Movement on an incline. Movement of a body along an inclined plane

V. M. Zrazhevsky

LABORATORY WORK NO.

ROLLING A SOLID BODY FROM AN INCLINED PLANE

Goal of the work: Verification of the law of conservation of mechanical energy when a rigid body rolls down an inclined plane.

Equipment: inclined plane, electronic stopwatch, cylinders of different masses.

Theoretical information

Let the cylinder have radius R and mass m rolls down an inclined plane forming an angle α with the horizon (Fig. 1). There are three forces acting on the cylinder: gravity P = mg, the force of normal pressure of the plane on the cylinder N and the friction force of the cylinder on the plane F tr. , lying in this plane.

The cylinder participates simultaneously in two types of motion: translational motion of the center of mass O and rotational motion relative to the axis passing through the center of mass.

Since the cylinder remains on the plane during movement, the acceleration of the center of mass in the direction of the normal to the inclined plane is zero, therefore

P∙cosα − N = 0. (1)

The equation for the dynamics of translational motion along an inclined plane is determined by the friction force F tr. and the gravity component along the inclined plane mg∙sinα:

ma = mg∙sinα − F tr. , (2)

Where a– acceleration of the center of gravity of the cylinder along an inclined plane.

The equation for the dynamics of rotational motion relative to an axis passing through the center of mass has the form

Iε = F tr. R, (3)

Where I– moment of inertia, ε – angular acceleration. Moment of gravity and relative to this axis is zero.

Equations (2) and (3) are always valid, regardless of whether the cylinder moves along the plane with sliding or without sliding. But from these equations it is impossible to determine three unknown quantities: F tr. , a and ε, one more additional condition is necessary.

If the friction force is sufficiently large, then the cylinder rolls along an inclined path without slipping. Then the points on the circumference of the cylinder must travel the same path length as the center of mass of the cylinder. In this case, linear acceleration a and angular acceleration ε are related by the relation

a = Rε.

(4) a/R From equation (4) ε =

. (5)

. After substitution into (3) we get F Replacing in (2)

. (6)

tr. on (5), we get

. (7)

From the last relation we determine the linear acceleration

. (8)

From equations (5) and (7) the friction force can be calculated: P = mg The friction force depends on the angle of inclination α, gravity I/and from attitude mR

When rolling without sliding, the static friction force plays a role. The rolling friction force, like the static friction force, has a maximum value equal to μ N. Then the conditions for rolling without sliding will be satisfied if

F tr. ≤ μ N. (9)

Taking into account (1) and (8), we obtain

, (10)

or, finally

. (11)

In the general case, the moment of inertia of homogeneous symmetrical bodies of revolution about an axis passing through the center of mass can be written as

I = kmR 2 , (12)

Where k= 0.5 for a solid cylinder (disk); k= 1 for a hollow thin-walled cylinder (hoop); k= 0.4 for a solid ball.

After substituting (12) into (11), we obtain the final criterion for a rigid body to roll off an inclined plane without slipping:

. (13)

Since when a solid body rolls on a solid surface, the rolling friction force is small, the total mechanical energy of the rolling body is constant. At the initial moment of time, when the body is at the top point of the inclined plane at a height h, its total mechanical energy is equal to potential:

W n = mgh = mgs∙sinα, (14)

Where s– the path traveled by the center of mass.

The kinetic energy of a rolling body consists of the kinetic energy of the translational motion of the center of mass with a speed υ and rotational motion with speed ω relative to an axis passing through the center of mass:

. (15)

When rolling without sliding, the linear and angular velocities are related by the relation

υ = Rω.

(16)

Let us transform the expression for kinetic energy (15) by substituting (16) and (12) into it:

. (18)

Movement on an inclined plane is uniformly accelerated:

. (19)

Let us transform (18) taking into account (4):

. (20)

Solving (17) and (19) together, we obtain the final expression for the kinetic energy of a body rolling along an inclined plane:

Description of installation and measurement method

You can study the rolling of a body on an inclined plane using the “plane” unit and the electronic stopwatch SE1, which are part of the modular educational complex MUK-M2.
U m The installation is an inclined plane 1, which can be installed at different angles α to the horizon using screw 2 (Fig. 2). Angle α is measured using scale 3. A cylinder 4 with mass

