The moment of inertia of the rod is homogeneous and thin. Using the Steiner theorem. Task example

In physics, the concept of body mass is used to describe the inertial qualities of translational or linear motion. If the movement is considered around a certain axis of rotation, then a slightly different physical characteristic is used - the moment of inertia. In this article, we consider what this quantity is and how to calculate the moment of inertia of a thin rod.

Rotation and moment of inertia

The moment of inertia is easiest to enter for a material point. When it, having a mass M, rotates around an axis, describing a circle of radius R, then the moment of inertia for it is determined by the formula:

I = M * R ² .

Any real body, no matter how complex a geometric shape it may have, can be represented as a set of material points. This means that for the whole body or system of solids, the value of I can be calculated by integrating the expression above over the elementary masses dm. The general formula for determining the moment of inertia is:

I = ∫ _m (r ² * dm).

Through volume and density, this equality is written in the following form:

I = ∫ _V (ρ * r ² * dV).

It is often used to calculate I values ​​of specific geometric objects.

The physical meaning of the inertia of moment I lies in the fact that it determines how “difficult” it is for a given force that creates some torque to spin or stop the rotating system. In other words, I characterizes the inertial properties of the system under study.

The most famous example of using the moment of inertia is the flywheel of an internal combustion engine in cars. Due to the large value of I, the flywheel ensures smooth movement of the car, smoothing out any sharp impacts on the crankshaft. An example of a different nature, where it is also important to know the moment of inertia, is the law of conservation of angular momentum. It is used to rotate around the axis of artificial satellites in outer space of the Earth.

Thin shaft and axis of rotation

Next, the moment of inertia of the rod relative to the axes (different) will be considered. Calculations will be carried out for a thin rod, which has a uniform distribution of mass, that is, its density at all points is a constant value. By thin we mean a rod whose width (thickness) is much smaller than its length L. To designate its mass, we will use the letter M.

From the above formulas it follows that the value of I depends on the relative position of the body and the axis of rotation. For the rod, three main axes can be distinguished. One of them passes through the length of the entire rod. Since its thickness tends to zero, the moment of inertia for this position of the body will also tend to this value.

The other two axes are perpendicular to the length of the body in question. One of them passes through the center of mass, let's call it O ₁ , the second through the end of the rod, we denote it O ₂ . Relative to them, we calculate the value I.

Inertia moment relative to O1

First of all, we write out the general formula. We have:

I = ∫ _V (ρ * r ² * dV).

We denote the cross-sectional area of ​​the rod by the letter S. Obviously, it tends to zero, since the rod is thin. But this designation is convenient to introduce for further calculations.

Now, mentally divide the rod into an infinite number of small pieces, each of which will have a section S and thickness dl. Replacing r with l in the formula above, we obtain:

I = ∫ _L (ρ * S * l ² * dl).

It remains only to substitute the correct limits of integration and write down the final formula. Since the axis O ₁ passes through the middle of the rod, the limits of integration will be as follows:

I = ∫ _{-L / 2} ^{L / 2} (ρ * S * l ² * dl).

The result of calculating this integral is the following formula:

I = M * L 2/12.

Thus, the moment of inertia of a thin rod is determined by its mass and length.

Inertia moment relative to O2

Now consider the situation when the axis of rotation will pass through any of the ends of the rod and will be perpendicular to it. The corresponding formula can be obtained from the integral written above, if the integration limits are correctly substituted. However, we will take a slightly different path and determine the inertia of the moment using Steiner's theorem.

She says that if two axes are parallel to each other and one of them (axis O) passes through the center of mass of the body, then the moment of inertia about the second axis can be calculated using this equation:

I = I ₀ + M * h ² .

Here I ₀ is the moment of inertia of the rod relative to the axis O, h is the distance between the axes.

This formula can be successfully applied to our case. Since I ₀ we calculated in the previous paragraph of the article about the axis O ₁ , and the distance between O ₁ and O ₂ is L / 2, using the Steiner theorem we get the following result:

I = I ₀ + M * h ² = M * L 2/12 + M * L 2/4 = M * L 2/3.

Thus, for the rod, the value of I relative to the axis O ₂ is 4 times greater than relative to the axis O ₁ . This means that to give the same angular acceleration to the rod in the case of rotation around the axis O ₂ should be applied 4 times more torque than in the case of the axis O ₁ .

Task example

A thin rod 0.5 m long and 5 kg in weight was given. At a distance of 2/5 from its end is the axis of rotation perpendicular to the rod. What is the moment of inertia of the system?

To solve the problem, we use the Steiner theorem. The distance between the axes O ₁ and specified in the task is equal to:

h = 0.25 - 0.2 = 0.05 m.

Then we get the moment of inertia of the rod (homogeneous):

I = I ₀ + M * h ² = 5 * 0.5 2/12 + 5 * 0.05 ² = 0.117 kg * m ² .

In SI, the moment of inertia of the rod is measured in the indicated units.