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 n 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, FROM 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 FROM 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, since 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, FROM 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 the 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 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 FROM. 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 FROM 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 FROM- 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 FROM 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: . That's why 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 FROM 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 FROM), 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: . That's why , 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:

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A round disk of radius a, immersed in a liquid, rotates around an axis passing through the center of the disk perpendicular to its plane. Friction resistance is equal to ku per unit area at each point of the disk, where v is the speed of the point, k is a constant.

A round disk of radius AC r rolls without slipping on a horizontal plane (Fig.

A weightless round disk of radius R 4 m is connected by means of weightless slings with a load of weight Q. Remaining horizontal, the disk descends in still air (at temperature t 0 and pressure h6 760 mm Hg) with a constant speed v 1 m / sec.

On the surface of a round disk of radius a from the center to the edge, N turns of a thin wire spiral are laid.

In this problem, a circular disk of radius R is loaded with a normal load an - - p (compression) along two diametrically located arcs of length 2aR each. The geometric scheme and loading conditions are shown in fig. 4.14, from which it can be seen that both lines x 0 and y 0 serve as axes of symmetry.

Show that when a round disk of radius a rotates about its diameter in a liquid at infinity, then the kinetic energy of the liquid is equal to 8d5 (o2 / 45, where u is the angular velocity of rotation of the disk, and Q is the density of the liquid.

The charge ql is located on the axis of symmetry of a round disk of radius a at a distance a from the plane of the disk.

The turbine wheel is schematically represented as a round disk of radius R and mass M mounted on the vertical axis ADB (Fig.

As an example of other possible applications of the theory, consider the problem of two equal round disks of radius c rotating parallel to each other around their line of centers in an infinite fluid. Let us denote by 21 the distance between the disks and suppose that they rotate with the same angular velocity ω either in the same direction or in opposite directions. Then, depending on whether the first case or the second occurs, the median plane behaves either as a free surface or as a solid boundary.

Around point A, the crank rotates with a constant absolute angular velocity (Ob directed counterclockwise, a round disk of radius r. Determine the absolute velocities and accelerations of points 1, 2, 3, 4 of the disk and its instantaneous centers of velocities and accelerations.

It is assumed that evaporation occurs from all points of the evaporator at the same rate. The case of a two-dimensional evaporator, first solved by von Hippel, will be considered by us in the next section. Let us first consider a model of an evaporator in the form of a round disk of radius s, the evaporation surface of which is parallel to the flat surface of the substrate.

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Find the Earth's Linear Velocity v during its orbital motion. Average radius of the earth's orbit R\u003d 1.5 10 8 km.

Answer and solution

v≈ 30 km/s.

v = 2πR/(365 24 60 60).

An aircraft propeller with a radius of 1.5 m rotates during landing with a frequency of 2000 min -1 , the landing speed of the aircraft relative to the Earth is 162 km/h. Determine the speed of the point at the end of the propeller. What is the trajectory of this point?

Answer and solution

v≈ 317 m/s. The point at the end of the propeller describes a helix with pitch h≈ 1.35 m.

The aircraft propeller rotates at a frequency of:

λ = 2000/60 s -1 = 33.33 s -1 .

Linear speed of the point at the end of the propeller:

v lin = 2 πRλ≈ 314 m/s.

Aircraft landing speed v= 45 m/s.

The resulting speed of the point at the end of the propeller is equal to the sum of the vectors of the linear speed during the rotation of the propeller and the speed of the aircraft during landing:

v cut = ≈ 317 m/s.

The step of the helical trajectory is equal to:

h = v/λ ≈ 1.35 m.

Disc radius R rolls without slipping at a constant speed v. Find the locus of points on the disk that currently have speed v.

Answer

The locus of points on a disk that have speed v at the moment, is the arc radius R, whose center lies at the point of contact of the disk with the plane, i.e. at the instantaneous center of rotation.

Cylindrical roller radius R placed between two parallel bars. Reiki move in one direction with speeds v 1 and v 2 .

Determine the angular speed of rotation of the roller and the speed of its center if there is no slippage. Solve the problem for the case when the velocities of the rails are directed in different directions.

Answer

; .

Rolls on a horizontal plane without sliding at a constant speed v with hoop radius R. What are the velocities and accelerations of the various points of the hoop relative to the Earth? Express the speed as a function of the angle between the vertical and the straight line drawn between the point of contact of the hoop with the plane and the given point of the hoop.