. The use of two rollers of different weights is provided. The rollers are fixed at the top point of the inclined plane using an electromagnet 5, which is controlled using

Work order

1. Loosen screw 2 (Fig. 2), set the plane at a certain angle α to the horizontal. Place roller 4 on an inclined plane.

2. Switch the toggle switch for controlling the electromagnets of the mechanical unit to the “flat” position.

3. Set the stopwatch SE1 to mode 1.

4. Press the start button of the stopwatch. Measure the rolling time.

5. Repeat the experiment five times. Record the measurement results in the table. 1.

6. Calculate the value of mechanical energy before and after rolling. Draw a conclusion.

7. Repeat steps 1-6 for other plane inclination angles.

Table 1

8. Repeat steps 1-7 for the second video. Record the results in the table. 2, similar to table. 1.

9. Draw conclusions based on all the results of the work.

Control questions

1. Name the types of forces in mechanics.

2. Explain the physical nature of friction forces.

3. What is the coefficient of friction? Its size?

4. What factors influence the coefficient of static, sliding, and rolling friction?

5. Describe the general nature of the motion of a rigid body during rolling.

6. What is the direction of the frictional moment when rolling on an inclined plane?

7. Write down a system of equations of dynamics when a cylinder (ball) rolls along an inclined plane.

8. Derive formula (13).

9. Derive formula (20).

10. Sphere and cylinder with the same masses m and equal radii R simultaneously begin to slide down an inclined plane from a height h. Will they simultaneously reach the bottom point ( h = 0)?

11. Explain the reason for the braking of a rolling body.

Bibliography

1. Savelyev, I.V. Course of general physics in 3 volumes. T. 1 / I.V. Savelyev. – M.: Nauka, 1989. – § 41–43.

2. Khaikin, S. E. Physical foundations of mechanics / S. E. Khaikin. – M: Nauka, 1971. – § 97.

3. Trofimova T. I. Physics course / T. I. Trofimova. – M: Higher. school, 1990. – § 16–19.

On the surface of the Earth gravity (gravity) is constant and equal to the product of the mass of the falling body and the acceleration of gravity: F g = mg

It should be noted that the acceleration of free fall is a constant value: g = 9.8 m/s 2, and is directed towards the center of the Earth. Based on this, we can say that bodies with different masses will fall to Earth equally quickly. How so? If you throw a piece of cotton wool and a brick from the same height, the latter will make its way to the ground faster. Don't forget about air resistance! For cotton wool it will be significant, since its density is very low. In an airless space, brick and wool will fall simultaneously.

The ball moves along an inclined plane 10 meters long, the angle of inclination of the plane is 30°. What will be the speed of the ball at the end of the plane?

The ball is affected only by the force of gravity Fg, directed downward perpendicular to the base of the plane. Under the influence of this force (component directed along the surface of the plane), the ball will move. What will be the component of gravity acting along the inclined plane?

To determine the component, it is necessary to know the angle between the force vector F g and the inclined plane.

Determining the angle is quite simple:

• the sum of the angles of any triangle is 180°;
• the angle between the force vector F g and the base of the inclined plane is 90°;
• the angle between the inclined plane and its base is α

Based on the above, the desired angle will be equal to: 180° - 90° - α = 90° - α

From trigonometry:

F g slope = F g cos(90°-α)

Sinα = cos(90°-α)

F g slope = F g sinα

It really is like this:

• at α=90° (vertical plane) F g tilt = F g
• at α=0° (horizontal plane) F g tilt = 0

Let's determine the acceleration of the ball from the well-known formula:

F g sinα = m a

A = F g sinα/m

A = m g sinα/m = g sinα

The acceleration of a ball along an inclined plane does not depend on the mass of the ball, but only on the angle of inclination of the plane.

Determine the speed of the ball at the end of the plane:

V 1 2 - V 0 2 = 2 a s

(V 0 =0) - the ball begins to move from place

V 1 2 = √2·a·s

V = 2 g sinα S = √2 9.8 0.5 10 = √98 = 10 m/s

Pay attention to the formula! The speed of the body at the end of the inclined plane will depend only on the angle of inclination of the plane and its length.

In our case, a billiard ball, a passenger car, a dump truck, and a schoolboy on a sled will have a speed of 10 m/s at the end of the plane. Of course, we don't take friction into account.

