Let us consider three mutually perpendicular axes, OX, OY, and OZ meeting at the origin O. A body is rotating with angular velocity \( \omega \) about the fixed point O.

So we can write,

\( \vec{\omega}={\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k} \),

where, \( {\omega}_x \), \( {\omega}_y \) and \( {\omega}_z \) are the components of the angular velocity along the OX, OY and OZ axes respectively.

Let \( \vec{r}_i \) be the position vector of the ith particle with respect to the origin O, So we can write,

\( \vec{r}_i=x_i\hat{i}+ y_i\hat{j}+ z_i\hat{k} \)

So the linear velocity of the ith particle is given by,

\( \vec{v}_i=\vec{\omega}\times\vec{r}_i=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \omega_x & \omega_y & \omega_z \\ x_i & y_i & j_i  \end{vmatrix} \)

\( or,\ \vec{v}_i= \hat{i} (\omega_yz_i-\omega_zy_i)+\hat{j}(\omega_zx_i-\omega_xz_i)+\hat{k}(\omega_xy_i-\omega_yx_i) \)

Let, \( m_i \) be the mass of the ith particle then the total kinetic energy of the body is given by,

\( T=\displaystyle{\frac{1}{2}\sum{m_i{v_i^2}}} \)

\( =\displaystyle{\frac{1}{2}}\sum{m_i}[{(\omega_yz_i-\omega_zy_i)}^2+{(\omega_zx_i-\omega_xz_i)}^2+{(\omega_xy_i-\omega_yx_i)}^2] \)

\( =\displaystyle{\frac{1}{2}}\sum{m_i}[\omega_y^2z_i^2+\omega_z^2y_i^2-2\omega_y\omega_z z_iy_i \\+\omega_z^2x_i^2+\omega_x^2z_i^2-2\omega_z \omega_xx_iz_i+\omega_x^2y_i^2+\omega_y^2x_i^2-2\omega_x \omega_yy_ix_i  ] \)

\( =\displaystyle{\frac{1}{2}}[\omega_x^2\sum{m_i}(y_i^2+z_i^2)+ \omega_y^2\sum{m_i}(z_i^2+x_i^2)+ \omega_z^2\sum{m_i}(x_i^2+y_i^2)\\ -2\omega_x\omega_y\sum{m_i}x_iy_i-2\omega_y\omega_z\sum{m_i}y_iz_i-2\omega_z\omega_x\sum{m_i}z_ix_i   ] \)

\( =\frac{1}{2}[I_{xx}\omega_x^2+I_{yy}\omega_y^2+I_{zz}\omega_z^2+2I_{xy}\omega_x\omega_y+2I_{yz}\omega_y\omega_z+2I_{zx}\omega_z\omega_x] \)

Here,

\( I_{xx}=\sum{m_i}(y_i^2+z_i^2) \), \( I_{yy}=\sum{m_i}(z_i^2+x_i^2) \), \( I_{zz}=\sum{m_i}(x_i^2+y_i^2) \), \( I_{xy}=-\sum{m_i}x_iy_i \), \( I_{yz}=-\sum{m_i}y_iz_i \) and \( I_{zx}=-\sum{m_i}z_ix_i \).

Let, \( \omega_x=l\omega \), \( \omega_y=m\omega \) and \( \omega_z=n\omega \) then the kinetic energy can be written as,

\( T=\displaystyle{\frac{1}{2}}\omega^2[I_{xx}l^2+I_{yy}m^2+I_{zz}n^2+2I_{xy}lm+2I_{yz}mn+2I_{zx}nl] \)

\( or,\ T=\displaystyle{\frac{1}{2}}I\omega^2 \)

where, \( I=[I_{xx}l^2+I_{yy}m^2+I_{zz}n^2+2I_{xy}lm+2I_{yz}mn+2I_{zx}nl] \)

If the axes are the principal axes of inertia, the \( I_{xy}=I_{yz}=I_{zx}=0 \), so now the kinetic energy will be,

\( T=\displaystyle{\frac{1}{2}}\omega^2[I_{xx}l^2+I_{yy}m^2+I_{zz}n^2] \)

Here, \( I_{xx} \), \( I_{yy} \) and \( I_{zz} \) are the principal moment of inertia.