Centripetal acceleration of the vehicle while driving. Centripetal acceleration when moving in a circle: concept and formulas

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 consequence, the type of formulas with which this movement is described. One of the most common trajectories is a circle. In this article, we will consider what centripetal acceleration is when moving in a circle.

Understanding full acceleration

Before characterizing centripetal acceleration when moving around a circle, let us consider the concept of full acceleration. It is believed under it physical quantity, which simultaneously describes the change in the value of the absolute and the velocity vector. In mathematical terms, this definition looks like this:

Acceleration is the full time derivative of speed.

As is known, the velocity v¯ of the body at each point of the trajectory is directed tangentially. This fact allows us to represent it as the product of the modulus v and the unit tangent vector u¯, that is:

Then the total acceleration can be calculated as follows:

a¯ = d (v * u¯) / dt = dv / dt * u¯ + v * du¯ / dt

The quantity a¯ is the vector sum of two terms. The first term is tangential (like the speed of a body) and is called tangential acceleration. It determines the rate of change of the speed module. The second term is normal acceleration... Let's consider it in more detail later in the article.

The expression obtained above 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 traversed by the body along the trajectory in time dt, re¯ is the unit vector directed to the center of the 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 quantity an¯ grows as the square of the velocity and decreases in inverse proportion to the radius, which distinguishes it from the tangential component. The latter is not equal to zero only if the speed module changes.
• 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 (straight-line displacement), then an¯ = 0, since r-> ∞.

Acceleration centripetal when moving in a circle

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 mentioned distance is its radius. If the speed of the body during rotation does not change in absolute value, then one speaks of an equally variable motion along a circle. In this case, the centripetal acceleration can be easily calculated 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 relationship between the angular and linear velocities:

Forces centripetal and centrifugal

At uniform movement the body around the circumference of the centripetal acceleration arises 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 untwists 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 centripetal force is the gravitational interaction between the Sun and the planets. It is it that makes all 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 perpendicularly to its speed.

Each person could pay attention to the fact that when the car turns, for example, to the left, the passengers are pressed against the right edge of the vehicle interior. This process is the result of the centrifugal force of the rotational movement. In fact, this force is not real, since it is due to the inertial properties of the body and its tendency to move along a straight trajectory.

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's movement 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 acceleration.

Aslamazov L.G. Motion in a circle // Kvant. - 1972. - No. 9. - S. 51-57.

By special agreement with the editorial board and editors of the Kvant magazine

To describe motion along a circle, along with linear velocity, the concept of angular velocity is introduced. If a point while 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 useful for describing rotation. solid around the axis. Although the linear velocities at 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.

Problem 1... Radius disc r rolls without slipping on a horizontal plane. The speed of the center of the disc is constant and equal to υ p. With what angular speed does the disc rotate?

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 point on the disk in contact with the plane is zero. This means that for the point A(Fig. 2) the speed of translational motion υ p is equal in magnitude and opposite in the direction of the linear speed of rotational motion υ bp = ω · r... From here we immediately get.

Objective 2. Find the velocities of the points V, WITH and D the same disk (Fig. 3).

Consider first the point V... The linear speed of its rotational motion is directed vertically upward and is equal to , that is, the value is equal to the speed of translational motion, which, however, is directed horizontally. Adding these two velocities vectorially, we find that the resulting velocity υ B is equal in size and forms an angle of 45º with the horizon. At the point WITH rotational and translational speeds are directed in one direction. The resulting speed υ C is equal to 2υ p and is directed horizontally. The speed of the point is found in a similar way 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, since the direction of the velocity vector changes. This acceleration is called centripetal... It is directed to 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 in 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 value of velocity over time Δ t).

Objective 3. Find Acceleration Points A, V, 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 translationally with a velocity 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 speed 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 velocity υ p), then in a stationary coordinate system the accelerations of the points of the disk will be the same.

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

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

R = m g, the reaction force of the road N and friction force F TP between the tires of the car and the road. Forces R and N directed vertically and equal in size: P = N... The frictional force that prevents the vehicle from slipping ("skidding") is directed towards the center of rotation and imparts centripetal acceleration:. Maximum friction force F tr max = k· N = k· m g, therefore, the minimum value of the radius of the circle along which movement with a speed υ is still possible is determined from the equation. Hence (m).

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

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

At the moment of a car rollover as a reaction force of the road N and the friction force F TP are attached to the "outer" wheel. When the car moves in a circle at a speed υ, a friction force acts on it. This force creates a moment relative to the vehicle's center of gravity. Maximum moment of reaction force of the road N = m g relative to the center of gravity is equal (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 roll over yet:

From where ≈ 30 m / s ≈ 110 km / h.

For a car to be able to move at this speed, a coefficient of friction is required (see the previous problem).

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

Problem 6... A motorcyclist drives on a horizontal road at a speed of υ = 70 km / h, making a turn with a radius R= 100 m. At what angle α to the horizon should it tilt in order not to fall?