Despite the different conditions of motion, the solution to problem 8 is fundamentally no different from the solution to problem 7. The only difference is that in problem 8 the forces acting on the body do not lie along one straight line, so the projections must be taken on two axes.

Task 8. A horse is pulling a sled weighing 230 kg, acting on it with a force of 250 N. How far will the sled travel before it reaches a speed of 5.5 m/s, moving from rest. The coefficient of sliding friction of the sled on the snow is 0.1, and the shafts are located at an angle of 20° to the horizon.

There are four forces acting on the sled: the traction (tension) force directed at an angle of 20° to the horizontal; gravity directed vertically downwards (always); the support reaction force directed perpendicular to the support from it, i.e. vertically upward (in this problem); sliding friction force directed against movement. Since the sled will move translationally, all applied forces can be transferred in parallel to one point - to center masses moving body (sleigh). We will also draw the coordinate axes through the same point (Fig. 8).

Based on Newton’s second law, we write the equation of motion:

.

Let's direct the axis Ox horizontally along the direction of movement (see Fig. 8), and the axis Oy– vertically up. Let's take the projections of the vectors included in the equation onto the coordinate axes, add an expression for the sliding friction force and obtain a system of equations:

Let's solve the system of equations. (The scheme for solving a system of equations similar to the system is usually the same: the support reaction force is expressed from the second equation and substituted into the third equation, and then the expression for the friction force is substituted into the first equation.) As a result, we obtain:

Let's rearrange the terms in the formula and divide its right and left sides by mass:

.

Since acceleration does not depend on time, we choose the formula for the kinematics of uniformly accelerated motion, containing speed, acceleration and displacement:

.

Considering that the initial speed is zero, and the scalar product of identically directed vectors is equal to the product of their modules, we substitute the acceleration and express the displacement module:

;

The resulting value is the answer to the problem, since during rectilinear motion the distance traveled and the module of displacement coincide.

Answer: the sled will travel 195 m.

1. Movement on an inclined plane

The description of the movement of small bodies on an inclined plane is not fundamentally different from the description of the movement of bodies vertically and horizontally, therefore, when solving problems on this type of movement, as in problems 7, 8, it is also necessary to write down the equation of motion and take projections of vectors onto the coordinate axes. When analyzing the solution to problem 9, it is necessary to pay attention to the similarity of the approach to describing various types of movement and to the nuances that distinguish the solution of this type of problem from the solution of the problems discussed above.

Task 9. A skier slides down a long, flat snow-covered slide, the angle of inclination to the horizon is 30°, and the length is 140 m. How long will the descent last if the coefficient of sliding friction of skis on loose snow is 0.21?

The movement of a skier along an inclined plane occurs under the influence of three forces: the force of gravity directed vertically downward; support reaction force directed perpendicular to the support; sliding friction force directed against the movement of the body. Neglecting the size of the skier compared to the length of the slide, Based on Newton’s second law, we write the equation of motion skier:

.

Let's select an axis Ox down along the inclined plane (Fig. 9), and the axis Oy– perpendicular to the inclined plane upwards. Let's take the projections of the equation vectors onto the selected coordinate axes, taking into account that the acceleration is directed downward along the inclined plane, and add to them an expression that determines the sliding friction force. We get a system of equations:

Let's solve the system of equations for acceleration. To do this, from the second equation of the system, we express the support reaction force and substitute the resulting formula into the third equation, and the expression for the friction force into the first. After reducing the mass we have the formula:

.

Acceleration does not depend on time, which means we can use the formula for the kinematics of uniformly accelerated motion, containing displacement, acceleration and time:

.

Taking into account the fact that the initial speed of the skier is zero, and the displacement module is equal to the length of the slide, we express time from the formula and, substituting acceleration into the resulting formula, we obtain:

;

Answer: descent time from the mountain 9.5 s.

Projection of forces. Movement on an inclined plane

Dynamics problems.

Newton's I and II laws.

Input and direction of axes.

Non-collinear forces.

Projection of forces on the axes.

Solving systems of equations.

The most typical problems in dynamics

Let's start with Newton's I and II laws.

Let's open a physics textbook and read it. Newton's first law: there are such inertial frames of reference in which... Let's close this tutorial, I don't understand either. Okay, I’m joking, I understand, but I’ll explain it more simply.

