« Physics - Grade 10 "

Why does the skater stretch along the axis of rotation to increase the angular velocity of rotation.
Should a helicopter rotate when its propeller rotates?

The questions asked suggest that if external forces do not act on the body or their action is compensated and one part of the body begins to rotate in one direction, then the other part must rotate in the other direction, just as when fuel is ejected from a rocket, the rocket itself moves in the opposite direction.

moment of impulse.

If we consider a rotating disk, it becomes obvious that the total momentum of the disk is zero, since any particle of the body corresponds to a particle moving with an equal speed in absolute value, but in the opposite direction (Fig. 6.9).

But the disk is moving, the angular velocity of rotation of all particles is the same. However, it is clear that the farther the particle is from the axis of rotation, the greater its momentum. Therefore, for rotational motion it is necessary to introduce one more characteristic, similar to an impulse, - the angular momentum.

The angular momentum of a particle moving in a circle is the product of the particle's momentum and the distance from it to the axis of rotation (Fig. 6.10):

The linear and angular velocities are related by v = ωr, then

All points of a rigid matter move relative to a fixed axis of rotation with the same angular velocity. A rigid body can be represented as a collection of material points.

The angular momentum of a rigid body is equal to the product of the moment of inertia and the angular velocity of rotation:

The angular momentum is a vector quantity, according to formula (6.3), the angular momentum is directed in the same way as the angular velocity.

The basic equation of the dynamics of rotational motion in impulsive form.

The angular acceleration of a body is equal to the change in angular velocity divided by the time interval during which this change occurred: Substitute this expression into the basic equation for the dynamics of rotational motion hence I(ω 2 - ω 1) = MΔt, or IΔω = MΔt.

Thus,

∆L = M∆t. (6.4)

The change in the angular momentum is equal to the product of the total moment of forces acting on the body or system and the time of action of these forces.

Law of conservation of angular momentum:

If the total moment of forces acting on a body or a system of bodies with a fixed axis of rotation is equal to zero, then the change in the angular momentum is also equal to zero, i.e., the angular momentum of the system remains constant.

∆L=0, L=const.

The change in the momentum of the system is equal to the total momentum of the forces acting on the system.

The spinning skater spreads his arms out to the sides, thereby increasing the moment of inertia in order to decrease the angular velocity of rotation.

The law of conservation of angular momentum can be demonstrated using the following experiment, called the "experiment with the Zhukovsky bench." A person stands on a bench with a vertical axis of rotation passing through its center. The man holds dumbbells in his hands. If the bench is made to rotate, then a person can change the speed of rotation by pressing the dumbbells to his chest or lowering his arms, and then spreading them apart. Spreading his arms, he increases the moment of inertia, and the angular velocity of rotation decreases (Fig. 6.11, a), lowering his hands, he reduces the moment of inertia, and the angular velocity of rotation of the bench increases (Fig. 6.11, b).

A person can also make a bench rotate by walking along its edge. In this case, the bench will rotate in the opposite direction, since the total angular momentum must remain equal to zero.

The principle of operation of devices called gyroscopes is based on the law of conservation of angular momentum. The main property of a gyroscope is the preservation of the direction of the axis of rotation, if external forces do not act on this axis. In the 19th century gyroscopes were used by navigators to navigate the sea.

Kinetic energy of a rotating rigid body.

The kinetic energy of a rotating solid body is equal to the sum of the kinetic energies of its individual particles. Let us divide the body into small elements, each of which can be considered a material point. Then the kinetic energy of the body is equal to the sum of the kinetic energies of the material points of which it consists:

The angular velocity of rotation of all points of the body is the same, therefore,

The value in brackets, as we already know, is the moment of inertia of the rigid body. Finally, the formula for the kinetic energy of a rigid body with a fixed axis of rotation has the form

In the general case of motion of a rigid body, when the axis of rotation is free, its kinetic energy is equal to the sum of the energies of translational and rotational motions. So, the kinetic energy of a wheel, the mass of which is concentrated in the rim, rolling along the road at a constant speed, is equal to

The table compares the formulas of the mechanics of the translational motion of a material point with similar formulas for the rotational motion of a rigid body.