The frictional force between the motorcycle and the road as it imparts centripetal acceleration to the rider. 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 Sin α = N· l Cos α, where l- distance OA from the center of gravity to the motorcycle track (see fig. 5).

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

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

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

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

Since we are not interested in this case in 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 should be directed to 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 equal, that is,

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

N Cos α - m g – F t p sin α = 0.

Substituting into these equations the maximum possible value of the friction force F tp = k N and excluding force N, we 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 the horizontal road.

Having dealt with the dynamics of rotation, let's move on to the tasks on rotary motion in the vertical plane.

Problem 8... Mass car m= 1.5 t moves at a speed υ = 70 km / h along the road shown in Figure 7. Road sections AB and Sun can be considered as arcs of circles of radius R= 200 m touching each other at a point V... Determine the force of pressure of the car on the road at points A and WITH... How the pressure force changes when the car passes the point V?

At the point A gravity acts on the car R = m g and the reaction force of the road N A... The vector sum of these forces should be directed to the center of the circle, that is, vertically downward, and create centripetal acceleration: from where (H). The force of the car's pressure 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 upward: and (H). Thus, at the point A the force of pressure is less than the force of gravity, and at the point WITH- more.

At the point V 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 to the direction to the center must exceed the reaction force of the road N B 1, and ... When driving on a concave section of the road, on the contrary, the reaction force of the road N B 2 is superior to gravity projection: .

From these equations we obtain that when passing through the point V the force of pressure of the car on the road changes abruptly by an amount 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 the car moves in a circle at a constant speed, the sum of the projections of all forces on the direction tangent to the circle should be equal to zero. In our case, the tangential component of the gravity force is balanced by the friction force between the wheels of the car and the road.

The magnitude of the frictional force is controlled by the torque applied to the wheels from the motor side. This moment tends to cause wheel slip in relation to the road. Therefore, there is a frictional force that prevents slippage and is proportional to the applied torque. The maximum value of the friction force is k N, where k- the coefficient of friction between the tires of the car and the road, N- the force of pressure on the road. When the car moves down, the friction force plays the role of a braking force, and when it moves up, on the contrary, the role of traction.

Problem 9... Vehicle weight m= 0.5 t, moving at a speed of υ = 200 km / h, makes a "loop" of radius R= 100 m (Fig. 8). Determine the force of pressure of the car on the road at the top of the loop A; at the point V, the radius vector of which makes an angle α = 30º with the vertical; at the point WITH, in which the vehicle speed 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 gravity and the reaction force of the road N A directed vertically downward. The sum of these forces creates a centripetal acceleration: ... That's why N.

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

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

It is easy to see that NB > N A; with an increase in the angle α, 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 the force of gravity is directed tangentially. When moving along the bottom of the loop, the reaction force will also exceed the maximum value H the reaction force has at the point D... Meaning , thus, is the minimum value of the reaction force.

Vehicle speed will be constant if the tangent gravity does not exceed the maximum frictional force k N at all points of 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 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 in a "loop" with a constant speed is possible.

Let us now consider the movement of a car in a "loop" with the engine off. As already noted, usually the frictional moment is opposed to the moment applied to the wheels from the motor side. When the car is moving with the engine turned 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 the force of gravity slows down or accelerates the movement of the car along the "loop". The centripetal acceleration will also change. It is created, as usual, by the resultant force of the reaction of the road and the projection of the force of gravity on the direction towards the center of the loop.

Problem 10... What is the slowest vehicle speed at the bottom of the loop D(see fig. 8) in order to perform it with the engine off? What will be the force of pressure of the car on the road at the point V? Loop radius R= 100 m, vehicle weight m= 0.5 t.

Let's see what the minimum speed a car can have at the top of the loop A to keep going in a circle?

Centripetal acceleration at this point on the road is created by the sum of gravity and the reaction force of the road. ... The lower the speed of the car, the less reaction force arises. N A... At value, this force vanishes. At a slower speed, gravity will exceed the value required to generate centripetal acceleration and the vehicle 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 A... Therefore, if the car at the top of the loop has speed, then it will not come off the loop anywhere.

Now let's determine what speed the car should have at the bottom of the loop D so that at the top of the loop A its 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 force of friction 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's equate the energy values ​​of the car in points A and D... In this case, we will measure 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 required speed υ D, we find: ≈ 70 m / s ≈ 260 km / h.

If the car enters the loop at this speed, then 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 V... Vehicle speed at point V 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 to the previous problem, for a given speed, we find the pressure force at the point B:

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

Exercises

1. Find the angular velocity artificial satellite Earth rotating in a circular orbit with a period of revolution T= 88 minutes Find the linear velocity 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 slats. Reiki moves with speeds υ 1 and υ 2. Determine the angular speed of rotation of the disk and the speed 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 points of the vertical diameter 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 aircraft body is inclined at an angle α = 5 °.