Newton's first law: if a body stands still or moves uniformly (without acceleration), the sum of the forces acting on it is zero.

Conclusion: If a body moves at a constant speed or stands still, the vector sum of forces will be zero.

Newton's II law: if a body moves uniformly accelerated or uniformly decelerated (with acceleration), the sum of the forces acting on it is equal to the product of mass and acceleration.

Conclusion: If a body moves with varying speed, then the vector sum of the forces that somehow influence this body (traction force, friction force, air resistance force) is equal to the mass of this body times the acceleration.

In this case, the same body most often moves differently (uniformly or with acceleration) in different axes. Let's consider just such an example.

Task 1. Determine the coefficient of friction of the tires of a car weighing 600 kg if an engine traction force of 4500 N causes an acceleration of 5 m/s².

In such problems, it is necessary to make a drawing and show the forces that act on the machine:

On X Axis: movement with acceleration

On the Y Axis: no movement (here the coordinate, as it was zero, will remain the same, the machine does not go up the mountains or down)

Those forces whose direction coincides with the direction of the axes will be plus, in the opposite case - minus.

Along the X axis: the traction force is directed to the right, just like the X axis, the acceleration is also directed to the right.

Ftr = μN, where N is the support reaction force. On the Y axis: N = mg, then in this problem Ftr = μmg.

We get that:

The friction coefficient is a dimensionless quantity. Therefore, there are no units of measurement.

Answer: 0.25

Problem 2. A mass of 5 kg, tied to a weightless inextensible thread, is lifted upward with an acceleration of 3 m/s². Determine the tension of the thread.

Let's make a drawing and show the forces that act on the load

T - thread tension force

On the X axis: no power

Let's figure out the direction of forces on the Y axis:

Let's express T (tension force) and substitute the numerical values:

Answer: 65 N

The most important thing is not to get confused with the direction of forces (along the axis or against), everything elsemake a calculator or everyone’s favorite column.

Not always all forces acting on a body are directed along the axes.

A simple example: a boy pulling a sled

If we also construct the X and Y axes, then the tension (traction) force will not lie on any of the axes.

To project the traction force onto the axes, recall a right triangle.

The ratio of the opposite side to the hypotenuse is the sine.

The ratio of the adjacent leg to the hypotenuse is the cosine.

Traction force on the Y axis - segment (vector) BC.

The traction force on the X axis is a segment (vector) AC.

If this is not clear, look at problem #4.

The longer the rope and, accordingly, the smaller the angle α, the easier it will be to pull the sled. Ideal when the rope is parallel to the ground, because the force that acts on the X axis is Fнcosα. At what angle is the cosine maximum? The larger this leg is, the stronger the horizontal force.

Task 3. The block is suspended by two threads. The tension force of the first is 34 N, the second- 21Н, θ1 = 45°, θ2 = 60°. Find the mass of the block.

Let's introduce the axes and project the forces:

We get two right triangles. Hypotenuses AB and KL are tension forces. LM and BC - projections on the X axis, AC and KM - on the Y axis.

Answer: 4.22 kg

Task 4. A block with a mass of 5 kg (mass is not needed in this problem, but so that everything is known in the equations, let’s take a specific value) slides off a plane that is inclined at an angle of 45°, with a coefficient of friction μ = 0.1. Find the acceleration of the block?

When there is an inclined plane, it is best to direct the axes (X and Y) in the direction of movement of the body. Some forces in this case (here it is mg) will not lie on any of the axes. This force must be projected so that it has the same direction as the taken axes.
ΔABC is always similar to ΔKOM in such problems (by right angle and angle of inclination of the plane).

Let's take a closer look at ΔKOM:

We obtain that KO lies on the Y axis, and the projection of mg onto the Y axis will be with a cosine. And the vector MK is collinear (parallel) to the X axis, the projection mg onto the X axis will be with a sine, and the vector MK is directed against the X axis (that is, it will be with a minus).

Do not forget that if the directions of the axis and the force do not coincide, it must be taken with a minus!