Let us determine the kinetic energy of a rigid body rotating around a fixed axis. Let's divide this body into n material points. Each point moves with a linear speed υ i =ωr i , then the kinetic energy of the point

or

The total kinetic energy of a rotating rigid body is equal to the sum of the kinetic energies of all its material points:

(3.22)

(J - moment of inertia of the body about the axis of rotation)

If the trajectories of all points lie in parallel planes (like a cylinder rolling down an inclined plane, each point moves in its own plane fig), this is flat motion. According to Euler's principle, plane motion can always be decomposed in an infinite number of ways into translational and rotational motion. If the ball falls or slides along an inclined plane, it only moves forward; when the ball rolls, it also rotates.

If a body performs translational and rotational motions at the same time, then its total kinetic energy is equal to

(3.23)

From a comparison of the formulas for kinetic energy for translational and rotational motions, it can be seen that the measure of inertia during rotational motion is the moment of inertia of the body.

§ 3.6 The work of external forces during the rotation of a rigid body

When a rigid body rotates, its potential energy does not change, therefore, the elementary work of external forces is equal to the increment in the kinetic energy of the body:

dA = dE or

Considering that Jβ = M, ωdr = dφ, we have α of the body at a finite angle φ equals

(3.25)

When a rigid body rotates around a fixed axis, the work of external forces is determined by the action of the moment of these forces about a given axis. If the moment of forces about the axis is equal to zero, then these forces do not produce work.

Examples of problem solving

Example 2.1. flywheel massm=5kg and radiusr= 0.2 m rotates around the horizontal axis with a frequencyν 0 =720 min -1 and stops when brakingt=20 s. Find the braking torque and the number of revolutions before stopping.

To determine the braking torque, we apply the basic equation for the dynamics of rotational motion

where I=mr 2 is the moment of inertia of the disk; Δω \u003d ω - ω 0, and ω \u003d 0 is the final angular velocity, ω 0 \u003d 2πν 0 is the initial one. M is the braking moment of the forces acting on the disk.

Knowing all the quantities, it is possible to determine the braking torque

Mr 2 2πν 0 = МΔt (1)

(2)

From the kinematics of rotational motion, the angle of rotation during the rotation of the disk to a stop can be determined by the formula

(3)

where β is the angular acceleration.

According to the condition of the problem: ω = ω 0 - βΔt, since ω=0, ω 0 = βΔt

Then expression (2) can be written as:

Example 2.2. Two flywheels in the form of disks of the same radii and masses were spun up to the speed of rotationn= 480 rpm and left to themselves. Under the action of the friction forces of the shafts on the bearings, the first one stopped aftert\u003d 80 s, and the second didN= 240 revolutions to stop. In which flywheel, the moment of the friction forces of the shafts on the bearings was greater and how many times.

We will find the moment of forces of thorns M 1 of the first flywheel using the basic equation of the dynamics of rotational motion

M 1 Δt \u003d Iω 2 - Iω 1

where Δt is the time of action of the moment of friction forces, I \u003d mr 2 - the moment of inertia of the flywheel, ω 1 \u003d 2πν and ω 2 \u003d 0 are the initial and final angular velocities of the flywheels

Then

The moment of friction forces M 2 of the second flywheel is expressed through the relationship between the work A of the friction forces and the change in its kinetic energy ΔE k:

where Δφ = 2πN is the angle of rotation, N is the number of revolutions of the flywheel.

Then where

O ratio will be

The friction torque of the second flywheel is 1.33 times greater.

