Kinematic energy of rotational motion. Kinetic energy of a rotating body
Kinetic energy of a rotating body is equal to the sum of the kinetic energies of all particles of the body:
The mass of a particle, its linear (circumferential) speed, proportional to the distance of this particle from the axis of rotation. Substituting into this expression and taking the angular velocity o common for all particles out of the sum sign, we find:
This formula for the kinetic energy of a rotating body can be reduced to a form similar to the expression for kinetic energy forward motion, if you enter the value of the so-called moment of inertia of the body. The moment of inertia of a material point is the product of the mass of the point and the square of its distance from the axis of rotation. The moment of inertia of a body is the sum of the moments of inertia of all material points of the body:
So, the kinetic energy of a rotating body is determined by the following formula:
Formula (2) differs from the formula that determines the kinetic energy of a body in translational motion in that instead of the mass of the body it includes the moment of inertia I and instead of the speed the group velocity
The large kinetic energy of a rotating flywheel is used in technology to maintain uniform running of the machine under suddenly changing loads. Initially, in order to bring a flywheel with a large moment of inertia into rotation, a significant amount of work is required from the machine, but when a large load is suddenly turned on, the machine does not stop and does the work using the reserve of kinetic energy of the flywheel.
Especially massive flywheels are used in rolling mills driven by an electric motor. Here is a description of one of these wheels: “The wheel has a diameter of 3.5 m and weighs. At a normal speed of 600 rpm, the reserve of kinetic energy of the wheel is such that at the moment of rolling the wheel gives the mill a power of 20,000 liters. With. Friction in the bearings is reduced to a minimum by the tale under pressure, and in order to avoid the harmful effects of centrifugal forces of inertia, the wheel is balanced so that a load placed on the circumference of the wheel brings it out of rest."
Let us present (without performing calculations) the values of the moments of inertia of some bodies (it is assumed that each of these bodies has the same density in all its areas).
The moment of inertia of a thin ring relative to an axis passing through its center and perpendicular to its plane (Fig. 55):
The moment of inertia of a circular disk (or cylinder) about an axis passing through its center and perpendicular to its plane (polar moment of inertia of the disk; Fig. 56):
The moment of inertia of a thin round disk relative to an axis coinciding with its diameter (equatorial moment of inertia of the disk; Fig. 57):
The moment of inertia of the ball relative to the axis passing through the center of the ball:
Moment of inertia of a thin spherical layer of radius about an axis passing through the center:
The moment of inertia of a thick spherical layer (a hollow ball having an outer surface radius and a cavity radius) about an axis passing through the center:
The moments of inertia of bodies are calculated using integral calculus. To give an idea of the progress of such calculations, let us find the moment of inertia of the rod relative to the axis perpendicular to it (Fig. 58). Let there be a cross section of the rod, density. Let us select an elementary small part of the rod, which has a length and is located at a distance x from the axis of rotation. Then its mass Since it is located at a distance x from the axis of rotation, its moment of inertia is integrated over the range from zero to I:
Moment of inertia of a rectangular parallelepiped relative to the axis of symmetry (Fig. 59)
Moment of inertia of the ring torus (Fig. 60)
Let us consider how the rotational energy of a body rolling (without sliding) along a plane is related to the energy of the translational motion of this body,
The energy of translational motion of a rolling body is equal to , where is the mass of the body and the speed of translational motion. Let denote the angular velocity of rotation of a rolling body and the radius of the body. It is easy to understand that the speed of translational motion of a body rolling without sliding is equal to the peripheral speed of the body at the points of contact of the body with the plane (during the time when the body makes one revolution, the center of gravity of the body moves a distance, therefore,
Thus,
Rotation energy
hence,
Substituting here the above values of the moments of inertia, we find that:
a) the energy of the rotational motion of a rolling hoop is equal to the energy of its translational motion;
b) the rotational energy of a rolling homogeneous disk is equal to half the energy of translational motion;
c) the rotational energy of a rolling homogeneous ball is the energy of translational motion.