Answer

v A=2 v C cos α . The acceleration of the rim points contains only a centripetal component equal to a c = v 2 /R.

The car is moving at a speed v= 60 km/h. With what frequency n its wheels rotate if they roll along the highway without slipping, and the outer diameter of the tires of the wheels is d= 60 cm? Find centripetal acceleration a tss outer layer of rubber on the tires of its wheels.

Answer

n≈ 8.84 s -1; a c ≈ 926 m / s 2.

A thin-walled cylinder is placed on a horizontal plane, rotating at a speed v 0 around its axis. What will be the speed of movement of the axis of the cylinder when the sliding of the cylinder relative to the plane stops?

Answer

v = v 0 /2.

Does the resultant of all forces applied to a body moving uniformly in a circle do work?

Answer

Load of mass m can slide without friction on a horizontal rod rotating about a vertical axis passing through one of its ends. The load is connected to this end of the rod by a spring, the coefficient of elasticity of which is k. At what angular speed ω Will the spring stretch to 50% of its original length?

Answer

Two point masses m 1 and m 2 are attached to the thread and are on a completely smooth table. The distances from them to the fixed end of the thread are l 1 and l 2 respectively.

The system rotates in a horizontal plane around an axis passing through the fixed end with an angular velocity ω . Find the tension forces of the sections of the thread T 1 and T 2 .

Answer

T 1 = (m 1 l 1 +m 2 l 2)ω 2 ; T 2 = m 2 ω 2 l 2 .

A man sits on the edge of a round horizontal platform with a radius R\u003d 4 m. With what frequency n the platform must rotate around a vertical axis so that a person cannot stay on it with a coefficient of friction k=0,27?

Answer

n= 6.75 min -1 .

body mass m located on a horizontal disk at a distance r from the axis. The disk starts spinning at a slow speed. Construct a graph of the dependence of the component of the friction force in the radial direction, acting on the body, on the angular velocity of disk rotation. At what value of the angular velocity of the disk will the body begin to slide?

Answer

Mass stone m=0.5 kg, tied to a rope length l=50 cm, rotates in a vertical plane. The tension in the rope when the stone passes the lowest point of the circle T\u003d 44 N. To what height h Will a stone rise above the lowest point of the circle if the rope is cut while its velocity is directed vertically upwards?

Answer

h≈ 2 m.

The athlete sends the hammer (the core on the cable) to a distance l\u003d 70 m along the trajectory that provides the maximum throw range. What strength T affects the hands of the athlete at the time of the throw? Hammer weight m=5 kg. Consider that the athlete accelerates the hammer, rotating it in a vertical plane around a circle with a radius R\u003d 1.5 m. Air resistance is not taken into account.

Answer

T≈ 2205 N.

Vehicle mass M\u003d 3 * 10 3 kg moves at a constant speed v\u003d 36 km / h: a) along a horizontal bridge; b) along the convex bridge; c) along a concave bridge. The radius of curvature of the bridge in the last two cases R\u003d 60 m. With what force does the car press on the bridge (in the last two cases) at the moment when the line connecting the center of curvature of the bridge with the car makes an angle α =10° with vertical?

Answer

a) F 1 ≈ 29400 N; b) F 2 ≈ 24,000 N; in) F 3 ≈ 34,000 N.

On a convex bridge, the radius of curvature of which R= 90 m, with speed v= 54 km/h a car of mass m\u003d 2 t. At the point of the bridge, the direction to which from the center of curvature of the bridge makes an angle with the direction to the top of the bridge α , the car presses with force F= 14 400 N. Determine the angle α .

Answer

α ≈ 8.5º.

Ball mass m= 100 g suspended from a thread of length l\u003d 1 m. The ball was spun so that it began to move in a circle in a horizontal plane. In this case, the angle made by the thread with the vertical, α = 60°. Determine the total work done in spinning the ball.

Answer

A≈ 1.23 J.

What is the maximum speed a car can travel on a curve with a radius of curvature? R\u003d 150 m, so that it does not “skid” if the friction coefficient of sliding tires on the road k = 0,42?

Answer

v≈ 89 km/h.

1. What should be the maximum coefficient of sliding friction k between the tires of the car and the asphalt so that the car can pass the rounding radius R= 200 m at speed v= 100 km/h?

2. A car with all-wheel drive, starting off, evenly picks up speed, moving along a horizontal section of the road, which is an arc of a circle α = 30° radius R= 100 m. With what maximum speed can the car drive onto a straight section of the track? Coefficient of wheel friction on the ground k = 0,3.