Example 2.3. Mass of a homogeneous solid disk m, masses of loads m 1 and m 2 (fig.15). There is no slip and friction of the thread in the axis of the cylinder. Find the acceleration of the masses and the ratio of the thread tensionsin the process of movement.

There is no slippage of the thread, therefore, when m 1 and m 2 will make translational motion, the cylinder will rotate about the axis passing through the point O. Let's assume for definiteness that m 2 > m 1.

Then the load m 2 is lowered and the cylinder rotates clockwise. Let us write down the equations of motion of the bodies included in the system

The first two equations are written for bodies with masses m 1 and m 2 performing translational motion, and the third equation is for a rotating cylinder. In the third equation, on the left is the total moment of forces acting on the cylinder (the moment of force T 1 is taken with a minus sign, since the force T 1 tends to turn the cylinder counterclockwise). On the right, I is the moment of inertia of the cylinder about the axis O, which is equal to

where R is the radius of the cylinder; β is the angular acceleration of the cylinder.

Since there is no thread slip,
. Taking into account the expressions for I and β, we get:

Adding the equations of the system, we arrive at the equation

From here we find the acceleration a cargo

It can be seen from the resulting equation that the thread tensions will be the same, i.e. =1 if the mass of the cylinder is much less than the mass of the weights.

Example 2.4. A hollow ball with mass m = 0.5 kg has an outer radius R = 0.08m and an inner radius r = 0.06m. The ball rotates around an axis passing through its center. At a certain moment, a force begins to act on the ball, as a result of which the angle of rotation of the ball changes according to the law
. Determine the moment of the applied force.

We solve the problem using the basic equation of the dynamics of rotational motion
. The main difficulty is to determine the moment of inertia of the hollow ball, and the angular acceleration β is found as
. The moment of inertia I of a hollow ball is equal to the difference between the moments of inertia of a ball of radius R and a ball of radius r:

where ρ is the density of the ball material. We find the density, knowing the mass of a hollow ball

From here we determine the density of the material of the ball

For the moment of force M we obtain the following expression:

Example 2.5. A thin rod with a mass of 300 g and a length of 50 cm rotates with an angular velocity of 10 s -1 in a horizontal plane around a vertical axis passing through the middle of the rod. Find the angular velocity if, during rotation in the same plane, the rod moves so that the axis of rotation passes through the end of the rod.

We use the law of conservation of angular momentum

(1)

(J i - moment of inertia of the rod relative to the axis of rotation).

For an isolated system of bodies, the vector sum of the angular momentum remains constant. Due to the fact that the distribution of the mass of the rod relative to the axis of rotation changes, the moment of inertia of the rod also changes in accordance with (1):

J 0 ω 1 = J 2 ω 2 . (2)

It is known that the moment of inertia of the rod about the axis passing through the center of mass and perpendicular to the rod is equal to

J 0 \u003d mℓ 2 / 12. (3)

According to the Steiner theorem

J = J 0 +m a 2

(J is the moment of inertia of the rod about an arbitrary axis of rotation; J 0 is the moment of inertia about a parallel axis passing through the center of mass; a- distance from the center of mass to the selected axis of rotation).

Let's find the moment of inertia about the axis passing through its end and perpendicular to the rod:

J 2 \u003d J 0 +m a 2 , J 2 = mℓ 2 /12 +m(ℓ/2) 2 = mℓ 2 /3. (4)

Let us substitute formulas (3) and (4) into (2):

mℓ 2 ω 1 /12 = mℓ 2 ω 2 /3

ω 2 \u003d ω 1 /4 ω 2 \u003d 10s-1/4 \u003d 2.5s -1

Example 2.6 . mass manm= 60 kg, standing on the edge of the platform with mass M = 120 kg, rotating by inertia around a fixed vertical axis with a frequency ν 1 =12min -1 , goes to its center. Considering the platform as a round homogeneous disk, and the person as a point mass, determine with what frequency ν 2 the platform will then rotate.

Given: m=60kg, M=120kg, ν 1 =12min -1 = 0.2s -1 .

