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My derivation is as follows.

  1. The total KE, $T_r$ for a rigid object purely rotating about an axis with angular velocity $\bf{ω}$ and with the $i$th particle rotating with velocity $ \textbf{v}_{(rot)i} = \textbf{r}_i \times \textbf{ω}$, (summing over the i'th particle) is $T_r = \frac{1}{2}m_i|\textbf{r}_i \times \textbf{ω}|^2$, as long as the origin passes through the axis of rotation.

  2. Let's decompose these using a coordinate system with 3 arbitrary orthogonal unit vectors, whose directions have subscript (1,2,3), and expand the brackets. The result can be shown to be $T_r = \frac{1}{2}I_{ij}ω_iω_j$ summing from $i,j = 1$ to $3$, and where $I_{ij}$ are elements of the moment/product of inertia tensor in the given coordinate system.

This seems like the standard expression for rotational kinetic energy. The only assumption was that the object has a constant rotation and that our chosen origin lies on the axis of rotation.

  1. Now let's boost the object with a velocity $\textbf{v}_o$. The total velocity is now $\textbf{v}_o + \textbf{v}_{(rot)i}$ so the total KE is $\frac{1}{2}Mv_o^2 + T_r + m_i \textbf{v}_o \cdot({\textbf{r}_i \times\textbf{ω})}$

It seems to me that the third term is not trivially zero. If it is, can anyone show this? If not, then why do we simply add rotation and translation energies in mechanics?

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  • $\begingroup$ Where are the summations of which you speak? $\endgroup$
    – Gert
    Jun 15 '20 at 18:22
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You're right that the third term does not generically vanish. The key element in decomposing the kinetic energy into rotational and translational parts is that you calculate the rotational kinetic energy about the center of mass.

If the center of mass of the object is at the coordinate origin, and $\mathbf r_i$ is the position of the $i^{th}$ mass, then everything proceeds as you suggest. The velocity of the $i^{th}$ mass is $\mathbf v_i = \mathbf r_i \times \boldsymbol \omega$, and so the total kinetic energy is

$$T = T_r = \sum_i\frac{1}{2} m_i (\mathbf r_i\times \boldsymbol \omega)^2$$

If we perform a boost, then we would have that $\mathbf v_i = \mathbf (r_i-\mathbf R) \times \boldsymbol \omega_{CM} + \mathbf v_0$, where $\boldsymbol \omega_{CM}$ is the angular velocity about the center of mass and $\mathbf R$ is the position of the center of mass. This would give us

$$T=\sum_i\frac{1}{2}m_i\left([\mathbf r_i -\mathbf R]\times\boldsymbol \omega_{CM} + \mathbf v_0\right)^2$$ $$= \sum_i\left\{ \frac{1}{2}m_i\big([\mathbf r_i - \mathbf R]\times \boldsymbol \omega_{CM}\big)^2 + \frac{1}{2}m_i \mathbf v_0^2 + m_i \mathbf v_0\cdot [\mathbf r_i-\mathbf R]\times \boldsymbol \omega_{CM}\right\}$$ The first term is the rotational kinetic energy about the center of mass. The second term is the translational kinetic energy, calculated as though the entire mass $M$ were concentrated at the center of mass position. The third term vanishes because if we sum over masses,

$$\sum_i (m_i\mathbf r_i - m_i \mathbf R) = M\mathbf R - M\mathbf R = 0$$.

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  • $\begingroup$ What about the KE of an object in a situation where the axis of rotation does not pass necessarily through the c.o.m? We know that for any rigid body, there must be an instantaneous axis of rotation, but we do not know that this passes through the centre of mass because there may be external forces acting. $\endgroup$
    – user86425
    Jun 15 '20 at 22:50
  • $\begingroup$ For example, take a gyroscope modelled as a spinning cone under gravity with its apex fixed on a table. The c.o.m lies on the axis of the cone however the axis of rotation keeps moving and the apex of the cone is therefore a suitable origin, but the centre of mass does not pass through the axis of rotation due to rotation about the vertical z axis. $\endgroup$
    – user86425
    Jun 15 '20 at 22:51
  • $\begingroup$ @lucky-guess I'm not sure I understand your question. If you don't calculate the rotational kinetic energy about an axis through the center of mass, then the total kinetic energy does not neatly decompose into a rotational part and a translational part. You're free to do it anyway, of course. $\endgroup$
    – J. Murray
    Jun 15 '20 at 23:25
  • $\begingroup$ it's just that the rotational part may not be well defined around the centre of mass, as in the case i previously discussed. But i guess then the third term is easily tractable as $M v_o \cdot{ (w \times{r_{c.o.m}})}$ $\endgroup$
    – user86425
    Jun 16 '20 at 13:44
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    $\begingroup$ @Dude156 Squaring stuff is how you find magnitudes. I used the fact that $(\mathbf A +\mathbf B)^2 = (\mathbf A +\mathbf B)\cdot (\mathbf A + \mathbf B) = \mathbf A\cdot \mathbf A + \mathbf B \cdot \mathbf B + 2\mathbf A \cdot \mathbf B$ $\endgroup$
    – J. Murray
    Nov 3 '20 at 13:14

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