Answer

1. k ≈ 0,4.

2. v≈ 14.5 m/s.

The train moves along a curve with a radius R= 800 m with speed v= 12 km/h. Determine how much the outer rail must be higher than the inner rail so that no lateral force occurs on the wheels. The horizontal distance between the rails is taken equal to d= 1.5 m.

Answer

∆h≈ 7.65 cm.

A motorcyclist is driving along a horizontal road at a speed of 72 km/h, making a turn with a radius of curvature of 100 m.

Answer

1. What is the maximum speed v a motorcyclist can ride on a horizontal plane, describing an arc with a radius R= 90 m if the coefficient of sliding friction k = 0,4?

2. At what angle φ should it deviate from the vertical direction?

3. What will be the maximum speed of a motorcyclist if he rides on an inclined track with an angle of inclination α = 30° with the same radius of curvature and coefficient of friction?

4. What should be the angle of inclination of the track α 0 so that the speed of the motorcyclist can be arbitrarily large?

Answer

1. v≈ 18.8 m/s. 2. φ ≈ 21.8°. 3. v max ≈ 33.5 m/s. four. α 0 = arctg(1/ k).

The aircraft makes a turn, moving along an arc of a circle at a constant speed v= 360 km/h. Define Radius R this circle, if the body of the aircraft is rotated around the direction of flight at an angle α = 10°.

Answer

R≈ 5780 m.

At the turn of the road with a radius R= 100 m the car moves uniformly. The vehicle's center of gravity is at a height h= 1 m, vehicle track width a= 1.5 m. Determine the speed v at which the vehicle may roll over. In the transverse direction, the car does not slip.

Answer

v≈ 26.1 m/s.

The driver, driving a car, suddenly noticed a fence in front of him, perpendicular to the direction of his movement. What is more profitable to do to prevent an accident: slow down or turn to the side?

Answer

Slow down.

In the carriage of a train moving uniformly along a curved track with a speed v= 12 km/h, the load is weighed on spring scales. Load weight m= 5 kg, and the radius of the curvature of the path R\u003d 200 m. Determine the reading of the spring balance (spring tension force T).

Answer

T≈ 51 N.

Find strength F unit separating cream (density ρ c \u003d 0.93 g / cm 3) from skimmed milk ( ρ m \u003d 1.03 g / cm 3) per unit volume, if separation occurs: a) in a stationary vessel; b) in a centrifugal separator rotating at a frequency of 6000 min -1 if the liquid is at a distance r= 10 cm from the axis of rotation.

Answer

a) F unit ≈ 980 N/m3;

b) F unit ≈ 3.94 10 5 N / m 3;

The aircraft makes a "dead loop" with a radius R= 100 m and moves along it at a speed v= 280 km/h. With what force F pilot's body mass M= 80 kg will put pressure on the aircraft seat at the top and bottom of the loop?

Answer

F in ≈ 4030 N, F n ≈ 5630 N.

Determine the pull force T rope giant steps, if the mass of a person M\u003d 70 kg and the rope during rotation forms an angle α \u003d 45 ° with the column. With what angular velocity will the giant steps rotate if the length of the suspension l= 5 m?

Answer

T≈ 990 N; ω ≈ 1.68 rad/s.

Find Period T rotation of a pendulum making circular motions in a horizontal plane. Thread length l. The angle formed by the thread with the vertical, α .

Answer

.

A weight suspended on a thread rotates in a horizontal plane so that the distance from the suspension point to the plane in which the rotation occurs is h. Find the frequency and rotation of the load, assuming it to be constant.

Answer

The result does not depend on the length of the suspension.

Chandelier mass m= 100 kg suspended from the ceiling on a metal chain, the length of which l= 5 m. Determine the height h, by which the chandelier can be deflected so that the chain does not break during subsequent swings? It is known that chain break occurs when the tension force T> 1960 N.

Answer

h≈ 2.5 m.

Ball mass m suspended from an inextensible thread. What is the minimum angle α min, it is necessary to deflect the ball so that during further movement the thread breaks if the maximum possible tension force of the thread is 1.5 mg?

Answer

α min ≈ 41.4°.

The pendulum is deflected to a horizontal position and released. At what angle α with the vertical, the tension force of the thread will be equal in magnitude to the force of gravity acting on the pendulum? The pendulum is considered mathematical.

