# Steiner’s theorem: explanation, applications, exercises

The **Steiner’s theorem** , also known as *the parallel axis theorem* to evaluate the moment of inertia of an elongated body, around an axis which is parallel to another object passing through the center of mass.

It was discovered by the Swiss mathematician Jakob Steiner (1796-1863) and states the following: let _{CM} be the moment of inertia of the object with respect to an axis passing through the center of mass CM and _{z} the moment of inertia with respect to another axis parallel to that

Knowing the distance D separating the two axes and the mass M of the body in question, the moment of inertia in relation to the incognito axis is:

*I _{z} = I _{CM} + MD ^{2}*

The moment of inertia indicates how easy it is for an object to rotate around a certain axis. It depends not just on the mass of the body, but on how it is distributed. For this reason, it is also known as *rotational inertia* , and its units are in the International System of Kg. m ^{2} .

The theorem shows that the moment of inertia *I _{z}* is always greater than the moment of inertia

*I*by an amount given by

_{CM}*MD*.

^{2}__applications__

__applications__

As an object is able to rotate around several axes, and in tables generally only the moment of inertia is given in relation to the axis passing through the centroid, Steiner’s theorem facilitates the calculation when it is necessary to rotate bodies on the axes. They don’t match that.

For example, a door usually does not rotate about an axis passing through the center of mass, but about a lateral axis, where the hinges adhere.

By knowing the moment of inertia, it is possible to calculate the kinetic energy associated with the rotation in that axis. If *K* is the kinetic energy, *I* the moment of inertia around the axis in question and *ω* the angular velocity, it follows that:

*K = ½ I.ω *^{2}

This equation is very similar to the very familiar formula of kinetic energy for an object of mass *M* moving at velocity *v* : *K = ½ Mv ^{2}* . And is that the moment of inertia or rotational inertia

*I*plays the same role in rotation as the mass

*M*in translation.

**Steiner’s theorem proof**

The moment of inertia of an extended object is defined as:

I = ^{2}* r ^{2} dm*

Where *dm* is an infinitesimal portion of mass and *r* is the distance between *dm* and the axis of rotation *z. *In Figure 2, this axis crosses the center of mass CM, but it can be anything.

Around another *z’* axis , the moment of inertia is:

I _{z} = ∫ ( *r ‘) ^{2} dm*

Now, according to the triangle formed by the vectors ** D** ,

**and**

*r***(see figure 2 on the right), there is a vector sum:**

*r ‘**r** + r ‘ = D *

*→*

**r’**=**D**–**r**The three vectors are in the object plane which can be *xy* . The origin of the coordinate system (0,0) is chosen in CM to facilitate the following calculations.

In this way, the square module of vector ** r ‘** is:

*(r ‘) ^{2} = (D _{x} – r _{x} ) ^{2} + (D _{y} – r _{y} ) ^{2} =*

*= D _{x }^{2} + D _{y }^{2} + r _{x }^{2} + r _{y} 2 -2D _{x} r _{x} – 2 D _{and} r _{y} =*

*= D ^{2} + r ^{2} – 2D _{x} r _{x} – 2 D _{y} r _{y}*

Now this development is substituted into the integral of the moment of inertia I _{z} and also the definition of density dm = ρ.dV is used:

The term M. D ^{2} that appears in Steiner’s theorem comes from the first integral, the second is the moment of inertia with respect to the axis passing through CM.

On the other hand, the third and fourth integrals are 0, since by definition they constitute the position of the CM, which was chosen as the origin of the coordinate system (0,0).

**solved exercises**

**solved exercises**

**Exercise -Fixed 1**

The rectangular door in figure 1 has a mass of 23 kg, 1.30 in width and 2.10 m in height. Determine the moment of inertia of the door in relation to the axis passing through the hinges, assuming that the door is thin and uniform.

**Solution**

From a table of moments of inertia for a rectangular plate of mass M and dimensions *a* and *b* , the moment of inertia around the axis passing through its center of mass is as follows: I _{CM} = (12/1) *H* ( *a *^{2} + *b *^{2} ).

A homogeneous gate will be assumed (an approximation, as the figure gate is probably not so much). In this case, the center of mass passes through its geometric center. In Figure 3, an axis that passes through the center of mass and is also parallel to the axis that passes through the hinges was drawn.

I _{CM} = (1/12) x 23 Kg x (1.30 ^{2} +2.10 ^{2} ) m ^{2} = 11.7 Kg.m ^{2}

Applying Steiner’s theorem to the green axis of rotation:

I = I _{CM} + MD ^{2} = 11.7 kg.m ^{2} + 23 kg x 0652 m ^{2} = 21.4 kg.

**– Exercise solved 2**

Find the moment of inertia of a thin homogeneous rod when it rotates with respect to an axis passing through one of its ends, see figure. Is it greater or less than the moment of inertia when it rotates around its center?

**Solution**

According to the moment of inertia table, the moment of inertia *I _{CM}* of a thin rod of mass

*M*and length

*L*is:

*I*

_{CM}= (1/12) ML^{2}And Steiner’s theorem states that when it is rotated around an axis passing through an end D = L / 2, it remains:

*I = I _{CM} + MD ^{2} = (1/12) ML ^{2} + M (L / 2) ^{2} *

*= (1/3) ML*

^{2}It is larger, though not just twice as large, but 4 times larger, as the other half of the rod (no shading in the figure) rotates describing a larger radius.

The influence of the distance to the axis of rotation is not linear but quadratic. A mass that is twice the distance from another will have a moment of inertia proportional to (2D) ^{2} = 4D ^{2} .