To find: v 1

Decision: According to the condition of the problem, the platform with the person rotates by inertia, i.e. the resulting moment of all forces applied to the rotating system is zero. Therefore, for the “platform-man” system, the law of conservation of momentum is fulfilled

I 1 ω 1 = I 2 ω 2

where
- the moment of inertia of the system when a person is standing on the edge of the platform (we took into account that the moment of inertia of the platform is equal to (R is the radius p
platform), the moment of inertia of a person at the edge of the platform is mR 2).

- the moment of inertia of the system when a person stands in the center of the platform (we took into account that the moment of a person standing in the center of the platform is equal to zero). Angular velocity ω 1 = 2π ν 1 and ω 1 = 2π ν 2 .

Substituting the written expressions into formula (1), we obtain

whence the desired rotational speed

Answer: v 2 =24 min -1 .

Consider an absolutely rigid body rotating about a fixed axis. Let's mentally break this body into infinitely small pieces with infinitely small sizes and masses. m v t., t 3 ,... at distances R v R 0 , R 3 ,... from the axis. Kinetic energy of a rotating body we find as the sum of the kinetic energies of its small parts:

- moment of inertia rigid body relative to the given axis 00,. From a comparison of the formulas for the kinetic energy of translational and rotational motions, it is obvious that moment of inertia in rotational motion is analogous to mass in translational motion. Formula (4.14) is convenient for calculating the moment of inertia of systems consisting of individual material points. To calculate the moment of inertia of solid bodies, using the definition of the integral, you can convert it to the form

It is easy to see that the moment of inertia depends on the choice of axis and changes with its parallel translation and rotation. Let us find the values ​​of the moments of inertia for some homogeneous bodies.

From formula (4.14) it is obvious that moment of inertia of a material point equals

where t - point mass; R- distance to the axis of rotation.

It is easy to calculate the moment of inertia for hollow thin-walled cylinder(or a special case of a cylinder with a small height - thin ring) radius R about the axis of symmetry. The distance to the axis of rotation of all points for such a body is the same, equal to the radius and can be taken out from under the sign of the sum (4.14):

Rice. 4.5

solid cylinder(or a special case of a cylinder with a small height - disk) radius R to calculate the moment of inertia about the axis of symmetry requires the calculation of the integral (4.15). It can be understood in advance that the mass in this case, on average, is concentrated somewhat closer to the axis than in the case of a hollow cylinder, and the formula will be similar to (4.17), but a coefficient less than one will appear in it. Let's find this coefficient. Let a solid cylinder have density p and height A. Let us divide it into hollow cylinders (thin cylindrical surfaces) with thickness dr(Fig. 4.5 shows a projection perpendicular to the axis of symmetry). The volume of such a hollow cylinder of radius r is equal to the surface area multiplied by the thickness: dV = 2nrhdr, weight: dm=2nphrdr, and the moment of inertia in accordance with formula (4.17): dj=

= r 2 dm = 2lr/?g Wr. The total moment of inertia of a solid cylinder is obtained by integrating (summing) the moments of inertia of hollow cylinders:

Similarly searched moment of inertia of a thin rod length L and the masses t, if the axis of rotation is perpendicular to the rod and passes through its middle. Let's break this

Taking into account the fact that the mass of a solid cylinder is related to the density by the formula t = nR 2 hp, we finally have moment of inertia of a solid cylinder:

Rice. 4.6

rod in accordance with fig. 4.6 pieces thick dl. The mass of such a piece is dm = mdl/L, and the moment of inertia in accordance with formula (4.6): dj = l 2 dm = l 2 mdl/L. The total moment of inertia of a thin rod is obtained by integrating (summing) the moments of inertia of the pieces:

Taking the elementary integral gives the moment of inertia of a thin rod of length L and the masses t

Rice. 4.7

The integral is taken somewhat more complicated when searching moment of inertia of a homogeneous ball radius R and mass /77 with respect to the axis of symmetry. Let a solid ball have density p. Let's break it down as shown in Fig. 4.7 for hollow thin cylinders thickness dr, whose symmetry axis coincides with the axis of rotation of the ball. The volume of such a hollow cylinder of radius G is equal to the surface area multiplied by the thickness:

where is the height of the cylinder h found using the Pythagorean theorem:

Then it is easy to find the mass of the hollow cylinder:

as well as the moment of inertia in accordance with the formula (4.15):

The total moment of inertia of a solid ball is obtained by integrating (summing) the moments of inertia of hollow cylinders:

Taking into account the fact that the mass of a solid ball is related to the density of the shape - 4 .

loy t = -npR A y we finally have the moment of inertia about the axis

symmetry of a homogeneous ball of radius R masses t:

Let's start by considering the rotation of the body around a fixed axis, which we will call the z-axis (Fig. 41.1). The linear speed of the elementary mass is where is the distance of the mass from the axis. Therefore, for the kinetic energy of an elementary mass, the expression is obtained

The kinetic energy of a body is composed of the kinetic energies of its parts:

The sum on the right side of this ratio is the moment of inertia of body 1 about the axis of rotation. Thus, the kinetic energy of a body rotating around a fixed axis is

Let an internal force and an external force act on the mass (see Fig. 41.1). According to (20.5), these forces will do work during the time

Carrying out a cyclic permutation of factors in mixed products of vectors (see (2.34)), we obtain:

where N is the moment of the internal force relative to the point O, N is the analogous moment of the external force.

Summing expression (41.2) over all elementary masses, we obtain the elementary work performed on the body during the time dt:

The sum of moments of internal forces is equal to zero (see (29.12)). Therefore, denoting the total moment of external forces through N, we arrive at the expression

(we used formula (2.21)).

Finally, taking into account that there is an angle through which the body rotates in time, we get:

The sign of the work depends on the sign, i.e., on the sign of the projection of the vector N onto the direction of the vector

So, when the body rotates, the internal forces do not perform work, while the work of external forces is determined by formula (41.4).

Formula (41.4) can be arrived at by using the fact that the work done by all the forces applied to the body goes to increase its kinetic energy (see (19.11)). Taking the differential of both sides of equality (41.1), we arrive at the relation

According to the equation (38.8) so, replacing through we will come to the formula (41.4).

Table 41.1

In table. 41.1, the formulas of the mechanics of rotational motions are compared with similar formulas of the mechanics of translational motion (the mechanics of a point). From this comparison it is easy to conclude that in all cases the role of mass is played by the moment of inertia, the role of force is the moment of force, the role of momentum is played by the moment of momentum, etc.

Formula. (41.1) we obtained for the case when the body rotates around a fixed axis fixed in the body. Now let's assume that the body rotates arbitrarily about a fixed point coinciding with its center of mass.

Let us rigidly connect the Cartesian coordinate system with the body, the origin of which will be placed at the center of mass of the body. The speed of the i-th elementary mass is Therefore, for the kinetic energy of the body, we can write the expression

where is the angle between the vectors Replacing a through and taking into account what we get:

We write the scalar products in terms of the projections of vectors on the axes of the coordinate system associated with the body:

Finally, by combining the terms with the same products of the components of the angular velocity and taking these products out of the signs of the sums, we get: so that formula (41.7) takes the form (compare with (41.1)). When an arbitrary body rotates around one of the main axes of inertia, say the axes and formula (41.7) goes into (41.10.

Thus. the kinetic energy of a rotating body is equal to half the product of the moment of inertia and the square of the angular velocity in three cases: 1) for a body rotating around a fixed axis; 2) for a body rotating around one of the main axes of inertia; 3) for a ball top. In other cases, the kinetic energy is determined by the more complex formulas (41.5) or (41.7).

