Seemingly correct and useful expression for kinetic energy of 3D rigid body not found anywhere

In all texts I've seen, the kinetic energy of a 3D rigid body with mass $$M$$, center of mass velocity $$\vec V$$, angular momentum $$\vec L$$, and angular velocity $$\vec \omega$$ is given as $$T=\dfrac12 MV^2 + \dfrac12 \vec\omega\cdot\vec L.$$

What I believe to be an equivalent and equally correct description, in terms of the perpendicular distance $$r_\perp$$ to the axis of rotation $$\vec \omega$$, namely $$T=\dfrac12MV^2+\dfrac12 \omega^2 \int_\mathbb M r_\perp^2 \, \mathrm dm \tag {⋆}$$ is not found anywhere.

I derive the equation as follows: first, the kinetic energy is $$\displaystyle T = \dfrac12\int_\mathbb M \vec v^2 \, \mathrm dm$$ where $$\vec v = \vec V + \vec \omega \times \vec r'$$ (with $$\vec r'$$ being a vector from the CoM to an arbitrary d$$m$$ on the body).

From there, we get, $$T=\dfrac12 MV^2 + \int_\mathbb M \vec V \cdot\vec \omega \times \vec r'\, \mathrm dm+\dfrac12\int_\mathbb M\left( \vec\omega \times\vec r'\right)^2\, \mathrm dm.$$

The middle term is just $$\displaystyle \vec V \cdot \vec \omega \times \int_\mathbb M \vec r' \, \mathrm dm=0$$ by the definition of the CoM, and the last term is $$\displaystyle \dfrac12 \int_\mathbb M\omega^2 r'^2 \sin^2(\theta) \, \mathrm dm=\dfrac12\omega^2 \int_\mathbb M r_\perp^2 \, \mathrm dm$$ with $$r_\perp \equiv r'\sin\theta$$ and thus, we get $$(\star)$$.

First, I'd like to know if this expression is correct. If so, why is it not listed anywhere in textbooks? I've seen a simplified version for 2-D rotation, but not 3-D.

• Have you checked Goldstein? Commented Jun 6, 2022 at 5:39
• @Aplateofmomos no... is it there? Commented Jun 6, 2022 at 5:41
• You describe the expression as 'useful'. What would you use it for? Commented Jun 6, 2022 at 7:20
• @EmilioPisanty computing KE. Instead of going through the inertia tensor calculation and taking 6 integrals, then finding $\vec L$, one can simply calculate $\displaystyle \int r^2_\perp \, \mathrm dm$ Commented Jun 6, 2022 at 17:04
• Your integral depends implicitly on the direction of $\vec{\omega}$, which will in general change as the body rotates. See my comment on @rob's answer below. Commented Jun 6, 2022 at 18:47

4 Answers

The integral

$$I= \int\mathrm dm\ r_\perp^2$$

is the moment of inertia about the given axis. You may be partway along the path to computing the moment of inertia tensor, which is useful describing rotation about an arbitrary axis, precessing or nutating rotations, and so on.

• I am aware of the connection. My question is that, in most books, they express KE in terms of the angular momentum (and thereby the inertia tensor), which IMO, is much more complicated to compute than $\displaystyle \int\mathrm dm\ r_\perp^2$... In 2-D, they introduce this integral simply as the moment of inertia about the axis of rotation, but that convention is completely abandoned in 3-D dynamics ... I am wondering why Commented Jun 6, 2022 at 17:07
• @user256872: In 3-D dynamics, the axis of rotation is not generally fixed (either in the body frame or in the space frame). This means that the quantity $r_\perp$ will change for a given point in the body as the body spins. This makes your integral less useful, since it would also be changing with time. Commented Jun 6, 2022 at 18:43
• And, of course, to account for this change in $\vec{\omega}$ with respect to time in the body frame you need... the full $3\times3$ inertia tensor. So there's no way around needing to calculate it. Commented Jun 6, 2022 at 18:49

Perhaps the expression is not common, I would be surprised if it had not been mentioned in some reference, because it is obvious to the eye. If one intends to use the principle of conservation of energy it is useful, but if one seeks to study one's own motion the expression may be impractical because of the following:

1. The axis of rotation is not usually constant in a general motion of a rigid solid.

2. that would imply that the integralwill not be constant either and that quantity cannot be computed if the motion is not known: $$\int_\mathbb{M} r_\bot^2\ \text{d}m \neq \text{constant}$$

3. That is the reason why Euler's equations of motion use a base that is co-mobile with the solid (precisely so that the inertia tensor is constant and is not an additional unknown).

