STEINER'S THEOREM: EXPLANATION, APPLICATIONS, EXERCISES - PHYSICAL - 2024

The Steiner 's theorem, also known as parallel axis theorem, to assess the moment of inertia of an extended body, about an axis that is parallel to another passing through the center of mass of the object.

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

Figure 1. A rectangular door rotating on its hinges has a moment of inertia that can be calculated by applying Steiner's theorem. Source: Pixabay.

Knowing the distance D that separates both axes and the mass M of the body in question, the moment of inertia with respect to the unknown axis is:

Moment of inertia indicates how easy it is for an object to rotate around a certain axis. It depends not only on the mass of the body, but on how it is distributed. For this reason it is also known as rotational inertia, being its units in the International System Kg. m ².

The theorem shows that the moment of inertia I _z is always greater than the moment of inertia I _CM by a quantity given by MD ².

Applications

Since an object is capable of rotating around numerous axes, and in the tables usually only the moment of inertia is given with respect to the axis passing through the centroid, Steiner's theorem facilitates the calculation when it is necessary to rotate bodies on axes that do not match this.

For example, a door commonly does not rotate about an axis through its 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 about said 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:

This equation is very similar to the very familiar formula for kinetic energy for an object of mass M moving at speed v: K = ½ Mv ². And it is that the moment of inertia or rotational inertia I plays the same role in the rotation as the mass M in the translation.

Proof of Steiner's theorem

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

I = ∫ 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, however it can be any.

Figure 2. An object extended in rotation around two parallel axes. Source: F. Zapata.

Around another z 'axis, the moment of inertia is:

I _z = ∫ (r ') ² dm

Now, according to the triangle formed by the vectors D, r and r ' (see figure 2 on the right), there is a vector sum:

r + r ' = D → r' = D - r

The three vectors lie on the plane of the object, which can be the xy. The origin of the coordinate system (0,0) is chosen in CM to facilitate the calculations that follow.

In this way the squared module of the vector r ' is:

Now this development is substituted in 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 that passes through CM.

For their part, the third and fourth integrals are worth 0, since by definition they constitute the position of the CM, which has been chosen as the origin of the coordinate system (0,0).

Solved exercises

-Solved exercise 1

The rectangular door in Figure 1 has a mass of 23 kg, 1.30 wide and 2.10 m high. Determine the moment of inertia of the door with respect to the axis passing through the hinges, assuming that the door is thin and uniform.

Figure 3. Schematic for Worked Example 1. Source: modified from Pixabay.

Solution

From a table of moments of inertia, for a rectangular plate of mass M and dimensions a and b, the moment of inertia with respect to the axis that passes through its center of mass is: I _CM = (1/12) M (a ² + b ²).

A homogeneous gate will be assumed (an approximation, since the gate in the figure is probably not so). In such a case, the center of mass passes through its geometric center. In figure 3 an axis that passes through the center of mass has been drawn and is also parallel to the axis that passes through the hinges.

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 for the green axis of rotation:

I = I _CM + MD ² = 11.7 Kg.m ² + 23 Kg x 0.652 m ² = 21.4 Kg.

-Solved exercise 2

Find the moment of inertia of a homogeneous thin rod when it rotates about an axis that passes through one of its ends, see figure. Is it greater or less than the moment of inertia when it rotates around its center? Why?

Figure 4. Scheme for the resolved example 2. Source: F. Zapata.

Solution

According to the table of moments of inertia, 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 that passes through one end D = L / 2 it remains:

It is greater, although not simply twice, but 4 times more, since the other half of the rod (not shaded 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 as another will have a moment of inertia proportional to (2D) ² = 4D ².

References

1. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill. 313-340.
2. Georgia State University. Rotational Motion. Recovered from: phys.nthu.edu.tw.
3. Parallel Axis Theorem. Recovered from: hyperphysics.phy-astr.gsu.edu.
4. Rex, A. 2011. Fundamentals of Physics. Pearson. 190-200.
5. Wikipedia. Parallel axis theorem. Recovered from: en.wikipedia.org