In the school physics course, much attention is paid to the description of the kinematics and dynamics of the translational motion of bodies in three-dimensional space. But rotational motion plays an equally important role in technology and nature. In this article, we consider what is meant by the moment of inertia of a body when it rotates around an axis.
Rotation dynamics
Before giving a definition of the moment of inertia of the body, we will explain why this quantity is needed and in what equations it appears. First of all, this is the main equation of the dynamics of rotation - the formula of moments. It is written like this:
M = I * Ξ±.
Here M, Ξ± and I are the moment of force, the angular acceleration and inertia moment, respectively. In fact, this equation can be called the second Newtonian law for rotational motion. It is easy to guess that the quantity I here plays the same role as the inertial mass in the case of translational motion.
In addition to the above equation, there is another important formula that is often used to solve problems on the rotation of bodies - this is the law of conservation of angular momentum. It is usually written in the following convenient form for practice:
I * Ο = const.
As you can see, here the inertia of the moment is also a key quantity, Ο is the angular velocity.
Moment of inertia of a solid
Now the time has come to give a definition of I. First we consider it for a material point. Its moment of inertia is called the product of mass and the square of the distance to the axis of rotation. If the mass is denoted by the letter m, and the distance to the axis from the point by the letter r, then the formula for I is written as follows:
I = m * r 2 .
As can be seen, I is expressed in kg * m 2 . Equality for a point can be used to determine the moment of inertia of a body about an axis. In this case, the following integral expression is used:
I = β« m (r 2 * dm).
This formula is used to calculate the values ββof I of absolutely any systems with different geometric shapes. The last equality is also used in solving practical problems in the following form:
I = β« V (Ο * r 2 * dV).
Where Ο is the density of the substance. Below in the article we show how to use integral equality to solve specific problems.
I value for cylinder
Each student imagines a cylinder figure. In truth, they are very different (elliptical, hyperbolic, inclined). Here we consider the simplest case. This is a circular straight cylinder, which is limited by a cylindrical surface and two identical circles. The axis of rotation of the figure passes through its center of mass and through the centers of both bases. We calculate the moment of body relative to it inertia.
We write the original formula:
I = β« V (Ο * r 2 * dV).
To apply it, imagine a cylinder in the form of thinly sliced ββround identical layers. Let us denote their thickness dl, the radius of the figure is R, and the height is L. Now we cut each thin layer of volume pi * R 2 * dl into an infinite number of rings, the thickness of each of which is equal to dr. After completing all the described mental geometric operations, we can write the formula for the elementary volume dV, that is, for the volume of one ring:
dV = 2 * pi * r * dr * dl.
As a result of this representation, the original expression for I is transformed into a double integral formula:
I = β« L β« R (Ο * r 2 * 2 * pi * r * dr * dl) = 2 * pi * Ο * L * R 4/4 = M * R 2/2.
Where the letter M denotes the mass of the entire cylinder.
Thus, we have obtained the final expression for the inertia of the cylinder moment. As you can see, it is determined only by the radius of the figure and its mass and does not depend on the length (height). The latter means that a similar formula can be used to determine the value of I for a disk of any thickness.
The value of I for the rod
Now we apply the formula to determine the moment of inertia of a thin rod. The fundamental point here is the fact that its thickness should be much less than the length L. The mass of the rod is denoted by the letter M. The moment of inertia is calculated for the position of the axis, which passes through the center of mass of the body and is perpendicular to it.
We start the calculation with the same formula as in the case of the cylinder:
I = β« V (Ο * r 2 * dV).
Mentally cut the entire rod into thin layers. We denote the cross-sectional area of ββeach of them S, and its thickness dl. Then we get the formula for dV:
dV = S * dl.
Now you can calculate the inertia of the moment of the body:
I = β« -L / 2 + L / 2 (Ο * S * l 2 * dl).
Note that each layer is located at a distance l from the axis of rotation, so we replaced the letter r. In addition, we pay attention to the limits of integration, which are of such importance because the axis passes exactly through the middle of the rod. As a result, we get:
I = β« -L / 2 + L / 2 (Ο * S * l 2 * dl) = Ο * S * l 3/3 | -L / 2 + L / 2 = M * L 2/12.
Using similar arguments and calculations, it can be shown that if the axis of rotation passes through any end of the rod, then its moment of inertia will be four times greater, that is:
I = M * L 2/3.
The physical meaning of quantity
We have already said a few words about what the moment of inertia of the body means from a physical point of view. Here we dwell in more detail on this issue.
If you carefully look at the formula for I, you can see that this value depends not only on the mass of the body, but also on its distribution, that is, on the shape of the body, as well as on its position relative to the axis of rotation.
A striking example is an ordinary mop or just a rod. Each person at least once in his life spun a mop around an axis running along its handle or perpendicular to it. In the first case, a slight movement of the palms is enough to give an angular acceleration to the squeegee, in the second case, you have to exert some hand force to unwind it. This fact is easy to explain. In the first case, the moment of inertia is practically zero, in the second - it has some finite value.