• all answers were good (+1) but this one answers my question best Commented Jun 6, 2022 at 21:54

Considering the sum of particles on a rigid body the mass moment of inertia tensor about some arbitrary point A is

$$\mathrm{I}_A = \int \left( \vec{r}\cdot\vec{r} - \vec{r} \odot \vec{r} \right) \,{\rm d}m$$

where $$\vec{r}$$ is the location of each particle relative to A, and $$\cdot$$ the inner product, and $$\odot$$ the outer product. This gives the perpendicular distance and cross products for the general case.

If $$\vec{r} = \pmatrix{x \\ y \\ z}$$ then the above is

$$\mathrm{I}_A = \int \begin{vmatrix} y^2+z^2 & -x y & - x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2 \end{vmatrix} \,{\rm d}m$$

The above is used in finding the angular momentum, summed at point A as

$$\vec{L}_A = \mathrm{I}_A \vec{\omega}$$

Considering that point A may have translational velocity $$\vec{v}_A$$, the kinetic energy of this rigid body is

$$KE = \tfrac{1}{2} \vec{v}_A \cdot \vec{p} + \tfrac{1}{2} \vec{\omega} \cdot \vec{L}_A \tag{1}$$

This equation is valid for all arbitrary points A, including the center of mass, point C

Now to expand the above in terms of the velocity vectors makes the above equation somewhat messy, since it involved the velocity of the arbitrary point and the velocity of the center of mass

$$KE=\tfrac{1}{2}\vec{v}_{A}\cdot m\vec{v}_{C}+\tfrac{1}{2}\vec{\omega}\cdot\left({\rm I}_{C}\vec{\omega}+\vec{c}\times m\left(\vec{\omega}\times\vec{c}\right)\right) \tag{2}$$

Here the vector $$\vec{c}$$ locates the center of mass relative to point A.

Only when the above is summed at the center of mass, with $$\vec{c}=0$$ does the above take the form you are driving to in your question.

$$KE=\tfrac{1}{2}\vec{v}_{C}\cdot m\vec{v}_{C}+\tfrac{1}{2}\vec{\omega}\cdot{\rm I}_{C}\vec{\omega} \tag{3}$$

Appendix

You can prove the equivalency between (1) and (3) by considering the following transformations

$$\begin{gathered} \vec{v}_{C}=\vec{v}_{A}+\vec{\omega}\times\vec{c}\\\vec{L}_{C}=\vec{L}_{A}+\vec{p}\times\vec{c} \end{gathered}$$

Here the vector $$\vec{c}$$ locates the center of mass relative to point A.

\require{cancel} \begin{aligned}KE & =\tfrac{1}{2}\vec{v}_{C}\cdot m\vec{v}_{C}+\tfrac{1}{2}\vec{\omega}\cdot{\rm I}_{C}\vec{\omega}\\ & =\tfrac{1}{2}\vec{v}_{C}\cdot\vec{p}+\tfrac{1}{2}\vec{\omega}\cdot\vec{L}_{C}\\ & =\tfrac{1}{2}\left(\vec{v}_{A}+\vec{\omega}\times\vec{c}\right)\cdot\vec{p}+\tfrac{1}{2}\vec{\omega}\cdot\left(\vec{L}_{A}+\vec{p}\times\vec{c}\right)\\ & =\tfrac{1}{2}\vec{p}\cdot\vec{v}_{A}+\cancel{\tfrac{1}{2}\left(\vec{\omega}\times\vec{c}\right)\cdot\vec{p}}+\tfrac{1}{2}\vec{\omega}\cdot\vec{L}_{A}+\cancel{\tfrac{1}{2}\vec{\omega}\cdot\left(\vec{p}\times\vec{c}\right)}\\ & =\tfrac{1}{2}\vec{v}_{A}\cdot\vec{p}+\tfrac{1}{2}\vec{\omega}\cdot\vec{L}_{A} \end{aligned}

The expression you gave is very common. You mostly get if when going through the proof of König's theorem for a continous system, which involves going to the center-of-mass frame (hence getting the relation between $$\vec{v}$$ and $$\vec{V}$$). Combine that with the definition of the moment of inertia, and you get this relation.