Angular momentum:

Let us consider a rigid body rotates with angular velocity \( \vec{\omega} \) about a fixed point O which is taken as origin . OX, OY and OZ are the three mutually perpendicular axes.

\( \vec{\omega}={\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k} \), where \( {\omega}_x \), \( {\omega}_y \) and \( {\omega}_z \) are the components along OX, OY and OZ axes respectively.

Let \( \vec{r_i}=x_i\hat{i}+y_i\hat{j}+z_i\hat{k} \) be the position vector of the ith particle of mass \( m_i \). The linear velocity of the ith particle is given by

\( \vec{v_i}=\vec{\omega}\times\vec{r_i} \)

Now the angular momentum of the ith particle about the fixed point O is given by

\( \vec{l_i}=\vec{r_i}\times(m_iv_i) \) \( =\vec{r_i}\times{m_i}(\vec{\omega}\times\vec{r_i}) \) \( =m_i\vec{r_i}\times(\vec{\omega}\times\vec{r_i}) \)

So the total angular momentum of the rigid body is given by

\( \vec{L}=\displaystyle{\sum_{i}}{l_i}\\=\displaystyle{\sum_{i}}m_i\vec{r_i}\times(\vec{\omega}\times\vec{r_i}) \)

Using the vector identity \( \vec{A}\times(\vec{B}\times\vec{C})=\vec{B}(\vec{A}\cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B}) \)

or, \( \vec{L}=\displaystyle{\sum_{i}}m_i[\vec{\omega}(\vec{r_i}\cdot\vec{r_i})-\vec{r_i}(\vec{r_i}\cdot\vec{\omega})] \)

or, \( \vec{L}=\displaystyle{\sum_{i}}m_i[\vec{\omega}{\vec{r_i}}^2-\vec{r_i}(\vec{r_i}\cdot\vec{\omega})] \)

Here \( {r_i}^2={x_i}^2+{y_i}^2+{z_i}^2 \)

and \( \vec{r_i}\cdot\vec{\omega}=(x_i\hat{i}+y_i\hat{j}+z_i\hat{k})\cdot({\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k})\\=x_i\omega_x+y_i\omega_y+z_i\omega_z \)

Now the total angular momentum

\( \vec{L}=\displaystyle{\sum_{i}}m_i\left[({\omega}_x\hat{i}+{\omega}_y\hat{j}+{\omega}_z\hat{k})({x_i}^2+{y_i}^2+{z_i}^2)\\-(x_i\hat{i}+y_i\hat{j}+z_i\hat{k})(x_i\omega_x+y_i\omega_y+z_i\omega_z )\right] \)

= \( \displaystyle{\sum_{i}}m_i\left[\hat{i}\left(\omega_x({y_i}^2+{z_i}^2)-\omega_y{x_i}{y_i}-\omega_z{x_i}{z_i}\right)\right]\\+\displaystyle{\sum_{i}}m_i\left[\hat{j}\left(\omega_y({x_i}^2+{z_i}^2)-\omega_x{x_i}{y_i}-\omega_z{y_i}{z_i}\right)\right]\\+\displaystyle{\sum_{i}}m_i\left[\hat{k}\left(\omega_z({x_i}^2+{y_i}^2)-\omega_x{z_i}{x_i}-\omega_y{z_i}{y_i}\right)\right] \)

= \( \hat{i}\left[\omega_x\displaystyle{\sum_{i}}m_i({y_i}^2+{z_i}^2)+\omega_y(-\displaystyle{\sum_{i}}m_i{x_i}{y_i})+\omega_z(-\displaystyle{\sum_{i}}m_i{x_i}{z_i})\right]\\+\hat{j}\left[\omega_y\displaystyle{\sum_{i}}m_i({x_i}^2+{z_i}^2)+\omega_x(-\displaystyle{\sum_{i}}m_i{x_i}{y_i})+\omega_z(-\displaystyle{\sum_{i}}m_i{y_i}{z_i})\right]\\+\hat{k}\left[\omega_z\displaystyle{\sum_{i}}m_i({x_i}^2+{y_i}^2)+\omega_x(-\displaystyle{\sum_{i}}m_i{z_i}{x_i})+\omega_y(-\displaystyle{\sum_{i}}m_i{z_i}{y_i})\right] \)

= \( \hat{i}\left[I_{xx}{\omega}_x+I_{xy}{\omega}_y+I_{xz}{\omega}_z\right]\\+\hat{j}\left[I_{yy}{\omega}_y+I_{yx}{\omega}_x+I_{yz}{\omega}_z\right]\\+\hat{k}\left[I_{zz}{\omega}_z+I_{zy}{\omega}_y+I_{zx}{\omega}_x\right] \)

or, \( \vec{L}=\hat{i}\left[I_{xx}{\omega}_x+I_{xy}{\omega}_y+I_{xz}{\omega}_z\right]\\+\hat{j}\left[I_{yx}{\omega}_x+I_{yy}{\omega}_y+I_{yz}{\omega}_z\right]\\+\hat{k}\left[I_{zx}{\omega}_x+I_{zy}{\omega}_y+I_{zz}{\omega}_z\right] \)

or, \( \vec{L}=\hat{i}L_x+\hat{j}L_y+\hat{k}L_z \)

where,

\( L_x=\left(I_{xx}{\omega}_x+I_{xy}{\omega}_y+I_{xz}{\omega}_z\right) \)

\( L_y=\left(I_{yx}{\omega}_x+I_{yy}{\omega}_y+I_{yz}{\omega}_z\right) \)

\( L_z=\left(I_{zx}{\omega}_x+I_{zy}{\omega}_y+I_{zz}{\omega}_z\right) \)

In the matrix form the total angular momentum can be represented as

\( \begin{bmatrix} L_x\\L_y\\L_z \end{bmatrix}=\begin{bmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I{zy}&I_{zz}\end{bmatrix}\ \begin{bmatrix} {\omega}_x\\{\omega}_y\\{\omega}_z \end{bmatrix} \)

If x, y and z axes are the principal axes of inertia at O then \( I_{xy}=I_{yz}=I_{zx}=0 \)

therefore,

\( L_x=I_{xx}{\omega}_x \),

\( L_y=I_{yy}{\omega}_y \) and

\( L_z=I_{zz}{\omega}_z \)

So the angular momentum vector does not lie along the axis of rotation, unless the axis is principal axis.