# Kinetic energy of a rotating body

On my own, I tried deriving the rotational kinetic energy for a rotating body with an arbitrary changing axis of rotation. What I did is use the formula:

$$T = \frac{1}{2}\omega^2 (\hat{n}^{T} \mathbf{I} \hat{n})$$

where $$\mathbf{I}$$ is the inertia tensor. I'll express everything with respect to the principle axis of rotation, so the tensor is diagonal. Also, $$\hat{n}^T = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$$ is the unit vector in spherical coordinates with respect to the principle axis of rotations basis. The formula I'd get is:

$$T = \frac{1}{2}\omega^2(\sin^2\theta \cos^2\phi I_{xx} + \sin^2 \theta \sin^2 \phi I_{yy} + \cos^2 \theta I_{zz})$$

My intuition tells me this is correct. I tried comparing it with the one expressed with euler angles however I can't figure out how to make them equivalent. The formula with euler angles contains three time derivatives $$\dot\theta, \dot\phi, \dot\psi$$, but the one above contains the derivative of one coordinate $$\omega$$, which I believe is $$\dot \psi$$. What am I doing incorrectly.

• Just like with translational motion , rotational motion is frame varrying. Commented Nov 3, 2023 at 17:39
• This is correct and I looked into it, however, I don't see how it would affect the equation. I am not taking the derivatives of the basis anywhere. Commented Nov 3, 2023 at 17:43

Using the basis-vectors of the inertial frame, rotational kinetic energy (assuming center of mass is fixed) is

$$T = \tfrac{1}{2} \boldsymbol{\omega}^\intercal \boldsymbol{L} = \tfrac{1}{2} \boldsymbol{\omega}^\intercal \mathbf{I} \,\boldsymbol{\omega}$$

where $$\boldsymbol{\omega}$$ is the rotational velocity vector, $$\boldsymbol{L}$$ the angular velocity vector, and $$\mathbf{I}$$ the mass moment of inertia tensor. All quantities are summed about the center of mass.

Now if you know the orientation (rotational transformation) of the body $$\mathrm{R}(t)$$ in terms of body-to-inertial matrix, and the body-fixed mass moment of inertia matrix $$\mathbf{I}_{\rm body}$$ then the above is

$$T = \tfrac{1}{2} \boldsymbol{\omega}^\intercal \left( \mathrm{R} \mathbf{I}_{\rm body} \mathrm{R}^\intercal \right) \boldsymbol{\omega}$$

• Now if you know that axis of rotation in the inertial frame, then $$\boldsymbol{\omega} = \omega \,\boldsymbol{\hat{n}}$$

Kinetic energy is thus calculated with

$$T = \tfrac{1}{2} \omega^2 \left( \boldsymbol{\hat{n}}^\intercal \mathrm{R} \mathbf{I}_{\rm body} \mathrm{R}^\intercal \boldsymbol{\hat{n}} \right)$$

Here the part in the parenthesis changes with time, as the orientation matrix changes. You can use Euler-Angles to encode the rotation, or use quaternions. Either way you need to calculated $$\mathrm{R}$$ first before doing any calculations on a rigid body.

• If you know that axis of rotation in the body frame, then $$\boldsymbol{\omega} = \omega \, \mathrm{R}\, \boldsymbol{\hat{n}}_{\rm body}$$

Kinetic energy is thus calculated with

$$T = \tfrac{1}{2} \omega^2 \left( \boldsymbol{\hat{n}}_{\rm body}^\intercal \mathbf{I}_{\rm body} \boldsymbol{\hat{n}}_{\rm body} \right)$$ Note that if the axis of rotation is fixed to the body, then the part in the parenthesis is a constant.

I think you are asking about the case where the axis of rotation is not fixed to the body, and KE is expressed in body-fixed coordinates (like above). In this case $$\boldsymbol{\hat{n}}_{\rm body} = \mathrm{R}^\intercal \boldsymbol{\hat{n}}$$

Specifically you are using $$\boldsymbol{\hat{n}} = \pmatrix{0 \\ 0 \\ 1}$$ and the spherical rotation matrix

$$\mathrm{R} = \mathrm{rot}_y(-\theta) \mathrm{rot}_z(-\phi) = \begin{pmatrix}\cos\phi\cos\theta & \sin\phi\cos\theta & -\sin\theta\\ -\sin\phi & \cos\phi & 0\\ \cos\phi\sin\theta & \sin\phi\sin\theta & \cos\theta \end{pmatrix}$$

Note that $${\rm R}^\intercal {\rm R} = 1$$ and that when the angles all are zero $${\rm R} = 1$$.

Such that $$\boldsymbol{\hat{n}}_{\rm body} = {\rm R}^\intercal \boldsymbol{\hat{n}} = \begin{pmatrix}\cos\phi\sin\theta\\ \sin\phi\sin\theta\\ \cos\theta \end{pmatrix}$$

Take the equation above and plug in the above to get

$$T = \tfrac{1}{2} \omega^2 \left( I_1 \cos^2 \phi \sin^2 \theta + I_2 \sin^2 \phi \sin^2\theta + I_3 \cos^2 \theta \right)$$

where $$I_1$$, $$I_2$$, and $$I_3$$ are the three diagonal terms of the body-fixed MMOI tensor (the principal MMOI values).

• Oh okay I see. I think this is very helpful. One more question. $T = \frac{1}{2} \mathbf{\omega}^T (R \mathbf{I_{\text{body}}} R^T) \mathbf{\omega}$ is the expression that, if I expand, is in terms of euler angles? I can see how euler angles enter through the rotational matrix, but I can't see where they would enter as time derivatives. Commented Nov 3, 2023 at 18:03
• @Habouz - see my updates above to clarify how to enter the Euler angles (hint: they go into $R$ and not into $n$). Commented Nov 3, 2023 at 18:12
• @Habouz - I made some corrections to the calculations above in order to match your post better. Commented Nov 3, 2023 at 18:34

You can say that the rotation axis is $$(dx,dy,dz)$$ then multiply it by the general 3d rotation matrix($$A=R_{x}R_{y} R_{z}$$) to check your answer.