horizontal road.

Having dealt with the dynamics of rotation, let's move on to problems for rotational motion in the vertical plane.

Task 8. mass car m= 1.5 t moves at a speed of υ = 70 km/h along the road shown in Figure 7. Road sections AB and sun can be considered arcs of circles of radius R= 200 m touching each other at a point AT. Determine the pressure force of the car on the road in points BUT and With. How does the pressure force change when a car passes a point AT?

At the point BUT gravity is acting on the car R = m g and road reaction force N A. The vector sum of these forces must be directed to the center of the circle, that is, vertically downwards, and create a centripetal acceleration: , whence (H). The pressure force of the car on the road is equal in magnitude and opposite in direction to the reaction force. At the point With the vector sum of the forces is directed vertically upwards: and (H). Thus, at the point BUT the force of pressure is less than the force of gravity, and at a point With- more.

At the point AT the car moves from a convex section of the road to a concave one (or vice versa). When driving on a convex section, the projection of gravity in the direction towards the center must exceed the reaction force of the road NB 1 , and . When driving on a concave section of the road, on the contrary, the reaction force of the road N B 2 outperforms the projection of gravity: .

From these equations we obtain that when passing through the point AT the pressure force of the car on the road changes abruptly by a value of ≈ 6·10 3 N. Of course, such shock loads act destructively both on the car and on the road. Therefore, roads and bridges always try to make their curvature change smoothly.

When a car moves along a circle at a constant speed, the sum of the projections of all forces on the direction tangent to the circle must be equal to zero. In our case, the tangential component of gravity is balanced by the force of friction between the wheels of the car and the road.

The magnitude of the friction force is controlled by the torque applied to the wheels by the motor. This moment tends to cause the wheels to slip relative to the road. Therefore, a friction force arises that prevents slippage and is proportional to the applied moment. The maximum value of the friction force is k N, where k is the coefficient of friction between the car's tires and the road, N- force of pressure on the road. When the car moves down, the friction force plays the role of a braking force, and when moving up, on the contrary, the role of the traction force.

Task 9. Vehicle mass m= 0.5 t, moving at a speed of υ = 200 km/h, makes a "dead loop" of radius R= 100 m (Fig. 8). Determine the pressure force of the car on the road at the top of the loop BUT; at the point AT, the radius vector of which makes an angle α = 30º with the vertical; at the point With where the speed of the car is directed vertically. Is it possible for a car to move along a loop at such a constant speed with a coefficient of tire friction on the road k = 0,5?

At the top of the loop, the force of gravity and the reaction force of the road N A directed vertically down. The sum of these forces creates a centripetal acceleration: . So N.

The pressure force of the car on the road is equal in magnitude and opposite in direction to the force N A.

At the point AT centripetal acceleration is created by the sum of the reaction force and the projection of gravity on the direction towards the center: . From here N.

It is easy to see that NB > N A; as the angle α increases, the reaction force of the road increases.

At the point With reaction force H; centripetal acceleration at this point is created only by the reaction force, and gravity is directed tangentially. When moving along the lower part of the loop, the reaction force will also exceed the maximum value H reaction force has at the point D. Meaning , thus, is the minimum value of the reaction force.

The speed of the car will be constant if the tangential component of gravity does not exceed the maximum friction force k N at all points in the loop. This condition is certainly satisfied if the minimum value exceeds the maximum value of the tangential component of the weight force. In our case, this maximum value is equal to m g(it is reached at the point With), and the condition is satisfied for k= 0.5, υ = 200 km/h, R= 100 m.

Thus, in our case, the movement of the car along the "dead loop" at a constant speed is possible.

Consider now the movement of the car along the "dead loop" with the engine off. As already noted, usually the moment of the friction force opposes the moment applied to the wheels by the motor. When the car is moving with the engine off, this moment is absent, and the friction force between the wheels of the car and the road can be neglected.

The speed of the car will no longer be constant - the tangential component of gravity slows down or speeds up the movement of the car along the "dead loop". The centripetal acceleration will also change. It is created, as usual, by the resultant reaction force of the road and the projection of gravity on the direction towards the center of the loop.

Task 10. What is the minimum speed the car should have at the bottom of the loop D(see Fig. 8) in order to make it with the engine off? What will be the pressure force of the car on the road at the point AT? Loop radius R= 100 m, vehicle weight m= 0.5 t.

Let's see what is the minimum speed the car can have at the top of the loop BUT to keep moving around the circle?

The centripetal acceleration at that point on the road is created by the sum of the force of gravity and the reaction force of the road . The lower the speed of the car, the lower the reaction force. N A. With a value, this force vanishes. At a lower speed, gravity will exceed the value needed to create centripetal acceleration, and the car will lift off the road. At speed, the reaction force of the road vanishes only at the top of the loop. Indeed, the speed of the car in other sections of the loop will be greater, and as it is easy to see from the solution of the previous problem, the reaction force of the road will also be greater than at the point BUT. Therefore, if the car at the top of the loop has speed , then it will not leave the loop anywhere.

Now we determine what speed the car should have at the bottom of the loop D to the top of the loop BUT his speed. To find the speed υ D you can use the law of conservation of energy, as if the car was moving only under the influence of gravity. The fact is that the reaction force of the road at each moment is directed perpendicular to the movement of the car, and, therefore, its work is zero (recall that the work Δ A = F·Δ s cos α, where α is the angle between the force F and direction of movement Δ s). The friction force between the wheels of the car and the road when driving with the engine off can be neglected. Therefore, the sum of the potential and kinetic energy of the car when driving with the engine off does not change.

