Rotational movement of bodies

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It is well known that any body can perform two types of movement - translational and healing. The first of them is studied in sufficient detail in the school physics course. But the medical movement was “unlucky.” Due to lack of time, very little attention is paid to it. Meanwhile, the healing movement is found in both nature and technology no less often than the progressive one. In many cases, the actual motion of bodies is a combination of both types of motion. In other cases, translational movement or even resting bodies have healing details. This primarily includes all devices with wheels - from wheelbarrows and carts to airplanes and helicopters. Just try to find a machine that doesn’t have anything in it! Such a common device as a watch is, one might say, “chocked” with wheels interlocking with each other.

Rotational motion is not uncommon in nature. It is enough to recall the complex healing movements of the planets, their satellites, the healing of the Sun and, probably, other stars. Let us note by the way that when applied to planets, the word “rotation” is used in relation to two different types of their movements. They say, for example, that the Earth is healed around the Sun, but they also talk about the Earth being healed around its axis. So, are these the same movements? Can it be said that both are caused by the same principles? It is known that the Earth moves around the Sun along a trajectory close to a circle because it is affected by the force of gravity towards the Sun. But can it be said that the same or a similar force forces the Earth to make healing movements around its axis? Obviously not. In fact, the Earth’s movements can be considered as a combination of translational and linear motion - translational motion around the Sun and linear motion around an axis passing through both earth’s poles.

This article talks about the medical movement of bodies, what distinguishes it from translational movement, and how it is similar to it.

Cinematika of the medical movement

First of all, let’s fill the reader with what they already know about the medical movement from the school physics course. Let’s start with a definition.

Rotational movement of the body is the type of movement in which all points of the body move in circles, the centers of which lie on the same straight line - the axis of healing.

If the axis of healing is fixed, then the body as a whole does not move, and the only thing that changes over time is the angle of rotation, which here plays the same role as the movement of some point of the body during its forward movement.

When a body is rotated through an angle φ, each point passes through a certain arc of a circle. If we denote the length of the arc by S, then there is a simple relationship between the values ​​of S and φ

$$S = \varphi r,$$

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where $r$ is the distance of the point from the rotation axis. Points located at different distances $r$ from the axis of rotation describe arcs of different lengths, but the angle $\varphi$ is the same for the entire body.

The task of the mechanics of rotational movement is to be able to achieve a certain angle of rotation of the body at any time, if the initial angle of rotation is known. To do this, you need to know the rapid change in the angle of rotation, that is, the corner speed, which is usually denoted by the letter $\omega$. For a body rotating at a constant angular velocity,

$$\omega = \frac{\varphi - \varphi_0}{t},$$

where $\varphi - \varphi_0$ is the change in the rotation angle over a period of time of $t$ seconds.

If the angular velocity is constant, then the value of the angle of rotation at any time can be easily found:

$$\varphi = \varphi_0 + \omega t;$$

if the initial rotation angle $\varphi_0$ is zero, then

$$\varphi = \omega t.$$

This expression is similar to the formula $S = vt$ for uniform translational motion. Consequently, the angular velocity in rotational motion plays the same role as linear velocity $v$ for translational motion.

Angular velocity, like linear velocity, may not be constant. Then, in order to find the speed at any time, you need to know the angular acceleration, which, like linear acceleration $a$, is defined as a rapid change in angular velocity. In uniform rotation

$$\alpha = \frac{\omega - \omega_0}{t},$$

where $\omega$ and $\omega_0$ are the values ​​of angular velocity at the end and at the beginning of a long time period of $t$ seconds.

It is easy to find the connection between angular velocity and acceleration and their linear “twins” $v$ and $a$. Since $\omega = \frac{\varphi}{t}$ and $\varphi = \frac{S}{r}$, then $\omega = \frac{S}{rt}$.

But $\frac{S}{t} = v$ therefore

$$\omega = \frac{v}{r} \text{ и } v = \omega r.$$

In the same way, since $\alpha = \frac{\omega - \omega_0}{t}$, and $\omega = \frac{v}{r}$ and $\omega_0 = \frac{v_0}{r}$, then $\alpha = \frac{v - v_0}{rt}$. But $\frac{v - v_0}{t}$ is equal to the linear acceleration of $a$, so

$$\alpha = \frac{a}{r} \text{ и } a = \alpha r.$$

Thus, the numerical values ​​of the linear quantities $S$, $v$ and $a$ for each point of the rotating body are equal to the corresponding values ​​of the angular quantities (they relate not to individual points, but to the entire body), multiplied by the radius $r$ of rotation of this point.

If you use the quantities $\varphi$, $\omega$ and $\alpha$, then you can solve problems related to the rotation of bodies, using well-known formulas of kine-

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The direction of vector $\vec{\alpha}$ coincides with the direction of vector $\omega$, but if the angular velocity decreases, then the directions of vectors $\vec{\alpha}$ and $\omega$ are opposite.

Dynamics of rotational motion

The dynamics of translational motion studies, as is known, the reasons for the appearance of accelerations in bodies and makes it possible to determine their magnitudes and directions. Rotational dynamics do the same for angular acceleration. If the acceleration of the translational motion of a body is caused by a force applied to it, then the reason for the acceleration of the angular acceleration of the body is the torque of the force applied to it. Readers are familiar with it from the 8th or 9th grade physics course. Let us recall that the numerical value of the moment of force about any axis is equal to the product of the force applied to the body by the length of the perpendicular drawn from the axis to the line of action of the force. For example, if a force $F$ is applied to the rim of a flat disk, shown in figure 2, the vector of which lies in the plane of the disk, then the moment of this force relative to the axis perpendicular to the plane of the disk and passing through the point $O$ is equal to $Fr$. If the axis passed not through the point $O$, but through the point $O_1$, then the moment of the same force relative to the new axis would be different: it would be equal to $Fr_1$. If the same force is applied as shown in figure 3, then its moment relative to the axis passing through point $O$ is equal to zero, since the line of action of the force passes through this point. Such a force cannot cause rotation around an axis and cannot change angular acceleration. But the same force can cause rotation of the disk around an axis passing through point $O_1$.

The moment of force is also a vector quantity. And the direction of the force moment vector is determined using the same expression. Namely, we agreed to consider the direction of the moment of force to be that in which the gimlet (screw) moves along the axis if its hand rotates in the same way as a body would rotate under the influence of a force applied to it. For example, the vector moment of force $F$ relative to the axis passing through the point $O$, at figure 2 is directed perpendicular to the drawing plane “from us”.

As already indicated, Newton’s laws of mechanics are also valid for rotational motion. From what has just been said about the moment of force, it follows, for example, that Newton’s law, when applied to rotational motion, states that a body rotates with a constant angular velocity or does not rotate at all if the moment of the force acting on it is

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was neglected, and the ball is so small compared to the length of the rod that it can be considered a material point.

Now all our difficulties will immediately disappear. In the formula

$$M = mr^2\alpha$$

$m$ is the mass of the ball, and $r$ is the distance from the ball to the axis of rotation. Since for a given ball and rod neither $m$ nor $r$ change, the product $mr^2$ is a constant quantity and can be denoted by one letter. Let us choose the letter $I$ for this value. Then the formula of Newton’s second law takes the form

$$M = I\alpha.$$

The similarity between this formula and the formula is striking

$$F = ma.$$

Only the force $F$ is replaced by the moment of force, and together with the linear acceleration $a$ the formula includes the angular acceleration $\alpha$, as it should be, since the furnace is not about translational, but about rotational motion. The role of mass in this formula is played by a completely new, unusual quantity, equal to the product of the mass by the square of its distance to the axis of rotation.

Just as the acceleration of a translationally moving body depends, in addition to force, also on the mass of the body being accelerated, so the acceleration of a rotating body depends not only on the moment of force, but also on the value $mr^2$, which is, so to speak, the rotational “double” of the mass. This quantity is called the moment of inertia of the body relative to the axis. For a rotational movement, it is not only mass that is important. It is also important how far this mass is removed from the axis of rotation. It is clear that a small mass at a large distance from the axis of rotation can receive the same angular acceleration at a given moment of force as a large mass located near the axis.

All this applies as if to a ball rotating on a thin rod. But in reality, the formula turns out to be true even when a disk, wheel, cylinder, and in general any “real” solid body, which also has a certain moment of inertia, rotates. To obtain the moment of inertia of a body relative to each axis, you need to mentally divide it into many small parts (in the limit of infinitely small), for each of them determine the product of its mass by the square of the distance from this axis and add all the products. This will be the moment of inertia of the whole body. It is clear that the moment of inertia of a body is greater, the greater the mass of those parts of it that are located away from the axis. In other words, the moment of inertia of a body depends on how the mass of the body is distributed over its volume. For different axes the moment of inertia of the same body will be different. For example, the moment of inertia of a rod of length $l$ and mass $m$, relative to the axis perpendicular to the rod and passing through its middle (Fig. 6), is equal to $$\frac{ml^2}{12}$$. And the moment of inertia of the same rod is

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The determined masses of bodies are a much simpler task than determining its rotational “double”—the moment of inertia. If the mass is determined by simple weighing, then “weighing” for rotational motion is a rather complex operation, which, moreover, for the same body, but for different axes of rotation, must be carried out separately, and “scales” for such “weighing” do not exist.

Momentum, angular momentum

So, for rotational motion, Newton’s second law is written as $M = I\alpha$, where $M$ is the moment of force, $I$ is the moment of inertia and $\alpha$ is the angular acceleration of the body.

No angular acceleration is expressed by the formula

$$\alpha = \frac{\omega - \omega_0}{t}.$$

Therefore, the formula of Newton’s second law can be written as

$$M = \frac{I\omega - I\omega_0}{t}.$$

In this form, this formula means that the moment of force is equal to the change in value $I\omega$ unit time. But, as is known, Newton’s second law can be written in the form

$$F = \frac{mv - mv_0}{t},$$

where $mv$ is a quantity called the momentum or impullas of the body. It is clear that implulys $mv$ also has an analogue in the rotational motion diaphragm and this analogue is the value $I\omega$. This follows from the fact that the moment of inertia $I$ plays the same role in rotational motion as mass does in translational motion, and the angular velocity $\omega$ replaces the linear velocity $v$. The quantity $L = I\omega$ is therefore called the angular momentum of the body or the angular momentum. This is also a vector quantity (the same as $\omega$, $F$, $\alpha$, $mv$, $M$), directed in the same way as the angular velocity vector. Similar to the other values ​​we have given that relate to rotational motion, the momentum of the impullas is different relative to different axes of rotation.

If the body has such a shape or size or such a radius of rotation that we can talk about the rotation of a point (for example, a ball on a thin rod, a planet as it rotates around the Sun), then the moment of impullas is easy to associate with impullas. Indeed, the moment of inertia $I$ is equal to $mr^2$, and the angular velocity $\omega$ is equal to $\frac{v}{r}$. That’s why

$$L = I\omega = \frac{mr^2v}{r} = rmv.$$

The moment of impuls of a point, therefore, is equal in magnitude to the product of the radius $r$ of the circle along which the point moves, and the impuls $mv$. It is clear that for a body of arbitrary shape such a simple correspondence is impossible.

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moment of inertia of the body relative to the vertical axis of rotation (the mass of the arms is close to this axis). Since this decrease in the moment of inertia $I$ is achieved not by external, but by internal forces, the moment of momentum $I\omega$ cannot change. Therefore, the angular velocity $\omega$ increases by the same number of times as the figure skater managed to reduce her moment of inertia. On the contrary, before completing the rotation, when the speed needs to be reduced, the skater spreads her arms wide, increasing the moment of inertia of her body. Dancers do the same when performing pruets, acrobats when they perform somersaults, etc. Interesting experiments of this kind can be carried out using the so-called Zhukovsky skama. It is a platform that can rotate around a vertical axis with little effort (Fig. 9). By standing on such a rotating platform, you can, by manipulating your hands, noticeably change the angular speed of rotation.

Conservation of angular momentum means conservation of not only the numerical value of the quantity $I\omega$. We must not forget that angular momentum is a vector quantity. Therefore, conservation of angular momentum also means conservation of its direction. But the moment of impulse is directed, as we know, along the axis of rotation. Therefore, any rotating body that is not affected by external moments of forces cannot change the position of the rotation axis. This is the “secret” of the stability of a rolling wheel, tops, etc. This is why discus throwers, throwing their projectile, cause it to rotate. This allows the disc to fly without tumbling. The ability of a rotating body to maintain the position of its axis of rotation (in the absence of moments of external forces) is widely used in technology. The use of so-called gyroscopic compasses, various stabilizers, etc. is based on this property. A gyroscopic compass is essentially a rotating body. If, at launch, the entrance was installed so that the weld was oriented to the north, then it will maintain this direction of the best magnetic compass, which is influenced by both the iron hull of the ship and changes in the Earth’s magnetic field. Gyroscopic stabilizers are sometimes installed on the hulls, which are huge multi-ton wheels driven by powerful engines. In an effort to maintain the position of the rotation axis, such a stabilizer changes the quality of the ship. Gyroscopic stabilizers find important applications in space flight technology and in automatic devices for various purposes.

Kinetic energy of rotating bodies

A rotating body is a moving body, since all its parts, except the point located on the rotation axis, move. Therefore it has kinetic energy. To express the kinetic energy, let’s start again with the rotation of a small ball on a thin rod, that is, with the rotation of a point. In this case, the kinetic energy can be written in the general form

$$K = \frac{mv^2}{2},$$

where $m$ is the mass of the ball, and $v$ is its linear speed. But since $v = \omega r$, the expression for $K$ can be written in the form

$$K = \frac{m\omega^2r^2}{2}.$$

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In conclusion, we again present for comparison some of the mechanics of translational motion, and similar formulas relating to rotational motion.

| Forward movement | Rotational movement |
| :— | :— |
| $\vec{F} = ma$ | $\vec{M} = I\alpha$ |
| $\frac{\Delta(mv)}{\Delta t} = \vec{F}$ | $\frac{\Delta(I\omega)}{\Delta t} = \vec{M}$ |
| $\Delta(mv) = 0$ at $\vec{F} = 0$ | $\Delta(I\omega) = 0$ at $\vec{M} = 0$ |
| $A = FS$ | $A = M\varphi$ |
| $K = \frac{mv^2}{2}$ | $K = \frac{I\omega^2}{2}$ |

Assignment and questions

1. In what units is the moment of inertia measured?
2. Two cylinders of identical diameters and masses roll down an inclined plane without piercing. One of them is lead and has a cork stop, the other is cork with a lead shell. Which cylinder will roll off the plane in less time?
3. A wheel whose moment of inertia is 0.01 SI units is expressed with an angular velocity of 4 rps. What must be the friction torque for the wheel to stop rotating after 5 revolutions?
4. Tides in the seas and oceans act on the rotating Earth like brake pads and create a braking friction force. Because of this action of the tides, in 100 years the length of a day on Earth will be 1 second longer than the next day. Calculate the tidal friction force.
5. A ship’s gyroscopic stabilizer with a radius of $R = 2$ m and a mass of $M = 120$ m is driven by a motor. What power is required to bring the stabilizer rotation speed to 13 rpm within 10 hours? Assume that the mass of the stabilizer is concentrated on its rim (in this case, the moment of inertia is equal to $MR^2$).
6. A homogeneous ball of radius $R$ and mass $M$ is initially launched along a plane so that it slides along it without rolling with a speed $v_0$. There is friction between the ball and the plane, the coefficient of which is $k$. How far will the ball travel before it stops piercing relative to the plane? What will be the speed of the ball at this moment?
7. Two windows of the spherical satellite are located at diametrically opposed points of the shell. The astronaut walks from the center of the ship along a radial ladder to one of the windows, then walks along the side of the ship to the second window, and finally returns along the radial ladder to his place in the center of the ship. At what angle does the ship rotate relative to the stars, if the mass of the ship is $M$, its radius is $R$, the mass of the astronaut is $m$ and we can assume that the entire mass of the ship is concentrated in its shell.
8. To replace the rotation of satellites, it is proposed to use the following devices. Two weights of mass $m$ each are attached to the side surface of the satellite on threads of length $l$. The loads are released simultaneously. They fly away from the satellite and at the moment when the threads are stretched, they follow from the hooks to which they are attached. At what length of the thread will the rotation speed of the satellite after this operation decrease by $n$ times?
   Assume that the satellite is a cylinder of mass $M$ and radius $R$; all the mass of the ship is concentrated in the side shell.