Consider first a rigid body rotating around a fixed axis OZ with an angular velocity ω (fig.5.6). Let's break the body into elementary masses. The linear velocity of an elementary mass is , where is its distance from the axis of rotation. Kinetic energy i-that elementary mass will be equal to

.

The kinetic energy of the whole body is made up of the kinetic energies of its parts, therefore

.

Considering that the sum on the right side of this relation represents the moment of inertia of the body about the axis of rotation, we finally obtain

. (5.30)

The formulas for the kinetic energy of a rotating body (5.30) are similar to the corresponding formulas for the kinetic energy of the translational motion of a body. They are obtained from the latter by the formal substitution .

In the general case, the motion of a rigid body can be represented as a sum of motions - translational with a speed equal to the speed of the center of mass of the body, and rotation with an angular velocity around the instantaneous axis passing through the center of mass. In this case, the expression for the kinetic energy of the body takes the form

.

Let us now find the work done by the moment of external forces during the rotation of a rigid body. Elementary work of external forces in time dt will be equal to the change in the kinetic energy of the body

Taking the differential from the kinetic energy of rotational motion, we find its increment

.

In accordance with the basic equation of dynamics for rotational motion

Taking into account these relations, we reduce the expression for elementary work to the form

where is the projection of the resulting moment of external forces on the direction of the axis of rotation OZ, is the angle of rotation of the body for the considered period of time.

Integrating (5.31), we obtain a formula for the work of external forces acting on a rotating body

If , then the formula is simplified

Thus, the work of external forces during the rotation of a rigid body about a fixed axis is determined by the action of the projection of the moment of these forces on a given axis.

Gyroscope

A gyroscope is a rapidly rotating symmetrical body, the axis of rotation of which can change its direction in space. So that the axis of the gyroscope can freely rotate in space, the gyroscope is placed in the so-called gimbal suspension (Fig. 5.13). The flywheel of the gyroscope rotates in the inner annular cage around the C 1 C 2 axis passing through its center of gravity. The inner cage, in turn, can rotate in the outer cage around the axis B 1 B 2 perpendicular to C 1 C 2 . Finally, the outer race can freely rotate in the strut bearings around the axis A 1 A 2 perpendicular to the axes C 1 C 2 and B 1 B 2 . All three axes intersect at some fixed point O, called the center of suspension or the fulcrum of the gyroscope. The gyroscope in the gimbal has three degrees of freedom and, therefore, can make any rotation around the center of the gimbal. If the suspension center of the gyroscope coincides with its center of gravity, then the resulting moment of gravity of all parts of the gyroscope relative to the suspension center is equal to zero. Such a gyroscope is called balanced.

Let us now consider the most important properties of the gyroscope, which have found wide application for it in various fields.

1) Sustainability.

With any rotation of the balanced gyroscope rack, its axis of rotation remains the same direction with respect to the laboratory frame of reference. This is due to the fact that the moment of all external forces, equal to the moment of friction forces, is very small and practically does not cause a change in the angular momentum of the gyroscope, i.e.

Since the angular momentum is directed along the axis of rotation of the gyroscope, its orientation must remain unchanged.

If an external force acts for a short time, then the integral that determines the increment of the angular momentum will be small

. (5.34)

This means that under short-term influences of even large forces, the movement of a balanced gyroscope changes little. The gyroscope, as it were, resists all attempts to change the magnitude and direction of its angular momentum. Connected with this is the remarkable stability that the motion of a gyroscope acquires after bringing it into rapid rotation. This property of the gyroscope is widely used to automatically control the movement of aircraft, ships, rockets and other vehicles.

If, however, the gyroscope is acted upon for a long time by a moment of external forces constant in direction, then the axis of the gyroscope is finally set in the direction of the moment of external forces. This phenomenon is used in the gyrocompass. This device is a gyroscope, the axis of which can freely rotate in a horizontal plane. Due to the daily rotation of the Earth and the action of the moment of centrifugal forces, the axis of the gyroscope rotates so that the angle between and becomes minimal (Fig. 5.14). This corresponds to the position of the gyroscope axis in the meridian plane.

