The moment of inertia of a solid and hollow cylinder: different positions of the axes of rotation

Knowing the moment of inertia of the body allows you to use the law of conservation of angular momentum or an expression to describe circular motion with angular acceleration. In this article, we will consider how to find the moment of inertia for a cylinder at different positions of the rotation axes.

Moment of Inertia: Mathematical Definition

The axial moment of inertia is introduced into physics through the study of the laws of rotational motion of bodies. For a material point with mass m rotating at a distance r from the axis, the moment of inertia will be equal to:

I = m * r ²

In the general case, for a body that has an arbitrary distribution of matter in space (any geometric shape), the value of I can be calculated as follows:

I = ∫r ² dm

In fact, this expression is a generalization of the previous one. It sums (integrates) the moments from each elementary particle dm, the distance to the axis from which is equal to r.

If we talk about the physical value of the considered quantity I, then it shows how "strongly" the system resists the influence of an external moment of force, which is trying to untwist it or, conversely, stop it.

Moment of inertia of the cylinder about an axis perpendicular to its bases

From the above formula, it can be understood that the quantity I is a characteristic of the entire rotating system, that is, it depends both on the shape of the body and the distribution of mass in it, and on the relative position of the axis.

In this section, we consider a simple case: it is necessary to determine the moment of inertia for a continuous cylinder, the axis of rotation of which is perpendicular to its bases and passes through the gravitational center of the figure.

To solve the problem, we apply the integral formula for I. In the process of integration, mentally divide the cylinder into thin rings of thickness dr. Each ring will have a volume: dV = 2 * pi * r * dr * h, here h is the height of the figure. Given that dm = ρ * dV, where ρ is the density of the cylinder, we obtain:

I = ∫r ² dm = ρ * ∫r ² dV = 2 * pi * ρ * h * ∫r ³ dr

This integral must be calculated for the limits from 0 to R, where R is the radius of the figure. Then we get:

I = 2 * pi * ρ * h * ∫ ^R ₀ r ³ dr = 2 * pi * ρ * h / 4 * (r ⁴ ) ∣ ^R ₀ = pi * ρ * h * R ^4/2

Using the formula for the mass of the cylinder through its volume and density, we arrive at the final expression:

I = m * R ^2/2 , where m = pi * ρ * h * R ²

We have obtained the inertia formula for the homogeneous cylinder moment. It shows that the value of I for this figure is 2 times less than for the material point of a similar mass, which rotates at a distance of the radius of the cylinder from the axis.

The moment of inertia of the hollow cylinder

Now leave the axis in the same place and find the value of I for the cylinder with a void inside (sleeve, pipe). Such a figure is described by two radii: external R ₁ and internal R ₂ . In this case, the integration takes exactly the same approach as for a solid cylinder, only the limits now vary from R ₂ to R ₁ . We have:

I = 2 * pi * ρ * h / 4 * (r 4 ) ∣ R 1 R2 = pi * ρ * h * R 4 / 2∣ R 1 R2 = pi * ρ * h / 2 * (R 1 4 -R 2 4 )

To further simplify this formula, we use the factorization of the expression in parentheses, we get:

I = pi * ρ * h * (R ₁ ² -R ₂ ² ) * (R ₁ ² + R ₂ ² ) / 2

Part of this expression, together with the first brackets, is the mass of the hollow cylinder, so we get the final formula:

I = m * (R ₁ ² + R ₂ ² ) / 2

This shows that the moment of inertia of the hollow cylinder is greater than this value for a continuous cylinder of the same mass and the same external radius by the value m * R ₂ 2/2. This result is not surprising, since the center of mass in the hollow cylinder is farther from the axis of rotation than in the solid.

The value of I for the cylinder, the axis of rotation of which runs parallel to the planes of its base

In such a system, the axis of rotation also passes through the center of mass of the cylinder, but now it lies, as it were, on its side (on a cylindrical surface, see the figure below).

The calculation for the moment of inertia of the cylinder for such a situation is a difficult task, since it requires the availability of additional knowledge to solve it. Nevertheless, we give the necessary mathematical calculations, so that readers have a more complete idea of ​​the integration in the calculation of I.

We begin to solve the problem. We break the continuous cylinder into separate disks of infinitely small thickness. To find out what moment of inertia this disk possesses relative to the axis that passes through it and is parallel to its bases, it is necessary to perform a separate integration. It gives the following result:

I _i = R ² * dm / 4

To find the value I _i for this disk relative to the already new axis, which is considered in the problem, it is necessary to use the Steiner theorem. We get:

I _i = R ² * dm / 4 + L ² * dm, here L is the distance from the axis to the thin disk.

Knowing that dm = pi * R ² * dL * ρ, we substitute in the integral formula for I and carry out integration over the limits (-L 0/2; + L _0/2 ), we have:

I = ∫ _m I _i = ∫ _m (R ² * dm / 4 + L ² * dm) = pi * R ² * ρ * ∫ ^{L0 / 2} _{-L0 / 2} (R ² * dL / 4 + L ² * dL )

The solution of this integral leads to the final formula:

I = m * (R 2/4 + L ₀ 2/12)

Problem solving example

We will solve an interesting problem of finding the axial moment of inertia of a cylinder. Let it lie on a cylindrical surface, and the axis of rotation is parallel to its base and passes through the end of the figure.

This situation is completely similar to that considered in the previous paragraph, only the axis does not cross the gravitational center of the cylinder, but the end of this figure. Nevertheless, to solve the problem, you can use the result of the previous paragraph of the article. We apply the aforementioned Steiner theorem, we obtain:

I = m * R 2/4 + m * L ₀ 2/12 + m * (L _0/2 ) ² = m * R 2/4 + m * L ₀ 2/3

Note that if R << L ₀ , then the first term can be neglected, and the formula reduces to the equality:

I = m * L ₀ 2/3

This moment of inertia corresponds to a rod with an axis of rotation at its end.