Rotational Kinetic Energy Revisited

If a point-mass is located at $ \underline{x}$ and is rotating about an axis-of-rotation $ \underline{\tilde{\omega}}$ with angular velocity $ \omega $, then the distance from the rotation axis to the mass is $ r= \vert\vert\,\underline{x}-{\cal P}_{\underline{\tilde{\omega}}}(\underline{...
...nderline{\tilde{\omega}}\underline{\tilde{\omega}}^T)\underline{x}\,\vert\vert $, or, in terms of the vector cross product, $ r= \vert\vert\,\underline{\tilde{\omega}}\times\underline{x}\,\vert\vert $. The tangential velocity of the mass is then $ r\omega$, so that the kinetic energy can be expressed as (cf. Eq.$ \,$(B.23))

$\displaystyle E_R \eqsp \frac{1}{2}m\left\Vert\,\underline{\tilde{\omega}}\times\underline{x}\,\right\Vert^2\omega^2 \eqsp \frac{1}{2}I\omega^2 \protect$ (B.25)

where

$\displaystyle I \eqsp m\left\Vert\,\underline{\tilde{\omega}}\times\underline{x}\,\right\Vert^2.
$

In a collection of $ N$ masses $ m_i$ having velocities $ \underline{v}_i$, we of course sum the individual kinetic energies to get the total kinetic energy.

Finally, we may also write the rotational kinetic energy as half the inner product of the angular-velocity vector and the angular-momentum vector:B.27

$\displaystyle E_R \eqsp \frac{1}{2}\, \underline{\omega}\cdot \underline{L}\eqsp \frac{1}{2} I \omega^2,
$

where the second form (introduced above in Eq.$ \,$(B.7)) derives from the vector-dot-product form by using Eq.$ \,$(B.20) and Eq.$ \,$(B.22) to establish that $ \underline{\omega}\cdot\underline{L}=\underline{\omega}\cdot \mathbf{I}\underl...
...^2 \underline{\tilde{\omega}}^T\mathbf{I}\underline{\tilde{\omega}}= I \omega^2$.


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