2). Gyroscopic effect.

If a pair of forces and is applied to a rotating gyroscope, tending to rotate it around an axis perpendicular to the axis of rotation, then it will rotate around the third axis, perpendicular to the first two (Fig. 5.15). This unusual behavior of the gyroscope is called the gyroscopic effect. It is explained by the fact that the moment of a pair of forces is directed along the O 1 O 1 axis and a change in the vector by a value over time will have the same direction. As a result, the new vector will rotate about the O 2 O 2 axis. Thus, the seemingly unnatural behavior of the gyroscope fully corresponds to the laws of the dynamics of rotational motion

3). Gyro precession.

The precession of a gyroscope is the conical movement of its axis. It occurs when the moment of external forces, remaining constant in magnitude, rotates simultaneously with the axis of the gyroscope, forming a right angle with it all the time. To demonstrate precession, a bicycle wheel with an extended axle, brought into rapid rotation (Fig. 5.16), can serve.

If the wheel is suspended by the extended end of the axle, then its axle will begin to precess around the vertical axis under the action of its own weight. A rapidly rotating top can also serve as a demonstration of precession.

Find out the reasons for the precession of the gyroscope. Consider an unbalanced gyroscope whose axis can freely rotate around a certain point O (Fig. 5.16). The moment of gravity applied to the gyroscope is equal in magnitude

where is the mass of the gyroscope, is the distance from the point O to the center of mass of the gyroscope, is the angle formed by the axis of the gyroscope with the vertical. The vector is directed perpendicular to the vertical plane passing through the axis of the gyroscope.

Under the action of this moment, the angular momentum of the gyroscope (its beginning is placed at point O) will receive an increment in time, and the vertical plane passing through the axis of the gyroscope will rotate by an angle. The vector is always perpendicular to , therefore, without changing in magnitude, the vector only changes in direction. In this case, after a while, the relative position of the vectors and will be the same as at the initial moment. As a result, the axis of the gyroscope will continuously rotate around the vertical, describing a cone. This movement is called precession.

Let us determine the angular velocity of precession. According to Fig.5.16, the angle of rotation of the plane passing through the axis of the cone and the axis of the gyroscope is equal to

where is the angular momentum of the gyroscope, and is its increment over time.

Dividing by , taking into account the above relations and transformations, we obtain the angular velocity of precession

. (5.35)

For gyroscopes used in technology, the angular velocity of precession is millions of times less than the rotational speed of the gyroscope.

In conclusion, we note that the phenomenon of precession is also observed in atoms due to the orbital motion of electrons.

Examples of applying the laws of dynamics

When rotating

1. Consider some examples of the law of conservation of angular momentum, which can be implemented using the Zhukovsky bench. In the simplest case, the Zhukovsky bench is a disc-shaped platform (chair) that can freely rotate around a vertical axis on ball bearings (Fig. 5.17). The demonstrator sits or stands on the bench, after which it is brought into rotational motion. Due to the fact that the friction forces due to the use of bearings are very small, the angular momentum of the system consisting of the bench and the demonstrator, relative to the axis of rotation, cannot change in time if the system is left to itself. If the demonstrator holds heavy dumbbells in his hands and spreads his arms to the sides, then he will increase the moment of inertia of the system, and therefore the angular velocity of rotation must decrease so that the angular momentum remains unchanged.

According to the law of conservation of angular momentum, we compose an equation for this case

where is the moment of inertia of the person and the bench, and is the moment of inertia of the dumbbells in the first and second positions, and are the angular velocities of the system.

The angular velocity of rotation of the system when breeding dumbbells to the side will be equal to

.

