Let's start by considering the rotation of the body around a motionless axis, which we will call the z axis (Fig. 41.1). The linear speed of an elementary mass is equal to where is the distance of the mass from the axis. Therefore, for the kinetic energy of the elementary mass we obtain the expression
Kinetic energy body is composed of the kinetic energies of its parts:
The sum on the right side of this relationship represents the moment of inertia of body 1 relative to the axis of rotation. Thus, the kinetic energy of a body rotating around fixed axis equal to
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 in time
Having carried out a cyclic rearrangement of factors in mixed products of vectors (see (2.34)), we obtain:
where N is the moment of internal force relative to point O, N is a similar moment of external force.
Having summed up expression (41.2) over all elementary masses, we obtain the elementary work performed on the body during time dt:
Sum of moments internal forces is equal to zero (see (29.12)). Therefore, denoting the total moment 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 over time, we obtain:
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 a body rotates, internal forces do no work, but the work of external forces is determined by formula (41.4).
Formula (41.4) can be arrived at by taking advantage of the fact that the work done by all forces applied to the body goes towards increasing its kinetic energy (see (19.11)). Taking the differential from both sides of equality (41.1), we arrive at the relation
According to equation (38.8) so, replacing through we arrive at formula (41.4).
Table 41.1
In table 41.1 the formulas of the mechanics of rotational motion are compared with similar formulas of mechanics forward motion(mechanics 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 played by the moment of force, the role of momentum is played by angular momentum, etc.
Formula. (41.1) we obtained for the case when the body rotates around a stationary axis fixed in the body. Now let us assume that the body rotates in an arbitrary manner relative to a fixed point coinciding with its center of mass.
We will rigidly associate a Cartesian coordinate system with the body, the origin of which will be placed at the center of mass of the body. Speed i elementary mass is equal to 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 that we get:
Let's write it down dot products through projections of vectors on the axes of the coordinate system associated with the body:
Finally, combining terms with identical products of angular velocity components and taking these products out of the signs of the sums, we obtain: so formula (41.7) takes the form (cf. (41.1)). When an arbitrary body rotates around one of the main axes of inertia, say the axis and, formula (41.7) becomes (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, kinetic energy is determined more clearly complex formulas(41.5) or (41.7).
Kinetic energy of rotation
Lecture 3. Rigid body dynamics
Lecture outline
3.1. Moment of power.
3.2. Basic Equations rotational movement. Moment of inertia.
3.3. Kinetic energy of rotation.
3.4. Moment of impulse. Law of conservation of angular momentum.
3.5. Analogy between translational and rotational motion.
Moment of power
Let us consider the motion of a rigid body around a fixed axis. Let the rigid body have a fixed axis of rotation OO ( Fig.3.1) and an arbitrary force is applied to it.
Rice. 3.1
Let us decompose the force into two components of force, the force lies in the plane of rotation, and the force is parallel to the axis of rotation. Then we will decompose the force into two components: – acting along the radius vector and – perpendicular to it.
Not every force applied to a body will rotate it. The forces create pressure on the bearings, but do not rotate it.
A force may or may not throw a body out of balance, depending on where in the radius vector it is applied. Therefore, the concept of moment of force about an axis is introduced. A moment of power relative to the axis of rotation is called the vector product of the radius vector and the force.
The vector is directed along the axis of rotation and is determined by the cross product rule or the right screw rule or the gimlet rule.
Modulus of moment of force
where α is the angle between the vectors and .
From Fig. 3.1. it's clear that .
r 0– the shortest distance from the axis of rotation to the line of action of the force is called the shoulder of the force. Then the moment of force can be written
M = F r 0 . (3.3)
From Fig. 3.1.
Where F– projection of the vector onto the direction, perpendicular to the vector radius vector. In this case, the moment of force is equal to
. (3.4)
If several forces act on a body, then the resulting moment of force is equal to the vector sum of the moments of the individual forces, but since all moments are directed along the axis, they can be replaced by an algebraic sum. The moment will be considered positive if it rotates the body clockwise and negative if it rotates counterclockwise. If all moments of forces () are equal to zero, the body will be in equilibrium.
The concept of torque can be demonstrated using a "capricious coil". The spool of thread is pulled by the free end of the thread ( rice. 3.2).
Rice. 3.2
Depending on the direction of the thread tension, the spool rolls in one direction or another. If pulled at an angle α , then the moment of force about the axis ABOUT(perpendicular to the figure) rotates the coil counterclockwise and it rolls back. In case of tension at an angle β the torque is directed counterclockwise and the reel rolls forward.
Using the equilibrium condition (), we can construct simple mechanisms, which are “transformers” of force, i.e. With less force you can lift and move different weights cargo. Levers, wheelbarrows, and blocks are based on this principle. various kinds, which are widely used in construction. To comply with the equilibrium condition in construction cranes To compensate for the moment of force caused by the weight of the load, there is always a system of counterweights that creates a moment of force of the opposite sign.
3.2. Basic equation of rotation
movements. Moment of inertia
Consider an absolutely rigid body rotating around a fixed axis OO(Fig.3.3). Let us mentally divide this body into elements with masses Δ m 1, Δ m 2, …, Δ m n. When rotated, these elements will describe circles with radii r 1,r 2 , …,r n. Forces act on each element accordingly F 1,F 2 , …,Fn. Rotation of a body around an axis OO occurs under the influence of the full torque M.
M = M 1 + M 2 + … + M n (3.4)
Where M 1 = F 1 r 1, M 2 = F 2 r 2, ..., M n = F n r n
According to Newton's II law, every force F, acting on an element of mass D m, causes acceleration of this element a, i.e.
F i = D m i a i (3.5)
Substituting the corresponding values into (3.4), we obtain
Rice. 3.3
Knowing the relationship between linear angular acceleration ε () and that the angular acceleration is the same for all elements, formula (3.6) will have the form
M = (3.7)
=I (3.8)
I– moment of inertia of the body relative to the fixed axis.
Then we will get
M = I ε (3.9)
Or in vector form
(3.10)
This equation is the basic equation for the dynamics of rotational motion. It is similar in form to equation II of Newton's law. From (3.10) the moment of inertia is equal to
Thus, the moment of inertia of a given body is the ratio of the moment of force to the angular acceleration it causes. From (3.11) it is clear that the moment of inertia is a measure of the inertia of a body with respect to rotational motion. The moment of inertia plays the same role as mass in translational motion. SI unit [ I] = kg m 2. From formula (3.7) it follows that the moment of inertia characterizes the distribution of masses of body particles relative to the axis of rotation.
So, the moment of inertia of an element of mass ∆m moving in a circle of radius r is equal to
I = r 2 D m (3.12)
I= (3.13)
In the case of a continuous mass distribution, the sum can be replaced by the integral
I= ∫ r 2 dm (3.14)
where integration is performed over the entire body mass.
This shows that the moment of inertia of a body depends on the mass and its distribution relative to the axis of rotation. This can be demonstrated experimentally ( Fig.3.4).
Rice. 3.4
Two round cylinders, one hollow (for example, metal), the other solid (wooden) with the same lengths, radii and masses begin to roll simultaneously. A hollow cylinder with big moment inertia, will lag behind the solid one.
The moment of inertia can be calculated if the mass is known m and its distribution relative to the axis of rotation. The simplest case is a ring, when all elements of the mass are located equally from the axis of rotation ( rice. 3.5):
I = 
(3.15)
Rice. 3.5
Let us present expressions for the moments of inertia of various symmetrical bodies of mass m.
1. Moment of inertia rings, hollow thin-walled cylinder relative to the axis of rotation coinciding with the axis of symmetry.
, (3.16)
r– radius of the ring or cylinder
2. For a solid cylinder and disk, the moment of inertia about the axis of symmetry
(3.17)
3. Moment of inertia of the ball about an axis passing through the center
(3.18)
r– radius of the ball
4. Moment of inertia of a thin rod with a long length l relative to an axis perpendicular to the rod and passing through its middle
(3.19)
l– length of the rod.
If the axis of rotation does not pass through the center of mass, then the moment of inertia of the body relative to this axis is determined by Steiner’s theorem.
(3.20)
According to this theorem, the moment of inertia about an arbitrary axis O’O’ ( ) is equal to the moment of inertia about a parallel axis passing through the center of mass of the body ( ) plus the product of body mass times the square of the distance A between axes ( rice. 3.6).
Rice. 3.6
Kinetic energy of rotation
Let us consider the rotation of an absolutely rigid body around a fixed axis OO with angular velocity ω (rice. 3.7). Let's break the solid body into n elementary masses ∆ m i. Each element of mass rotates along a circle of radius r i with linear speed (). Kinetic energy consists of the kinetic energies of individual elements.
(3.21)
Rice. 3.7
Let us recall from (3.13) that – moment of inertia relative to the OO axis.
Thus, the kinetic energy of a rotating body
E k = (3.22)
We considered the kinetic energy of rotation around a fixed axis. If a body is involved in two movements: translational and rotational motion, then the kinetic energy of the body consists of the kinetic energy of translational motion and the kinetic energy of rotation.
For example, a ball of mass m rolls; the center of mass of the ball moves translationally at a speed u (rice. 3.8).
Rice. 3.8
The total kinetic energy of the ball will be equal to
(3.23)
3.4. Moment of impulse. Conservation Law
angular momentum
Physical quantity equal to the product of the moment of inertia I to angular velocity ω , is called angular momentum (angular momentum) L relative to the axis of rotation.
– angular momentum is a vector quantity and its direction coincides with the direction of angular velocity.
Differentiating equation (3.24) with respect to time, we obtain
Where, M– total moment of external forces. In an isolated system there is no torque of external forces ( M=0) and
Since a solid body is special case systems material points, then the kinetic energy of a body when rotating around a fixed axis Z will be equal to the sum of the kinetic energies of all its material points, that is
All material points of a rigid body rotate in this case in circles with radii and with the same angular velocities. The linear speed of each material point of a rigid body is equal to . The kinetic energy of a solid body will take the form
The sum on the right side of this expression, in accordance with (4.4), represents the moment of inertia of this body relative to a given axis of rotation. Therefore, the formula for calculating the kinetic energy of a rigid body rotating relative to a fixed axis will take its final form:
. (4.21)
It is taken into account here that
Calculating the kinetic energy of a rigid body in the case of arbitrary motion becomes much more complicated. Let us consider plane motion when the trajectories of all material points of the body lie in parallel planes. The speed of each material point of a rigid body, according to (1.44), can be represented in the form
,
where as the instantaneous axis of rotation we choose the axis passing through the center of inertia of the body perpendicular to the plane of the trajectory of any point of the body. In this case, in the last expression it represents the speed of the center of inertia of the body, the radii of the circles along which points of the body rotate with angular velocity around an axis passing through its center of inertia. Since with such a movement ^, the vector equal to lies in the plane of the point’s trajectory.
Based on the above, the kinetic energy of a body during its flat movement equal to
.
By squaring the expression in parentheses and taking the constant quantities for all points of the body out of the sum sign, we obtain
It is taken into account here that ^.
Let's consider each term on the right side of the last expression separately. The first term, by virtue of obvious equality, is equal to
The second term is equal to zero, since the sum determines the radius vector of the center of inertia (3.5), which in in this case lies on the axis of rotation. Taking into account (4.4), the last term will take the form . Now, finally, kinetic energy during arbitrary but plane motion of a rigid body can be represented as the sum of two terms:
, (4.23)
where the first term represents the kinetic energy of a material point with a mass equal to the mass of the body and moving with the speed of the center of mass of the body;
the second term represents the kinetic energy of a body rotating around an axis (moving at speed) passing through its center of inertia.
Conclusions: So, the kinetic energy of a rigid body during its rotation around a fixed axis can be calculated using one of the relations (4.21), and in the case of plane motion using (4.23).
Control questions.
4.4. In what cases does (4.23) transform into (4.21)?
4.5. What will the formula for the kinetic energy of a body look like when it moves in a plane if the instantaneous axis of rotation does not pass through the center of inertia? What is the meaning of the quantities included in the formula?
4.6. Show that the work done by internal forces during rotation of a rigid body is zero.
Let us consider an absolutely rigid body rotating about a fixed axis. Let's mentally break this body into infinitesimal 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 equal to area surface multiplied by 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:
Let's define kinetic energy solid, 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 a inclined plane, each point moves in its own plane (Fig), 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.