Editing Angular momentum (section)

=== Relation to Newton's second law of motion ===

While angular momentum total conservation can be understood separately from [[Newton's laws of motion]] as stemming from [[Noether's theorem]] in systems symmetric under rotations, it can also be understood simply as an efficient method of calculation of results that can also be otherwise arrived at directly from Newton's second law, together with laws governing the forces of nature (such as Newton's third law, [[Maxwell's equations]] and [[Lorentz force]]). Indeed, given initial conditions of position and velocity for every point, and the forces at such a condition, one may use Newton's second law to calculate the second derivative of position, and solving for this gives full information on the development of the physical system with time.<ref>Tenenbaum, M., & Pollard, H. (1985). Ordinary differential equations en elementary textbook for students of mathematics. Engineering and the Sciences.</ref> Note, however, that this is no longer true in [[quantum mechanics]], due to the existence of [[Spin (physics)|particle spin]], which is an angular momentum that cannot be described by the cumulative effect of point-like motions in space.

As an example, consider decreasing of the [[moment of inertia]], e.g. when a [[figure skating|figure skater]] is pulling in their hands, speeding up the circular motion. In terms of angular momentum conservation, we have, for angular momentum ''L'', moment of inertia ''I'' and angular velocity ''ω'':
<math display="block"> 0 = dL = d (I\cdot \omega) = dI \cdot \omega + I \cdot d\omega </math>

Using this, we see that the change requires an energy of:
<math display="block"> dE = d \left(\tfrac{1}{2} I\cdot \omega^2\right) = \tfrac{1}{2} dI \cdot \omega^2 + I \cdot \omega \cdot d\omega = -\tfrac{1}{2} dI \cdot \omega^2</math>
so that a decrease in the moment of inertia requires investing energy.

This can be compared to the work done as calculated using Newton's laws. Each point in the rotating body is accelerating, at each point of time, with radial acceleration of:
<math display="block">-r\cdot \omega^2</math>

Let us observe a point of mass ''m'', whose position vector relative to the center of motion is perpendicular to the z-axis at a given point of time, and is at a distance ''z''. The [[centripetal force]] on this point, keeping the circular motion, is:
<math display="block">-m\cdot z\cdot \omega^2</math>

Thus the work required for moving this point to a distance ''dz'' farther from the center of motion is:
<math display="block">dW = -m\cdot z\cdot \omega^2\cdot dz = -m\cdot \omega^2\cdot d\left(\tfrac{1}{2} z^2\right)</math>

For a non-pointlike body one must integrate over this, with ''m'' replaced by the mass density per unit ''z''. This gives:
<math display="block">dW = - \tfrac{1}{2}dI \cdot \omega^2</math>
which is exactly the energy required for keeping the angular momentum conserved.

Note, that the above calculation can also be performed per mass, using [[kinematics]] only. Thus the phenomena of figure skater accelerating tangential velocity while pulling their hands in, can be understood as follows in layman's language: The skater's palms are not moving in a straight line, so they are constantly accelerating inwards, but do not gain additional speed because the accelerating is always done when their motion inwards is zero. However, this is different when pulling the palms closer to the body: The acceleration due to rotation now increases the speed; but because of the rotation, the increase in speed does not translate to a significant speed inwards, but to an increase of the rotation speed.