Dependence of the moment of inertia on the position of the axis of rotation. Let the rod (Fig. 61) with the center of gravity at point C rotate with angular velocity (o around the axis O, perpendicular to the plane of the drawing. Let us assume that over a certain period of time it has moved from position A B to and the center of gravity has described an arc. This movement of the rod can be considered as if the rod first translationally (i.e., remaining parallel to itself) moved to position and then rotated around C to position Let us denote (the distance of the center of gravity from the axis of rotation) by a, and the angle by When the rod moves from position A B to position, the movement of each of its particles is the same as the movement of the center of gravity, i.e., it is equal to or around an axis passing through O can be decomposed into two parts.
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 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 is the moment of inertia of the body relative to 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), this flat movement. According to Euler's principle, plane motion can always be decomposed into translational and rotational motion in countless ways. If a ball falls or slides along an inclined plane, it moves only translationally; when the ball rolls, it also rotates.
If a body performs translational and rotational motion simultaneously, 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 is clear that the measure of inertia during rotational motion is the moment of inertia of the body.
§ 3.6 Work of external forces during 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 
Taking into account that Jβ = M, ωdr = dφ, we have α of the body at a finite angle φ is equal to
(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 relative to this axis. If the moment of forces relative to the axis is 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 a horizontal axis with frequencyν 0 =720 min -1 and when braking it stops behindt=20 s. Find the braking torque and the number of revolutions before stopping.
To determine the braking torque, we apply the basic equation of the dynamics of rotational motion
where I=mr 2 – moment of inertia of the disk; Δω =ω - ω 0, and ω =0 is the final angular velocity, ω 0 =2πν 0 is the initial. M is the braking moment of forces acting on the disk.
Knowing all the quantities, you can 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 before stopping can be determined by the formula
(3)
where β is the angular acceleration.
According to the conditions 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 identical radii and masses were spun up to a rotation speedn= 480 rpm and left to our own devices. Under the influence of the friction forces of the shafts on the bearings, the first stopped throught=80 s, and the second one didN= 240 rpm to stop. Which flywheel had a greater moment of friction between the shafts and bearings and by how many times?
We will find the moment of forces of the thorn M 1 of the first flywheel using the basic equation of the dynamics of rotational motion
M 1 Δt = Iω 2 - Iω 1
where Δt is the time of action of the moment of friction forces, I=mr 2 is the moment of inertia of the flywheel, ω 1 = 2πν and ω 2 = 0 – the initial and final angular velocities of the flywheels
Then 
The moment of friction forces M 2 of the second flywheel will be expressed through the connection 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 from
ABOUT 
the ratio will be equal
The frictional moment of the second flywheel is 1.33 times greater.
Example 2.3.
Mass of a homogeneous solid disk m, mass of loads m 1
and m 2
(Fig. 15). There is no slipping or friction of the thread in the cylinder axis. Find the acceleration of the loads and the ratio of the thread tensions
in the process of movement.
There is no slipping of the thread, therefore, when m 1 and m 2 make translational motion, the cylinder will rotate about the axis passing through point O. Let us 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 undergoing translational motion, and the third equation is written 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 rotate the cylinder counterclockwise). On the right I is the moment of inertia of the cylinder relative to the O axis, 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 slippage, then 
. Taking into account the expressions for I and β, we obtain:
Adding the equations of the system, we arrive at the equation
From here we find the acceleration a cargo
From the resulting equation it is clear 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 loads.
Example 2.4.
A hollow ball with mass m = 0.5 kg has an outer radius R = 0.08 m and an inner radius r = 0.06 m. 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 a hollow ball, and we find the angular acceleration β 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. Finding the density by knowing the mass of a hollow ball
From here we determine the density of the ball material
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 is the moment of inertia of the rod relative to the axis of rotation).
For an isolated system of bodies, the vector sum of 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 relative to the axis passing through the center of mass and perpendicular to the rod is equal to
J 0 = mℓ 2 /12. (3)
According to Steiner's theorem
J =J 0 +m A 2
(J is the moment of inertia of the rod relative to an arbitrary axis of rotation; J 0 is the moment of inertia relative to a parallel axis passing through the center of mass; A- distance from the center of mass to the selected axis of rotation).
Let us find the moment of inertia about the axis passing through its end and perpendicular to the rod:
J 2 =J 0 +m A 2, J 2 = mℓ 2 /12 +m(ℓ/2) 2 = mℓ 2 /3. (4)
Let's substitute formulas (3) and (4) into (2):
mℓ 2 ω 1 /12 = mℓ 2 ω 2 /3
ω 2 = ω 1 /4 ω 2 =10s-1/4=2.5s -1
Example 2.6 . Man of massm=60kg, standing on the edge of a platform with mass M=120kg, rotating by inertia around a fixed vertical axis with frequency ν 1 =12min -1 , moves to its center. Considering the platform to be a round homogeneous disk and the person to be 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 .
Find:ν 1
Solution: According to the conditions 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-person” system the law of conservation of angular momentum is satisfied
I 1 ω 1 = I 2 ω 2
Where 
- moment of inertia of the system when a person stands on the edge of the platform (take into account that the moment of inertia of the platform is equal to 
(R – radius n 
platform), the moment of inertia of a person at the edge of the platform is mR 2).
- moment of inertia of the system when a person stands in the center of the platform (take into account that the moment of a person standing in the center of the platform is zero). Angular velocity ω 1 = 2π ν 1 and ω 1 = 2π ν 2.
Substituting the written expressions into formula (1), we obtain
where does the desired rotation speed come from?
Answer: ν 2 =24min -1.
Tasks
1. Determine how many times the effective mass is greater than the gravitating mass of a train weighing 4000 tons, if the mass of the wheels is 15% of the mass of the train. Consider the wheels to be discs with a diameter of 1.02 m. How will the answer change if the diameter of the wheels is half as large?
2. Determine the acceleration with which a wheel pair weighing 1200 kg rolls down a hill with a slope of 0.08. Consider wheels to be disks. Rolling resistance coefficient 0.004. Determine the adhesion force between wheels and rails.
3. Determine the acceleration with which a wheel pair weighing 1400 kg rolls up a hill with a slope of 0.05. Resistance coefficient 0.002. What should the coefficient of adhesion be so that the wheels do not slip? Consider wheels to be disks.
4. Determine with what acceleration a car weighing 40 tons rolls down a hill with a slope of 0.020, if it has eight wheels weighing 1200 kg and a diameter of 1.02 m. Determine the adhesion force of the wheels to the rails. Resistance coefficient 0.003.
5. Determine the pressure force of the brake pads on the tires if a train weighing 4000 tons brakes with an acceleration of 0.3 m/s 2 . The moment of inertia of one wheel pair is 600 kg m 2, the number of axles is 400, the sliding friction coefficient of the pad is 0.18, and the rolling resistance coefficient is 0.004.
6. Determine the braking force acting on a four-axle car weighing 60 tons on the braking platform of a hump if the speed on a track of 30 m decreased from 2 m/s to 1.5 m/s. The moment of inertia of one wheel pair is 500 kg m 2.
7. The locomotive’s speed gauge showed an increase in train speed within one minute from 10 m/s to 60 m/s. It is likely that the drive wheel pair has slipped. Determine the moment of forces acting on the armature of the electric motor. The moment of inertia of the wheelset is 600 kg m 2, the armature is 120 kg m 2. The gear ratio is 4.2. The pressure force on the rails is 200 kN, the sliding friction coefficient of the wheels on the rail is 0.10.
11. KINETIC ENERGY OF ROTATIONAL
MOVEMENTS
Let us derive the formula for the kinetic energy of rotational motion. Let the body rotate with angular velocity ω
relative to a fixed axis. Any small particle of a body undergoes translational motion in a circle with a speed where r i – distance to the axis of rotation, radius of the orbit. Particle kinetic energy
masses m i equal to 
. The total kinetic energy of a system of particles is equal to the sum of their kinetic energies. Let us sum up the formulas for the kinetic energy of particles of a body and take out half the square of the angular velocity, which is the same for all particles, as the sum sign, 
. The sum of the products of the particle masses by the squares of their distances to the axis of rotation is the moment of inertia of the body relative to the axis of rotation 
.
So, the kinetic energy of a body rotating relative to a fixed axis is equal to half the product of the moment of inertia of the body relative to the axis and the square of the angular velocity of rotation:
With the help of rotating bodies, mechanical energy can be stored. Such bodies are called flywheels. Usually these are bodies of revolution. The use of flywheels in the pottery wheel has been known since ancient times. In internal combustion engines, during the power stroke, the piston imparts mechanical energy to the flywheel, which then performs work on rotating the engine shaft for three subsequent strokes. In dies and presses, the flywheel is driven into rotation by a relatively low-power electric motor, accumulates mechanical energy during almost a full revolution and, at a short moment of impact, releases it to the stamping work.
There are numerous attempts to use rotating flywheels to drive vehicles: cars, buses. They are called mahomobiles, gyromobiles. Many such experimental machines were created. It would be promising to use flywheels to accumulate energy during braking of electric trains in order to use the accumulated energy during subsequent acceleration. Flywheel energy storage is known to be used on New York City subway trains.
« Physics - 10th grade"
Why does a skater stretch along the axis of rotation to increase the angular velocity of rotation?
Should a helicopter rotate when its rotor 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 should 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 velocity, 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 further a particle is from the axis of rotation, the greater its momentum. Consequently, for rotational motion it is necessary to introduce another characteristic similar to impulse - 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):
Linear and angular velocities are related by the relation v = ωr, then
All points of a solid object move relative to a fixed axis of rotation with the same angular velocity. A solid 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:
Angular momentum is a vector quantity; according to formula (6.3), angular momentum is directed in the same way as the angular velocity.
The basic equation for the dynamics of rotational motion in pulse form.
The angular acceleration of a body is equal to the change in angular velocity divided by the period of time during which this change occurred: Substitute this expression into the basic equation of the dynamics of rotational motion 
hence I(ω 2 - ω 1) = MΔt, or IΔω = MΔt.
Thus,
ΔL = MΔt. (6.4)
The change in angular momentum is equal to the product of the total moment of forces acting on a body or system and the duration of action of these forces.
Law of conservation of angular momentum:
If the total moment of forces acting on a body or system of bodies having a fixed axis of rotation is equal to zero, then the change in angular momentum is also zero, i.e., the angular momentum of the system remains constant.
ΔL = 0, L = const.
The change in the momentum of the system is equal to the total momentum of the forces acting on the system.
A rotating skater spreads his arms out to the sides, thereby increasing the moment of inertia to reduce the angular velocity of rotation.
The law of conservation of angular momentum can be demonstrated using the following experiment, called the “Zhukovsky bench experiment.” A person stands on a bench that has a vertical axis of rotation passing through its center. A man holds dumbbells in his hands. If the bench is made to rotate, the person can change the speed of rotation by pressing the dumbbells to the chest or lowering the arms and then raising them. By spreading his arms, he increases the moment of inertia, and the angular speed of rotation decreases (Fig. 6.11, a), lowering his arms, he reduces the moment of inertia, and the angular speed 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 should 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 rotation axis if external forces do not act on this axis. In the 19th century Gyroscopes were used by sailors for orientation at 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 parentheses, as we already know, is the moment of inertia of the rigid body. Finally, the formula for the kinetic energy of a rigid body having 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 motion. Thus, 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 for the mechanics of translational motion of a material point with similar formulas for the rotational motion of a rigid body.
Let us 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,... located at distances R v R 0 , R 3,... from the axis. Kinetic energy of a rotating body we find it as the sum of the kinetic energies of its small parts:
- moment of inertia of a rigid body relative to a given axis 00,. From a comparison of the formulas for the kinetic energy of translational and rotational motions, it is obvious that the 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 transform it to the form
It is easy to notice that the moment of inertia depends on the choice of the axis and changes with its parallel translation and rotation. Let's 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 the special case of a cylinder with a low height - thin ring) radius R relative to 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 sum sign (4.14):
Rice. 4.5
Solid cylinder(or a special case of a cylinder with low height - disk) radius R to calculate the moment of inertia relative to the axis of symmetry requires calculating the integral (4.15). You can understand 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 it will contain a coefficient less than unity. 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) of thickness dr(Figure 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:
Search in the same way moment of inertia of a thin rod length L and masses T, if the axis of rotation is perpendicular to the rod and passes through its middle. Let's break this one down
Taking into account the fact that the mass of a solid cylinder is related to 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 equal to 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 masses T
Rice. 4.7
It is somewhat more difficult to take the integral when searching moment of inertia of a homogeneous ball radius R and mass /77 relative to the axis of symmetry. Let a solid ball have density p. Let's break it down in accordance with Fig. 4.7 for hollow thin cylinders thick dr, the axis of symmetry of which coincides with the axis of rotation of the ball. The volume of such a hollow cylinder of radius G 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 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 form-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:
