Exercise 2.1: Rotational kinetic energy

Show that the rotational kinetic energy can also be written as:

$$ T = \frac{1}{2} I\omega^2\\ $$

where $I$ is the moment of inertia about the line through the center of mass with direction $\hat{\omega}$ and $\omega$ is the instantaneous rate of rotation.


Let $\hat{\omega}$ be one of the basis vectors of the coordinate system, $\hat{e_0}$, and with its origin at the center of mass of the body. For a constituent particle in the body, $(\xi_\alpha^1)^2 + (\xi_\alpha^2)^2 = \xi_\alpha^\perp$ will equal to the distance from the particle to the line $\hat{e_0}$ which is the same as the line $\hat{\omega}$. Therefore, the inertia component, $I_{00}$ is:

$$ \begin{align*} I_{00} &= \sum_\alpha m_\alpha \left( (\xi_\alpha^1)^2 + (\xi_\alpha^2)^2 \right) \\ &= \sum_\alpha m_\alpha \left( \xi_\alpha^\perp \right)^2\\ &= I\\ \end{align*} $$

where $I$ is the moment of inertia about the line $\hat{\omega}$. Also, by definition, the components of the angular velocity vector, $\vec{\omega}$, are:

$$ \begin{align*} \omega^0 &= \omega\\ \omega^1 &= 0\\ \omega^2 &= 0\\ \end{align*} $$

since we defined $\vec{\omega}$ as being in the direction $\hat{\omega} = \hat{e}_0$. Therefore the kinetic energy is:

$$ \begin{align*} T &= \frac{1}{2} \sum_{ij} \omega^i\omega^j I_{ij} \\ &= \frac{1}{2} \left( \omega^0\omega^0 I_{00} + \omega^0\cancel{\omega^1} I_{01} + \omega^0 \cancel{\omega^2} I_{02} + ... \right) \\ &= \frac{1}{2} I_{00} \omega^0 \omega^0 \end{align*} $$

All the other terms in the expression for $T$ cancel out as the components $\omega^1$ and $\omega^2$ are equal to zero. Therefore,

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

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