5. Weight load m= 100 g, suspended on a thread of length l= 1 m, rotates evenly in a circle in the 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 thread tension.

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. What is the minimum coefficient of friction between the tires of the car and the cylinder surface that is possible?

7. Weight m suspended on 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 with 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. The angular velocity of an artificial Earth satellite ≈ 0.071 rad / s. The linear velocity of the satellite υ = ω R... where R- orbital radius. 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 bars 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 have taken 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 r M- distance from point M to the center of the disc 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. We denote by ρ the distance from any point of the diameter VA to the point A contact of the disc with the plane. Then it is obvious that r M = ρ – R and r N = R – ρ = –(ρ – R). where R is the radius of the disc. Therefore, the speed of any point on the diameter VA is found by the formula: υ ρ = υ + ω R). Since the disk rolls without slipping, then for the speed υ ρ we obtain υ ρ = ω · ρ. Hence it follows that the ends of the velocity vectors are on the straight line outgoing from the point A 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 motion of points located on the diameter VA, can be considered at any given moment as a simple rotation around a fixed point A 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 A.

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

4. The plane is acted upon (see Fig. 3) by gravity R = m g and lift N, directed perpendicular to the plane of the wings (since the plane is moving at a constant speed, the thrust force and the force of frontal air resistance balance each other). Effective forces R

6. The car is acted upon (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 moves in a horizontal circle, the forces R and F tp balance each other, and the strength 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 with a radius l(fig. 6). The centripetal acceleration of the load (υ is the speed of the load) is created by the difference in the values ​​of the tension force of the thread T and the projection of gravity m g thread direction: ... That's why , where β is the angle formed by the thread with the vertical. As the weight descends, its speed will increase, and the angle β will decrease. The thread tension will become maximum at an angle β = 0 (at the moment when the thread is vertical): ... The maximum speed of the load υ 0 is found along 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 m ax = m g· (3 - 2 cos α). By the condition of the problem T m ax = 2m g... Equating these expressions, we find cos α = 0.5 and, therefore, α = 60 °.

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

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

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

Acceleration is known to be determined by the formula

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

Suppose that the body moves in a circle with a radius and that at some moment 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 sec, the body is at point B, and its speed is now

directed tangentially to the circle at point B. The modulus of speed 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 to a point.

Let us first find out how this acceleration is directed.

Let's draw the 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 tangency, 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 between the vectors and Its direction - this is the direction of the vector acceleration. We already know how vectors are subtracted (see § 6). To find the difference between the vectors and position them so that they originate from one point (Fig. 68), and connect their ends, directing the arrow from the subtracted to the reduced one (from the end of the vector to the end of the vector Vector is the difference of vectors Therefore, the acceleration is directed along the vector. What can you say about this direction?

The triangle (see fig. 68) is isosceles. Apex angle equal to the angle between the radii and (Fig. 67), since they are formed by mutually perpendicular sides. Points A and B are located 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 one, 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. This means that the acceleration is perpendicular to the speed. But the speed is directed tangentially to the circle at point A, and the tangent is perpendicular to the radius. This means that the acceleration is directed along the radius to the center of the circle. It is therefore called centripetal acceleration.

With a uniform movement of a body around a circle, the acceleration at any of its points is perpendicular to the speed of movement and is directed to the center of the circle.

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

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

The acceleration of a body moving along a circle is equal to the product of its linear velocity and the angular velocity of rotation of the radius drawn to the body.

It is more convenient to present the formula for centripetal acceleration in such a way 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 ratio (is the radius of the circle), then, substituting this expression into the formula, we get:

But therefore, the formula for centripetal acceleration can also be written like this:

With uniform motion around the circumference, the body moves with

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

Consequently, 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 moment in time (the initial position of the body must, of course, be known). Thus, the main task of mechanics will be solved.

Recall that acceleration with uniform motion along a circle is of interest to us because any motion along a curved trajectory is a motion along arcs of circles of different radii.

Now we can say that with uniform motion at any point of the curvilinear trajectory, the body moves with acceleration directed towards the center of the circle of which this 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 shows the vectors of centripetal acceleration at various points of the trajectory.

Task. The plane, leaving the dive, 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 aircraft's acceleration at its lowest point if its speed is 800 km / h, and compare this value with the acceleration due to gravity.

4. Grinding wheel, radius of which is equal to 10 cm, during rotation makes 1 revolution in 0.2 sec. Find the speed of the points farthest from the axis of rotation.

5. The car moves along the curve of the road with a radius of 100 m at a speed of 54 km / h. What is the magnitude of the vehicle's centripetal acceleration?

6. The period of revolution of the first spacecraft-satellite "Vostok" around the Earth was equal to 90 minutes. The average height of the satellite ship 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 a car if its wheels with a radius of 30 cm make 10 revolutions per second?

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