# Why does my book consider moment of inertia as a scalar when it is a tensor?

I found in the internet that the moment of inertia of a rotating body is a tensor quantity. But in my book it is considered as a scalar quantity. Won't doing this give wrong results? So how does it work?

Can MOI be considered as a scalar quantity in special cases only? What are the limitations of considering this as a scalar? What are some situations in which considering MOI as scalar quantity will be wrong and I will have to consider it as a tensor?

• Which book? Which page? Commented Apr 25, 2020 at 16:47
• @Qmechanic I live in Bangladesh. It's a very not famous book. Also it's written in my native language. Commented Apr 25, 2020 at 16:48
• Any author should in principle be credited, independently of language. Commented Apr 25, 2020 at 17:48

It can be treated as a scalar when the body is constrained to move on a single rigid axis. In this case you have a line of points through the body that are fixed to not move. All other points move around this axis. Due to the body being rigid every point will exhibit circular motion around this axis with the same infinitesimal angular displacement. The body's rotation motion is now 1-dimensional. This is how most lower level undergraduate or high school books would introduce the topic since tensors and matrices might be beyond the level of the student. This makes learning the basic concepts a bit easier. In upper level courses and texts the tensor treatment is standard, the constrained case being a special case.

• Even if it's rotating around a fixed axis, you might still need to treat it as a tensor quantity. If the fixed axis isn't a principal axis, then $\vec{L}$ and $\vec{\omega}$ are not parallel, and you can't simply write $\vec{L} = I \vec{\omega}$ for a scalar $I$. Commented Apr 26, 2020 at 17:53
• @MichaelSeifert Is correct. A single fixed axis is not sufficient. Commented Apr 27, 2020 at 1:34

The moment of inertia is a scalar when calculated as seen here, which is the same as stated in your book.

Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation. For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, $$I = mr^2$$

When all the possible axes of rotation are taken into account, a tensor can be defined.

Inertia tensor

For the same object, different axes of rotation will have different moments of inertia about those axes. In general, the moments of inertia are not equal unless the object is symmetric about all axes. The moment of inertia tensor is a convenient way to summarize all moments of inertia of an object with one quantity. It may be calculated with respect to any point in space, although for practical purposes the center of mass is most commonly used.

• What is meant by an object that is symmetric about all axes? Does this mean that it can be cobsidered scalar about some special axes? Commented Apr 25, 2020 at 17:07
• Think of a sphere, any axis that goes to the center of mass gives the samemoment of inertia. Although , like antular momentum, moment of inertia can be defined even when the axis is outside the body.too. Commented Apr 25, 2020 at 17:34

Actually in general case it is a tensor. But may be they assume that there is some spherical symmetry in the task? Every moment of inertia tensor, as a symmetric tensor, may be brought to diagonal form, in case, when all eigenvalues are equal, the problem is sprherically symmetric, and in that case they may treat at as a scalar.

Let angular momentum $$L$$ has components $$L_1,L_2,L_3$$, is a linear combination of angular velocity $$ω$$ with components $$ω_1,ω_2,ω_3$$. The proportionality factor is moment of inertia (MOI) tensor $$I$$. In matrix form :
$$\left( \begin{array}{c} L_1 \\ L_2 \\ L_3 \\ \end{array} \right)=\left( \begin{array}{ccc} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \\ \end{array} \right) \left( \begin{array}{c} \omega _1 \\ \omega _2 \\ \omega _3 \\ \end{array} \right)$$ or $$L_i=I_{ij} ω_j$$

In special case, where off diagonal components of MOI $$I$$ are zero, MOI is a vector quantity :
$$\left( \begin{array}{c} L_1 \\ L_2 \\ L_3 \\ \end{array} \right)=\left( \begin{array}{ccc} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \\ \end{array} \right) \left( \begin{array}{c} \omega _1 \\ \omega _2 \\ \omega _3 \\ \end{array} \right)=\left( \begin{array}{ccc} I_1 & I_2 & I_3 \\ \end{array} \right) \left( \begin{array}{c} \omega _1 \\ \omega _2 \\ \omega _3 \\ \end{array} \right)$$ or $$L_i=I_i ω_i$$

In the simplest case, where $$I_1=I_2=I_3=I$$, then MOI is a scalar quantity :
$$\left( \begin{array}{c} L_1 \\ L_2 \\ L_3 \\ \end{array} \right)=\left( \begin{array}{ccc} I & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & I \\ \end{array} \right) \left( \begin{array}{c} \omega _1 \\ \omega _2 \\ \omega _3 \\ \end{array} \right)=I \left( \begin{array}{c} \omega _1 \\ \omega _2 \\ \omega _3 \\ \end{array} \right)$$ or $$L_i=Iω_i$$

Reference:
S.T. Thornton & J.B. Marion, Classical Dynamics of Particles and Systems, 5th ed., Brooks/Cole-Thomson Learning, 2004, Ch. 11.