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I've been doing a bit of work on a personal project of mine lately and have found myself rather stuck on what first appeared to be a fairly straightforward mechanics problem. The system in question is a rod that is allowed to pivot freely on its end about the end of an actuator and is viewed from an inertial frame at rest with respect to whatever hardware is driving the actuator (the "lab frame", so to speak). The actuator has no prescribed motion and so the axis of rotation will in general have a non-zero acceleration. I understand that the kinetic energy of an object which is translating and rotating about an axis through its centre of mass can be decomposed into the sum of the translational kinetic energy of its centre of mass and its rotational kinetic energy about its centre of mass: $$T=\frac{1}{2}mv_{CM}^2+\frac{1}{2}I_{CM}\omega_{CM}^2.$$ I also understand that if the axis of rotation is at rest with respect to an inertial reference frame, then the kinetic energy is simply $$T=\frac{1}{2}I_{axis}\omega_{axis}^2.$$ That's all well and good and is fairly basic classical mechanics. However I have no idea how to treat this much more general case in which the axis of rotation is accelerating and not through the centre of mass of the rigid rod. If the system had a point-mass rather than the extended rod then I could simply write down the coordinates of said point mass and determine its kinetic energy directly but that won't work here either. The person answering this question offers the following expression for the kinetic energy of an object that is translating and rotating about a fixed point: $$T=\frac 12mV^2+\frac 12\omega^2I_n+m\vec R_{cm}'\cdot(\vec V\times\vec\omega)$$ However, as they present this equation without any sort of proof or derivation or a reference of some kind it is difficult for me to feel assured that it will apply to the specific system under investigation.

Any help clearing up my confusion would be greatly appreciated!

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The correct equation for kinetic energy with general 3D motion is

$$ T = \tfrac{1}{2} \boldsymbol{v}_{\rm CM} \cdot m\, \boldsymbol{v}_{\rm CM} + \tfrac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_{\rm CM} \boldsymbol{\omega} \tag{1}$$

where $m$ is the mass, $\boldsymbol{v}_{\rm CM}$ is the translational velocity vector of the center of mass, $\boldsymbol{\omega}$ is the rotational velocity vector of the body and $\mathbf{I}_{\rm CM}$ is the mass moment of inertia matrix at the center of mass.

But given the definition of translational momentum $\boldsymbol{p} = m\, \boldsymbol{v}_{\rm CM}$ as well as rotational momentum $\boldsymbol{L}_{\rm CM} = \mathbf{I}_{\rm CM} \boldsymbol{\omega}$, you can see that kinetic energy is also defined as $$ T = \tfrac{1}{2} \boldsymbol{v}_{\rm CM} \cdot \boldsymbol{p} + \tfrac{1}{2} \boldsymbol{\omega} \cdot \boldsymbol{L}_{\rm CM} \tag{2} $$

Lemma - The above calculation yields the exact same result if the motion of the body at a different point is considered, and both translational velocity and angular momentum is transformed accordingly.

Proof - Consider a reference point A away from the center of mass. The following is true from standard mechanics if the center of mass is at $\boldsymbol{r}_{\rm CM}$ relative to point A.

$$\begin{aligned} \boldsymbol{v}_A & = \boldsymbol{v}_{\rm CM} - \boldsymbol{\omega} \times \boldsymbol{r}_{\rm CM} \\ \boldsymbol{L}_{A} & = \boldsymbol{L}_{\rm CM} + \boldsymbol{r}_{\rm CM} \times \boldsymbol{p} \end{aligned} \tag{3} $$

Now to show that $T$ calculated at A is the same as calculated at CM

$$ \begin{aligned} T & = \tfrac{1}{2} \boldsymbol{v}_A \cdot \boldsymbol{p} + \tfrac{1}{2} \boldsymbol{\omega} \cdot \boldsymbol{L}_A \\ & = \tfrac{1}{2}\left(\boldsymbol{v}_{{\rm CM}}-\boldsymbol{\omega}\times\boldsymbol{r}_{{\rm CM}}\right)\cdot\boldsymbol{p}+\tfrac{1}{2}\boldsymbol{\omega}\cdot\left(\boldsymbol{L}_{{\rm CM}}+\boldsymbol{r}_{{\rm CM}}\times\boldsymbol{p}\right) \\ &=\tfrac{1}{2}\boldsymbol{v}_{{\rm CM}}\cdot\boldsymbol{p}-\tfrac{1}{2}\left(\boldsymbol{\omega}\times\boldsymbol{r}_{{\rm CM}}\right)\cdot\boldsymbol{p}+\tfrac{1}{2}\boldsymbol{\omega}\cdot\boldsymbol{L}_{{\rm CM}}+\tfrac{1}{2}\boldsymbol{\omega}\cdot\left(\boldsymbol{r}_{{\rm CM}}\times\boldsymbol{p}\right)\\&=\tfrac{1}{2}\boldsymbol{v}_{{\rm CM}}\cdot\boldsymbol{p}+\tfrac{1}{2}\boldsymbol{\omega}\cdot\boldsymbol{L}_{{\rm CM}}+\tfrac{1}{2}\boldsymbol{\omega}\cdot\left(\boldsymbol{r}_{{\rm CM}}\times\boldsymbol{p}\right)-\tfrac{1}{2}\boldsymbol{p}\cdot\left(\boldsymbol{\omega}\times\boldsymbol{r}_{{\rm CM}}\right)\\&=\tfrac{1}{2}\boldsymbol{v}_{{\rm CM}}\cdot\boldsymbol{p}+\tfrac{1}{2}\boldsymbol{\omega}\cdot\boldsymbol{L}_{{\rm CM}}\;\;\checkmark \end{aligned} \tag{4} $$

Note that $\boldsymbol{\omega}\cdot\left(\boldsymbol{r}_{{\rm CM}}\times\boldsymbol{p}\right)=\boldsymbol{p}\cdot\left(\boldsymbol{\omega}\times\boldsymbol{r}_{{\rm CM}}\right)$ from the vector triple product identities.

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  • $\begingroup$ So the fact that the object is constrained to rotate about the reference point A which, in general, may or may not coincide with the centre of mass is of no consequence, then? $\endgroup$ – Wormaldson Apr 26 at 5:19
  • $\begingroup$ Well it is because it makes the calculation easier since $\boldsymbol{v}_A = 0$ $\endgroup$ – John Alexiou Apr 26 at 19:46
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rigid body is rotating about arbitrary axis $\vec{n}$ and translating in 3D space, what is the kinetic energy ?

enter image description here

Kinetic energy

$$T_A=\frac{1}{2}\,m\,\vec{v}^T_A\,\vec{v}_A+\frac{1}{2}\vec{\omega}^T_A\,\Theta_A\,\vec{\omega}_A$$

with :

$$\vec{v}_A=\begin{bmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \\ \end{bmatrix}$$

$$\Theta_A=\Theta_C-m\,\widetilde{\vec{u}}\,\widetilde{\vec{u}}$$

$$\vec{u}=\vec{r}_A-\vec{r}_C$$

$$\vec{\omega}_A=\dot{\varphi}\,\vec{n}_N$$

where :

$\vec{n}_N=\frac{\vec{n}}{||\vec{n}||}$

and

$\widetilde{\vec{u}}=\left[ \begin {array}{ccc} 0&-u_z&u_y\\ u_z&0&-u_x \\ -u_y&x&0\end {array} \right] $

you have 4 generalized coordinates $x\,,y\,,z\,,\varphi$

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