Answer

α = arccos(⅓).

Load of mass m, tied to an inextensible thread, rotates in a vertical plane. Find the maximum difference in the tension forces of the thread.

Answer

The gymnast "twirls the sun" on the crossbar. Gymnast weight m. Assuming that all his mass is concentrated at the center of gravity, and the speed at the top point is zero, determine the force acting on the gymnast's hands at the bottom point.

Answer

One weight is suspended from an inextensible thread of length l, and the other - on a rigid weightless rod of the same length. What minimum speeds must be given to these weights so that they rotate in a vertical plane?

Answer

For thread v min = ; for rod v min = .

Ball mass M hung on a thread. In the taut state, the thread was placed horizontally and the ball was released. Derive the dependence of the thread tension T from the corner α , which currently forms a thread with a horizontal direction. Check the derived formula by solving the problem for the case when the ball passes through the equilibrium position, with α = 90°.

Answer

T = 3mg sin α ; T = 3mg.

Mathematical pendulum length l and weight M taken to a corner φ 0 from the equilibrium position and told him the initial speed v 0 directed perpendicular to the thread upwards. Find the tension in the string of the pendulum T depending on the angle φ vertical threads.

Answer

.

A weight suspended on a thread is taken aside so that the thread assumes a horizontal position, and released. What angle with the vertical α does the drink form at the moment when the vertical component of the weight's velocity is greatest?

Answer

Identical elastic balls with mass m, suspended on threads of equal length to one hook, are deflected in different directions from the vertical by an angle α and let go. The balls hit and bounce off each other. What is the strength F, acting on the hook: a) at the extreme positions of the threads; b) at the initial and final moments of impact of the balls; c) at the moment of the greatest deformation of the balls?

Answer

a) F = 2mg cos 2 α ;

b) F = 2mg(3 - 2cos α );

in) F = 2mg.

To a mathematical pendulum with a flexible inextensible thread of length l impart a horizontal velocity from the equilibrium position v 0 . Determine the maximum lift height h when moving in a circle, if v 0 2 = 3gl. What trajectory will the pendulum ball follow after it has reached its maximum lifting height? h on a circle? Determine the maximum height H achieved with this movement of the pendulum.

Answer

; along a parabola; .

A small ball is suspended at a point BUT on a thread of length l. At the point O on distance l/2 below point BUT a nail is driven into the wall. The ball is withdrawn so that the thread is in a horizontal position, and released. At what point in the trajectory does the tension of the thread disappear? How far will the ball move? What is the highest point the ball will rise to?

Answer

On the l/6 below the suspension point; along a parabola; on 2 l/27 below the suspension point.

A vessel having the shape of an expanding truncated cone with a bottom diameter D= 20 cm and the angle of inclination of the walls α = 60°, rotates around the vertical axis 00 one . At what angular velocity of rotation of the vessel ω a small ball lying at its bottom will be thrown out of the vessel? Friction is ignored.

Answer

ω > ≈13 rad/s.

Sphere with radius R= 2 m rotates uniformly around the axis of symmetry with a frequency of 30 min -1 . Inside the sphere is a ball of mass m= 0.2 kg. Find Height h, corresponding to the equilibrium position of the ball relative to the sphere, and the reaction of the sphere N.

Answer

h≈ 1 m; N≈ 0.4 N.

Inside a conical surface moving with acceleration a, the ball rotates in a circle with a radius R. Define period T circular motion of the ball. Cone apex angle 2 α .

Answer

.

A small body of mass m slides down an inclined slope, turning into a dead loop with a radius R.

Friction is negligible. Determine: a) what should be the smallest height h slope so that the body makes a full loop without falling out; b) what pressure F at the same time, it produces a body on the platform at a point whose radius vector makes an angle α with vertical.

Answer

a) h = 2,5R; b) F = 3mg(1 - cos α ).

The conveyor belt is inclined to the horizon at an angle α . Determine the minimum speed of the tape v min, at which the ore particle lying on it is separated from the surface of the belt at the place where it runs onto the drum, if the radius of the drum is equal to R.

Answer

v min = .

A small body slides down from the top of the sphere. At what height h from the vertex the body will come off the surface of the sphere with a radius R? Ignore friction.

Answer

h = R/3.

Find the kinetic energy of the hoop mass m rolling at a speed v. There is no slippage.

Answer

K = mv 2 .

