Here is a good exercise in vector identities and matrices, leading to some
important general results: (a) For a rigid body made up of particles of mass
\(m_{\alpha},\) spinning about an axis through the origin with angular velocity
\(\omega,\) prove that its total kinetic energy can be written as
$$
T=\frac{1}{2} \sum m_{\alpha}\left[\left(\omega
r_{\alpha}\right)^{2}-\left(\omega \cdot \mathbf{r}_{\alpha}\right)^{2}\right]
$$
Remember that \(\mathbf{v}_{\alpha}=\boldsymbol{\omega} \times
\mathbf{r}_{\alpha} .\) You may find the following vector identity useful: For
any two vectors a and b,
$$
(\mathbf{a} \times \mathbf{b})^{2}=a^{2} b^{2}-(\mathbf{a} \cdot
\mathbf{b})^{2}
$$
(If you use the identity, please prove it.) (b) Prove that the angular
momentum \(\mathbf{L}\) of the body can be written as
$$
\mathbf{L}=\sum m_{\alpha}\left[\omega
r_{\alpha}^{2}-\mathbf{r}_{\alpha}\left(\omega \cdot
\mathbf{r}_{\alpha}\right)\right]
$$
For this you will need the so-called \(B A C-C A B\) rule, that \(\mathbf{A}
\times(\mathbf{B} \times \mathbf{C})=\mathbf{B}(\mathbf{A} \cdot
\mathbf{C})-\mathbf{C}(\mathbf{A} \cdot \mathbf{B})\)
(c) Combine the results of parts (a) and (b) to prove that
$$
T=\frac{1}{2} \omega \cdot \mathbf{L}=\frac{1}{2} \tilde{\omega} \mathbf{I}
\omega
$$
Prove both equalities. The last expression is a matrix product; \(\omega\)
denotes the \(3 \times 1\) column of numbers \(\omega_{x}, \omega_{y},
\omega_{z},\) the tilde on \(\tilde{\omega}\) denotes the matrix transpose (in
this case a row), and I is the moment of inertia tensor. This result is
actually quite important; it corresponds to the much more obvious result that
for a particle, \(T=\frac{1}{2} \mathbf{v} \cdot \mathbf{p} .\) (d) Show that
with respect to the principal axes, \(T=\frac{1}{2}\left(\lambda_{1}
\omega_{1}^{2}+\lambda_{2} \omega_{2}^{2}+\lambda_{3} \omega_{3}^{2}\right)\)
as in Equation (10.68).
