Classical angular momentum

Angular momentum L is the rotational analogue of linear momentum p. For a particle rotating in a plane at radius r about O (see diagram below), its velocity component v_\perp that is perpendicular to r is

v_\perp=\frac{2\pi r}{T}=r\omega\; \; \; \; \; \; \; \; 58

where T is the period (time taken to complete a revolution) and \omega=\frac{2\pi}{T} is the angular velocity.

The rotational kinetic energy KE_{rot} of the particle is


where I=mr^{2} is the moment of inertia.



What is moment of inertia and why is it equal to mr^{2}?


The moment of inertia is the rotational equivalent of a particle’s inertia in linear motion. For a particle in linear motion, its inertia is quantified by its mass. For a particle in rotational motion, its inertia is dependent on both its mass and the distribution of that mass relative to O (i.e. dependent on m and r for a point mass). Since \frac{1}{2}mv_{\perp}^{\, \, 2}=\frac{1}{2}mr^{2}\omega^{2}, where \omega is the angular velocity of the rotating particle, we define I=mr^{2} such that there is a correspondence between \omega and v_{\perp}, and between I and m.


Consequently, angular momentum, which is the rotational equivalent of linear momentum p=mv, is defined as:

L=I\omega=mr^{2}\frac{v_{\perp}}{r}=rmv_{\perp}\; \; \; \; \; \; \; \; 59

Since a particle’s orbit may be circular or non-circular (see diagram above), its angular momentum is generally expressed as:


or equivalently,


Therefore, angular momentum is a pseudo-vector with a direction indicated by the right-hand rule.


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