Tuesday 24 December 2019

rotational kinematics - Can the direction of angular momentum and angular velocity differ?


While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:


Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not?



Answer



Consider a thin rectangular block with width $w$, height $h$ resting along the xy plane as shown below.


block1


The mass of the block is $m$. The mass moment of inertia (tensor) of the block about point A is


$$ {\bf I}_A = m \begin{vmatrix} \frac{h^2}{3} & -\frac{w h}{4} & 0 \\ -\frac{w h}{4} & \frac{w^2}{3} & 0 \\ 0 & 0 & \frac{w^2+h^2}{3} \end{vmatrix} $$



This was derived from the definition (as seen on https://physics.stackexchange.com/a/244969/392)


If this block is rotating along the x axis with a rotational velocity $$ \boldsymbol{\omega} = \begin{pmatrix} \Omega \\ 0 \\ 0 \end{pmatrix} $$ then the angular momentum about point A is


$${\bf L}_A = m \Omega\,\begin{pmatrix} \frac{h^2}{3} \\ -\frac{w h}{4} \\ 0 \end{pmatrix} $$


As you can see, there is a component of angular momentum in the y direction. The angular momentum vector forms an angle $\psi = -\tan^{-1} \left( \frac{3 w}{4 h} \right)$


In the figure below you see the direction of angular momentum, and the circle about which the center of mass is going to orbit due to precession.


Block2


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