A thin hoop without slipping rolls into a hemisphere-shaped pit. At what depth h is the force of normal pressure of the hoop on the wall of the pit equal to its gravity? Pit radius R, hoop radius r.

Answer

h = (R - r)/2.

A small hoop rolls without slipping on the inner surface of a large hemisphere. At the initial moment, the hoop rested at its upper edge. Determine: a) the kinetic energy of the hoop at the lowest point of the hemisphere; b) what proportion of the kinetic energy falls on the rotational movement of the hoop around its axis; c) normal force pressing the rim to the lower point of the hemisphere. The mass of the hoop is m, hemisphere radius R.

Answer

a) K = mgR; b) 50%; in 2 mg.

Water flows through a pipe located in a horizontal plane and having a rounding radius R= 2 m. Find the lateral water pressure. Pipe diameter d= 20 cm. M= 300 tons of water.

Answer

p\u003d 1.2 10 5 Pa.

The body slips from the point BUT exactly AT along two curved inclined surfaces passing through points A and AT once along a convex arc, the second - along a concave one. Both arcs have the same curvature and the coefficient of friction is the same in both cases.

In what case is the speed of the body at a point B more?

Answer

In the case of movement along a convex arc.

A rod of negligible mass, length l with two small balls m 1 and m 2 (m 1 > m 2) at the ends it can rotate about an axis passing through the middle of the rod perpendicular to it. The rod is brought to a horizontal position and released. Determine angular velocity ω and force of pressure F on the axis at the moment the rod with balls passes the equilibrium position.

Answer

; .

A small ring of mass m. The ring without friction begins to slide in a spiral. With what force F the ring will press on the spiral after it passes n full turns? Turn radius R, the distance between adjacent turns h(turn pitch). Count h ≪ R.

Answer

.

A closed metal chain lies on a smooth horizontal disk, being loosely placed on a centering ring coaxial with the disk. The disk is set in rotation. Taking the shape of the chain as a horizontal circle, determine the tension force T along the chain if its mass m= 150 g, length l= 20 cm and the chain rotates with a frequency n= 20 s -1 .

Answer

T≈ 12 N.

Reactive plane m= 30 tons flies along the equator from west to east with a speed v= 1800 km/h. By how much will the lift force acting on the plane change if it flies at the same speed from east to west?

Answer

ΔF under ≈ 1.74 10 3 N.

Absolute point acceleration directional

1) Tangent to the path

2) Normal to the trajectory

3) Tangent to the speed hodograph

4) Normal to the velocity hodograph

5) Same as speed

Answer 1

The bionormal component of the acceleration of a point moving in space is

1) the second time derivative of another coordinate

2) the square of the speed divided by the radius of curvature

3) derivative of the speed modulus with respect to time

5) derivative of speed with respect to time

Answer 4

26. Vanya and Manya ride bicycles parallel to each other at a constant speed of 3m/s. The distance from Vanya to Mani is H=3 (they travel perpendicular to the trajectories). The fly flew off Vanya's nose and flew towards Mani's nose at a constant speed of 1 m/s

1) not to see the Manino nose fly

2) the fly will fly to the manna in 1 second

3) the fly will arrive after a time equal to the square root of 5 seconds

4) the fly will fly to the manna in 4 seconds

5) the fly will fly to the manna in 5 seconds

Answer in question

Vanya and Manya ride bicycles parallel to each other at a constant speed V. The distance from Vanya to Manya is H (they ride perpendicular to the trajectories). The fly flew off Vanya's nose and flew towards Mani's nose with a constant speed Vm. The fly flew to Mani's nose after a time equal to

4)H/Root(Vm*Vm-V*V)

5) H/root of (Vm*Vm+V*V)

Answer in question

The Angular Acceleration Vector is equal to

1) The vector product of the angular velocity vector by the radius vector

2) Derivative of the angular velocity vector with respect to time

3) The second derivative of the angular velocity vector with respect to time

4) Time derivative of the angle of rotation

5) The product of the distance from a point to the axis of rotation by the square of the angular velocity of the body

Answer 2

In a gear train of 3 consecutive gears with radii R1 R2 R3, the first wheel has an angular velocity W 1

Angular speed of 3 wheels?

1) Increases with increasing R2

2) Decreases with increasing R2

3) Does not depend on R2

4) Directly proportional to R3

5) Directly proportional to R1

Answer 3

29. A gymnast rotates on a horizontal bar with an angular velocity W=1 and an angular acceleration e=1. An insect runs along it towards the crossbar with a speed of Vr=0.5. At the moment when the insect is at a distance of 1m from the crossbar, its absolute acceleration will be equal to?

1) 3.165 m/s/s

4) 1.407 m/s/s

5) 2.236 m/s/s

Answer in question

thirty. . The gymnast rotates on the horizontal bar with an angular velocity W=4.000000 and an angular acceleration e=8000000. An insect runs along it towards the crossbar with the speed Vr=2.000000. At the moment when the insect is at a distance of 1m from the crossbar, its absolute acceleration will be equal to?

Answer 4

The head of a fighter without rules received a heel strike from his colleague and began to make a complex movement

- Straight line parallel to the floor, caused by contact with the enemy.

-Rotational with the fighter himself around the point of his standing

Is the cariolis acceleration directed?

2) In the direction of the blow from the enemy

3) Sideways (right, left)

4) In the direction opposite to the blow (towards the opponent)

Answer 3

24. The load is suspended on a thread and oscillates. At the moment of the greatest deviation of the load, its acceleration:

2) not visible

Answer 5

To determine the MHC (MCS), it is necessary and sufficient to know

5) The value of all accelerations of a flat figure

37. A girl jumps from a cliff with her head into a whirlpool with a radius of 30 m. The angular velocity of water rotation in the whirlpool is W=1/c. The head enters the water vertically. The speed of the girl at the moment of touching the water is 2 m / s / s. The absolute acceleration of the head in this case will be approximately.

Answer in question

15. The movement of a point is described by the equations y=t, x= cos kt. At the moment t=3/1416/2

1) tangential acceleration is 0, normal is not

2) the total acceleration of the point is 0

3) the total acceleration of the point is greater than 0

4) tangential and normal acceleration of the point is 0

5) speed is 0

6) speed less than 0

Answer 2

16. The movement of a point is described by the equations x=-3 sinkt, y=-3coskt

1) the point moves in a circle

2) the point moves along an ellipse

3) the point moves in a straight line

4) the point moves along a parabola

5) the point moves along the hyperbola

Answer 1

17. The movement of a point is described by the equations x=2 sinkt, y=-2coskt

1) tangential acceleration of a point is always 0

2) tangential acceleration of a point is 0 only if 2kt=3.1416

3) tangential acceleration of a point is 0 only if kt=3.1416

4) tangential acceleration is always positive

5) tangential acceleration of the point is 0

Answer 3

18. The movement of a point is described by the equations x=2sin kt, y=2coskt

the normal acceleration of a point is always directed towards the origin

1) normally, the acceleration of a point is always directed towards the origin

2) normal point acceleration is directed to the origin only if 2kt=3.1416

3) normal point acceleration is directed to the origin only if kt=3.1416

4) normally the acceleration of the point is equal to kt

5) normal point acceleration is 2.5

Answer 1

A girl with a string bag rushes home with a speed V along an arc of a circle of radius R. A fly crawls vertically upward along the string bag with a speed U.

Cariolis acceleration of the fly is equal.

Answer 5

If the speed of two points of a plane figure at some point in time is equal to 0, then

1) The projections of the acceleration of the points on the line passing through the points are equal to each other

2) The figure is at rest

3) The projections of the acceleration of the points on the line passing through the points are equal to 0

4) Acceleration points are equal to each other

5) Acceleration points 0

Answer 2

32. A square with a diagonal equal to 2d moves in its own plane. The equation of motion of the center of the square Us=at Xc=bt. The angle of rotation is described by the equation Ф=Wt. The acceleration of the upper right corner of the square is

3) root of (a*a+b*b+W*d)

5) (a*W+b*W+W*W*d)

Answer in question

33. A square with a diagonal equal to 2d with side 2 moves in its plane. The equation of motion of the center of the square Us=at Xc=bt. The angle of rotation is described by the equation Ф=Wt. At the moment when t=0 the maximum speed of the points of the square =?

4) root of (a*a+b*b+W*W*d*d)

5) root of ((a-W*c)*(a-W*c)+(b+W*c)*(b+W*c))

Answer in question

A round disk of radius R is rolling. A point moves along the rim of the disk with a constant modulo velocity V1. Is the absolute acceleration of a point equal to?

2) (V*V+V1*V1)/R

3) square root of (V*V*V*V+V1*V1*V1*V1)/R

4) (V+V1)* (V+V1)/R

5) not visible