The work done by a person when moving dumbbells can be determined through a change in the kinetic energy of the system

2. Let's give one more experiment with Zhukovsky's bench. The demonstrator sits or stands on a bench and is given a rapidly rotating wheel with a vertically directed axis (Fig. 5.18). The demonstrator then turns the wheel 180 0 . In this case, the change in the angular momentum of the wheel is entirely transferred to the bench and the demonstrator. As a result, the bench, together with the demonstrator, comes into rotation with an angular velocity determined on the basis of the law of conservation of angular momentum.

The angular momentum of the system in the initial state is determined only by the angular momentum of the wheel and is equal to

where is the moment of inertia of the wheel, is the angular velocity of its rotation.

After turning the wheel at an angle of 180 0, the moment of momentum of the system will already be determined by the sum of the moment of momentum of the bench with the person and the moment of momentum of the wheel. Taking into account the fact that the momentum vector of the wheel has changed its direction to the opposite, and its projection on the vertical axis has become negative, we obtain

,

where is the moment of inertia of the "man-platform" system, is the angular velocity of rotation of the bench with the person.

According to the law of conservation of angular momentum

and .

As a result, we find the speed of rotation of the bench

3. Thin rod mass m and length l rotates with an angular velocity ω=10 s -1 in a horizontal plane around a vertical axis passing through the middle of the rod. Continuing to rotate in the same plane, the rod moves so that the axis of rotation now passes through the end of the rod. Find the angular velocity in the second case.

In this problem, due to the fact that the distribution of the mass of the rod relative to the axis of rotation changes, the moment of inertia of the rod also changes. In accordance with the law of conservation of the angular momentum of an isolated system, we have

Here - the moment of inertia of the rod about the axis passing through the middle of the rod; - the moment of inertia of the rod about the axis passing through its end and found by Steiner's theorem.

Substituting these expressions into the law of conservation of angular momentum, we obtain

,

.

4. Rod length L=1.5 m and weight m 1=10 kg is hinged at the upper end. A bullet hits the center of the rod with a mass m2=10 g, flying horizontally at a speed of =500 m/s, and gets stuck in the rod. Through what angle will the rod deviate after the impact?

Let's imagine in Fig. 5.19. system of interacting bodies "rod-bullet". The moments of external forces (gravity, axis reaction) at the moment of impact are equal to zero, so we can use the law of conservation of angular momentum

The angular momentum of the system before impact is equal to the angular momentum of the bullet relative to the suspension point

The angular momentum of the system after an inelastic impact is determined by the formula

,

where is the moment of inertia of the rod relative to the point of suspension, is the moment of inertia of the bullet, is the angular velocity of the rod with the bullet immediately after the impact.

Solving the resulting equation after substitution, we find

.

Let us now use the law of conservation of mechanical energy. Let us equate the kinetic energy of the rod after the bullet hits it with its potential energy at the highest point of the ascent:

,

where is the height of the center of mass of the given system.

Having carried out the necessary transformations, we obtain

The deflection angle of the rod is related to the value by the ratio

.

Having carried out the calculations, we obtain =0,1p=18 0 .

5. Determine the acceleration of the bodies and the tension of the thread on the Atwood machine, assuming that (Fig. 5.20). The moment of inertia of the block about the axis of rotation is I, block radius r. Ignore the mass of the thread.

Let's arrange all the forces acting on the loads and the block, and compose the dynamics equations for them

If there is no slippage of the thread along the block, then the linear and angular acceleration are related by the relation

Solving these equations, we get

Then we find T 1 and T 2 .

6. A thread is attached to the pulley of the Oberbeck cross (Fig. 5.21), to which a load of mass M= 0.5 kg. Determine how long it takes for a load to fall from a height h=1 m to the bottom position. Pulley radius r\u003d 3 cm. Four weights of mass m=250g each at a distance R= 30 cm from its axis. Neglect the moment of inertia of the cross itself and the pulley compared to the moment of inertia of the weights.