From the Y axis we express N and substitute it into the equation of the X axis, we find the acceleration:

Answer: 6.36 m/s²

As you can see, the mass in the numerator can be taken out of brackets and reduced with the denominator. Then it is not necessary to know it; it is possible to get an answer without it.
Yes Yes, under ideal conditions (when there is no air resistance, etc.), both the feather and the weight will roll (fall) in the same time.

Task 5. A bus slides down a hill at a slope of 60° with an acceleration of 8 m/s² and a traction force of 8 kN. The coefficient of friction between tires and asphalt is 0.4. Find the mass of the bus.

Let's make a drawing with forces:

Let's introduce the X and Y axes. Project mg onto the axes:

Let's write Newton's second law for X and Y:

Answer: 6000 kg

Task 6. A train moves along a curve of radius 800 m at a speed of 72 km/h. Determine how much the outer rail should be higher than the inner one. The distance between the rails is 1.5 m.

The most difficult thing is to understand which forces act where, and how the angle affects them.

Remember, when you drive in a circle in a car or on a bus, where does it push you? This is why the tilt is needed so that the train does not fall on its side!

Corner α specifies the ratio of the difference in the height of the rails to the distance between them (if the rails were horizontal)

Let's write down what forces act on the axis:

The acceleration in this problem is centripetal!

Let's divide one equation by another:

Tangent is the ratio of the opposite side to the adjacent side:

Answer: 7.5 cm

As we found out, solving such problems comes down to arranging the directions of forces, projecting them onto axes and solving systems of equations, which is almost a mere trifle.

To reinforce the material, solve several similar problems with hints and answers.

The movement of a body along an inclined plane is a classic example of the movement of a body under the action of several non-directional forces. The standard method for solving problems of this kind of motion is to expand the vectors of all forces into components directed along the coordinate axes. Such components are linearly independent. This allows us to write Newton's second law for components along each axis separately. Thus, Newton's second law, which is a vector equation, turns into a system of two (three in the three-dimensional case) algebraic equations.

The forces acting on the block are
case of accelerated downward movement

Consider a body that is sliding down an inclined plane. In this case, the following forces act on it:

• Gravity m g , directed vertically downwards;
• Ground reaction force N , directed perpendicular to the plane;
• Sliding friction force F tr, directed opposite to the speed (up along the inclined plane when the body slides)

When solving problems in which an inclined plane appears, it is often convenient to introduce an inclined coordinate system, the OX axis of which is directed downward along the plane. This is convenient, because in this case you will have to decompose only one vector into components - the gravity vector m g , and the friction force vector F tr and ground reaction forces N already directed along the axes. With this expansion, the x-component of gravity is equal to mg sin( α ) and corresponds to the “pulling force” responsible for the accelerated downward movement, and the y-component is mg cos( α ) = N balances the ground reaction force, since there is no body movement along the OY axis.
Sliding friction force F tr = µN proportional to the ground reaction force. This allows us to obtain the following expression for the friction force: F tr = µmg cos( α ). This force is opposite to the "pulling" component of gravity. Therefore for body sliding down , we obtain expressions for the total resultant force and acceleration:

F x = mg(sin( α ) – µ cos( α ));
a x = g(sin( α ) – µ cos( α )).

It's not hard to see what if µ < tg(α ), then the expression has a positive sign and we are dealing with uniformly accelerated motion down an inclined plane. If µ >tg( α ), then the acceleration will have a negative sign and the movement will be equally slow. Such movement is possible only if the body is given an initial speed down the slope. In this case, the body will gradually stop. If provided µ >tg( α ) the object is initially at rest, it will not begin to slide down. Here the static friction force will completely compensate for the “pulling” component of gravity.

When the friction coefficient is exactly equal to the tangent of the angle of inclination of the plane: µ = tg( α ), we are dealing with mutual compensation of all three forces. In this case, according to Newton's first law, the body can either be at rest or move at a constant speed (in this case, uniform motion is only possible downward).

The forces acting on the block are
sliding on an inclined plane:
case of slow motion upward

However, the body can also drive up an inclined plane. An example of such motion is the movement of a hockey puck up an ice slide. When a body moves upward, both the frictional force and the “pulling” component of gravity are directed downward along the inclined plane. In this case, we are always dealing with uniformly slow motion, since the total force is directed in the direction opposite to the speed. The expression for acceleration for this situation is obtained in a similar way and differs only in sign. So for body sliding up an inclined plane , we have.