Let us equate the values ​​of the energy of the car at the points BUT and D. In this case, we will count the height from the level of the point D, that is, the potential energy of the car at this point will be considered equal to zero. Then we get

Substituting here the value for the desired speed υ D, we find: ≈ 70 m/s ≈ 260 km/h.

If the car enters the loop at this speed, it will be able to complete it with the engine off.

Let us now determine with what force the car will press on the road at the point AT. Vehicle speed at point AT again it is easy to find from the law of conservation of energy:

Substituting the value here, we find that the speed .

Using the solution of the previous problem, for a given speed, we find the pressure force at the point B:

Similarly, you can find the pressure force at any other point of the "dead loop".

Exercises

1. Find the angular velocity of an artificial Earth satellite rotating in a circular orbit with a period of revolution T= 88 min. Find the linear speed of this satellite, if it is known that its orbit is located at a distance R= 200 km from the Earth's surface.

2. Disk radius R placed between two parallel bars. The rails move at speeds υ 1 and υ 2. Determine the angular velocity of the disc and the velocity of its center. There is no slippage.

3. The disc rolls on a horizontal surface without slipping. Show that the ends of the velocity vectors of the vertical diameter points are on the same straight line.

4. The plane moves in a circle with a constant horizontal speed υ = 700 km/h. Define Radius R this circle if the body of the aircraft is inclined at an angle α = 5°.

5. Mass load m\u003d 100 g, suspended on a thread of length l= 1 m, rotates uniformly in a circle in a horizontal plane. Find the period of rotation of the load if, during its rotation, the thread is deflected vertically by an angle α = 30°. Also determine the tension of the thread.

6. The car moves at a speed υ = 80 km/h along the inner surface of a vertical cylinder of radius R= 10 m in a horizontal circle. At what minimum coefficient of friction between the tires of the car and the surface of the cylinder is this possible?

7. Mass load m suspended from an inextensible thread, the maximum possible tension of which is 1.5 m g. At what maximum angle α can the thread be deflected from the vertical so that the thread does not break during further movement of the load? What will be the tension of the thread at the moment when the thread makes an angle α/2 with the vertical?

Answers

I. Angular velocity of an artificial Earth satellite ≈ 0.071 rad/s. Linear velocity of the satellite υ = ω· R. where R is the radius of the orbit. Substituting here R = R 3 + h, where R 3 ≈ 6400 km, we find υ ≈ 467 km/s.

2. Two cases are possible here (Fig. 1). If the angular velocity of the disk is ω, and the velocity of its center is υ, then the velocities of the points in contact with the rails will be respectively equal to

in case a) υ 1 = υ + ω R, υ 2 = υ - ω R;

in case b) υ 1 = υ + ω R, υ 2 = ω R – υ.

(We assumed for definiteness that υ 1 > υ 2). Solving these systems, we find:

a)

b)

3. Speed ​​of any point M lying on the segment OV(see Fig. 2) is found by the formula υ M = υ + ω· rM, where rM- distance from point M to the center of the disk O. For any point N belonging to the segment OA, we have: υ N = υ – ω· rN, where r N- distance from point N to the center. Denote by ρ the distance from any point of the diameter VA to the point BUT contact of the disk with the plane. Then it is obvious that rM = ρ – R and r N = R – ρ = –(ρ – R). where R is the disk radius. Therefore, the speed of any point on the diameter VA is found by the formula: υ ρ = υ + ω (ρ – R). Since the disk rolls without slipping, then for the speed υ ρ we obtain υ ρ = ω · ρ. It follows from this that the ends of the velocity vectors are on the straight line emanating from the point BUT and inclined to the diameter VA at an angle proportional to the angular velocity of rotation of the disk ω.

The proved statement allows us to conclude that the complex movement of points located on the diameter VA, can be considered at any given moment as a simple rotation around a fixed point BUT with an angular velocity ω equal to the angular velocity of rotation around the center of the disk. Indeed, at each moment the velocities of these points are directed perpendicular to the diameter VA, and are equal in magnitude to the product of ω and the distance to the point BUT.

It turns out that this statement is true for any point on the disk. Moreover, it is a general rule. With any movement of a rigid body, at every moment there is an axis around which the body simply rotates - the instantaneous axis of rotation.

4. The plane is affected (see Fig. 3) by gravity R = m g and lifting force N, directed perpendicular to the plane of the wings (since the aircraft is moving at a constant speed, the thrust force and the drag force of the air balance each other). Resultant force R

6. The car is affected (Fig. 5) by gravity R = m g, the reaction force from the side of the cylinder N and friction force F tp . Since the car is moving in a horizontal circle, the forces R and F tp balance each other, and the force N creates centripetal acceleration. The maximum value of the friction force is related to the reaction force N ratio: F tp = k N. As a result, we obtain a system of equations: , from which the minimum value of the friction coefficient is found

7. The load will move in a circle of radius l(Fig. 6). The centripetal acceleration of the load (υ - the speed of the load) is created by the difference in the values ​​of the thread tension force T and gravity projections m g thread direction: . So , where β is the angle formed by the thread with the vertical. As the load descends, its speed will increase and the angle β will decrease. The thread tension will become maximum at the angle β = 0 (at the moment when the thread is vertical): . The maximum speed of the load υ 0 is found from the angle α, by which the thread is deflected, from the law of conservation of energy:

Using this ratio, for the maximum value of the thread tension, we obtain the formula: T max = m g(3 – 2 cos α). According to the task T m ax = 2m g. Equating these expressions, we find cos α = 0.5 and, therefore, α = 60°.

Let us now determine the tension of the thread at . The speed of the load at this moment is also found from the law of conservation of energy:

Substituting the value of υ 1 into the formula for the tension force, we find: