The equation of kinetic energy of a rotating body. Kinetic energy during rotational motion

« 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.

angular momentum solid 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.

In this way,

∆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 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 mechanics forward movement material point with similar formulas for the rotational motion of a rigid body.

Since the solid is special case system of material points, then the kinetic energy of the body during rotation 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 along 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 rigid body takes the form

The sum on the right side of this expression, in accordance with (4.4), is the moment of inertia of this body about the given axis of rotation. Therefore, the formula for calculating the kinetic energy of a rigid body rotating relative to a fixed axis will take the final form:

. (4.21)

It is taken into account here that

The calculation of the kinetic energy of a rigid body in the case of arbitrary motion becomes much more complicated. Consider a 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 as

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 some point of the body. In this case, in the last expression is the speed of the center of inertia of the body, - the radii of the circles along which the points of the body rotate with an angular velocity around the axis passing through the center of its inertia. Since with such a movement ^, then the vector equal to lies in the plane of the trajectory of the point.

Based on the above, the kinetic energy of the body during its plane motion is equal to

Raising the expression in parentheses to the square and taking out the constant values for all points of the body beyond the sum sign, we obtain

Here it is taken into account that ^.

Consider each term on the right side of the last expression separately. The first term, due to the 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 this case lies on the axis of rotation. The last term, taking into account (4.4), takes the form . Now, finally, the kinetic energy for an arbitrary, but plane motion of a rigid body can be represented as the sum of two terms:

, (4.23)

where the first term is the kinetic energy of a material point with mass, equal to the mass body and moving with the speed that the center of mass of the body has;

the second term is the kinetic energy of a body rotating about 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 a plane motion using (4.23).

Test questions.

4.4. In what cases does (4.23) go over to (4.21)?

4.5. What will the formula for the kinetic energy of a body look like during its plane motion 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 it's work internal forces during rotation of a rigid body is zero.

The expression for the kinetic energy of a rotating body, taking into account that the linear velocity of an arbitrary material point that makes up the body, relative to the axis of rotation is equal, has the form

where is the moment of inertia of the body about the chosen axis of rotation, its angular velocity about this axis, the moment of momentum of the body about the axis of rotation.

If the body performs a translational rotational motion, then the calculation of the kinetic energy depends on the choice of the pole, relative to which the motion of the body is described. The end result will be the same. So, if for a round body rolling at a speed v without slipping with a radius R and a coefficient of inertia k, the pole is taken at its CM, at point C, then its moment of inertia, and the angular velocity of rotation around the axis С. Then the kinetic energy of the body

If the pole is taken at the point O of contact between the body and the surface through which the instantaneous axis of rotation of the body passes, then its moment of inertia about the axis O becomes equal to . Then the kinetic energy of the body, taking into account that the angular velocities of rotation of the body relative to parallel axes are the same and the body performs a pure rotation around the O axis, will be equal to . The result is the same.

The theorem on the kinetic energy of a body performing a complex motion will have the same form as for its translational motion: .

Example 1 A body of mass m is tied to the end of a thread wound on a cylindrical block of radius R and mass M. The body is raised to a height h and released (Fig. 65). After an inelastic jerk of the thread, the body and the block immediately begin to move together. What heat will be released during a jerk? What will be the acceleration of the movement of the body and the tension of the thread after the jerk? What will be the speed of the body and the distance traveled by it after the jerk of the thread after time t?

Given: M, R, m, h, g, t. Find: Q -?, a -?, T -?, v -?, s -?

Solution: The speed of the body before the thread is pulled. After the thread is jerked, the block and the body will begin to rotate about the axis of the block O and will behave like bodies with moments of inertia about this axis equal to and . Their total moment of inertia about the axis of rotation.

The jerk of the thread is a fast process and during the jerk the law of conservation of the angular momentum of the block-body system takes place, which, due to the fact that the body and the block immediately after the jerk begin to move together, has the form: . Where does the initial angular velocity of rotation of the block come from , and the initial linear velocity of the body .

The kinetic energy of the system due to the conservation of its angular momentum immediately after the thread jerk is equal to . The heat released during the jerk according to the law of conservation of energy

The dynamic equations of motion of the bodies of the system after a jerk of the thread do not depend on their initial speed. For a block, it looks like or , and for the body . Adding these two equations, we get . Where does the acceleration of the movement of the body come from. Thread tension force

The kinematic equations of motion of the body after the jerk will have the form where all parameters are known.

Answer: . .

Example 2. Two round bodies with inertia coefficients (hollow cylinder) and (ball) located at the base of an inclined plane with an inclination angle α report the same initial speeds directed upwards along an inclined plane. To what height and in what time will the bodies rise to this height? What are the accelerations of the body's rise? How many times do the heights, times and accelerations of the rise of bodies differ? Bodies move along an inclined plane without slipping.

Given: . Find:

Solution: The body is affected by: gravity m g, reaction of the inclined plane N, and the force of adhesion friction (Fig. 67). The work of the normal reaction and the adhesion friction force (there is no slippage and no heat is released at the point of adhesion of the body and the plane.) Are equal to zero: , therefore, to describe the motion of bodies, it is possible to apply the law of conservation of energy: . Where .

We find the times and accelerations of the motion of bodies from the kinematic equations . Where , . The ratio of heights, times and accelerations of the rise of bodies:

Answer: , , , .

Example 3. A bullet of mass , flying at speed , hits the center of a ball of mass M and radius R, attached to the end of a rod of mass m and length l, suspended at point O by its second end, and flies out of it with speed (Fig. 68). Find the angular velocity of rotation of the rod-ball system immediately after the impact and the angle of deflection of the rod after the impact of the bullet.

Given: . Find:

Solution: The moments of inertia of the rod and the ball relative to the point O of the suspension of the rod according to the Steiner theorem: and . Total moment of inertia of the rod-ball system . The impact of a bullet is a fast process, and the law of conservation of the angular momentum of the bullet-rod-ball system takes place (the bodies begin to rotate after the collision): . Where does the angular velocity of the rod-ball system come from immediately after the impact?

The position of the CM of the rod-ball system relative to the suspension point O: . The law of conservation of energy for the CM of the system after impact, taking into account the law of conservation of the angular momentum of the system upon impact, has the form . Where is the height of the CM of the system after the impact . The deflection angle of the rod after impact is determined by the condition .

Answer: , , .

Example 4. To a round body of mass m and radius R, with a coefficient of inertia k, rotating with an angular velocity , a block is pressed with a force N (Fig. 69). After what time will the cylinder stop and how much heat will be released when the shoe rubs against the cylinder during this time? The coefficient of friction between the pad and the cylinder is .

Given: Find:

Solution: The work of the friction force until the body stops according to the kinetic energy theorem is equal to . Heat released during rotation .

The equation of rotational motion of the body has the form . Where angular acceleration its slow rotation . Time of rotation of the body before it stops.

Answer: , .

Example 5. round body mass m and radius R with the coefficient of inertia k are untwisted to the angular velocity counterclockwise and placed on a horizontal surface that joins a vertical wall (Fig. 70). After what time will the body stop and how many revolutions will it make before stopping? What will be the heat released during the friction of the body on the surface during this time? The coefficient of friction of the body on the surface is .

Given: . Find:

Solution: The heat released during the rotation of the body until it stops is equal to the work of the friction forces, which can be found from the theorem on the kinetic energy of the body. We have .

Reaction of the horizontal plane. The friction forces acting on the body from the horizontal and vertical surfaces are equal: and .From the system of these two equations we obtain and .

Taking into account these relations, the equation of the rotational motion of the body has the form

Answer: , , , .

Example 6. A round body with a coefficient of inertia k rolls down without slipping from the top of a hemisphere of radius R, standing on a horizontal surface (Fig. 71). At what height and with what speed will it break away from the hemisphere and with what speed will it fall onto a horizontal surface?

Given: k, g, R. Find:

Solution: forces acting on the body . Work and 0, (there is no slippage and heat is not released at the point of coupling of the hemisphere and the ball), therefore, to describe the motion of the body, it is possible to apply the law of conservation of energy. Newton's second law for the CM of a body at the point of its separation from the hemisphere, taking into account that at this point it has the form , whence . The law of conservation of energy for the initial point and the separation point of the body has the form . Whence the height and speed of separation of the body from the hemisphere are equal, .

After the separation of the body from the hemisphere, only its translational kinetic energy changes, therefore the law of conservation of energy for the points of separation and fall of the body to the ground has the form . Where, taking into account, we get . For a body sliding on the surface of a hemisphere without friction, k=0 and , , .

Answer: , , .

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

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 of 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 disk rotation to 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. (four)

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 .

Find: v 1

Solution: 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 .

The main dynamic characteristics of rotational motion are the angular momentum about the rotation axis z:

and kinetic energy

In the general case, the energy during rotation with angular velocity is found by the formula:

, where is the inertia tensor .

In thermodynamics

By exactly the same reasoning as in the case of translational motion, equipartition implies that at thermal equilibrium the average rotational energy of each particle of a monatomic gas is: (3/2)k B T. Similarly, the equipartition theorem allows one to calculate the root-mean-square angular velocity of molecules.

The equation of kinetic energy of a rotating body. Kinetic energy during rotational motion

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

Examples of problem solving

In thermodynamics

see also

See what "Energy of rotational motion" is in other dictionaries: