1.3. Dynamics term

We need to now derive (H). In Figure 1, x is the center of the cuboid as considered previously and y is an arbitrary point in the cuboid. If we consider a particle of mass m moving with velocity v, its angular momentum is given by:

⟨H⟩ = ⟨r⟩ × mv

Here, ⟨r⟩ is the position vector of the particle relative to the point O about which we are measuring the angular momentum.

In this chunk, we are defining angular momentum for a particle. Angular momentum is a measure of the rotational motion of an object around a point. In our case, the momentum depends on the mass (m) of the particle, its velocity (v), and its distance (the position vector ⟨r⟩) from the point of reference (O). The cross product indicates that angular momentum is dependent not just on the particle's motion, but also on its position relative to the point O.

Think of a spinning figure skater. As they pull their arms in, they spin faster due to conservation of angular momentum. The mass of their arms and the speed at which they spin are crucial—similar to how a particle's mass and position contribute to angular momentum.

To find the angular momentum of the mass m in the cuboid about its center, we need to integrate over all the particles that it contains. As we are writing the equation in the center of mass frame, so the momentum of the particle is simply mass dm times the velocity of the particle v(y) relative to the center of mass, i.e., v(y).

Here, we discuss how to compute the total angular momentum for all particles within a cuboid. By integrating (summing) the angular momentum contributions from every particle, we capture the total effect. The velocity of each particle is considered relative to the center of mass, ensuring a consistent reference point across all particles. This concept is crucial in understanding how massive bodies behave in systems partly due to their collective dynamics.

Imagine a classroom of students spinning around a merry-go-round. Each student’s position and how fast they are moving (their velocity) affects how the entire group spins. Calculating the total spin (angular momentum) involves considering each student’s location and motion around the center.

Taking the time derivative in control mass setting, we get:

d/dt(⟨H⟩) = ρ(ΔV)

In this section, we address how we can calculate changes in angular momentum over time when looking at a fixed amount of mass (control mass). By analyzing how the angular momentum changes as time progresses, we derive an equation that models the system's dynamics. Since the mass is constant, we avoid complications arising from changing mass.

Picture a car driving in a circle. The speed and direction of the car can change over time (angular momentum), but as long as it stays the same car (constant mass), we can accurately track how it moves around the circle.

Comparing this with the derivation of the torque due to body force term, we see that this is exactly similar to that term with just b replaced with ρa(y). So, upon integrating this dynamic term using Taylor’s expansion for acceleration, we would get a similar result as that of the body force contribution to torque.

This chunk emphasizes the relationship between dynamic terms (involving acceleration) and static forces (like body forces). By applying Taylor's expansion (a mathematical tool to approximate functions) to acceleration, we can simplify our analysis. This analogy helps in understanding how previously derived equations for static conditions relate to those involving motion.

Consider a boxer throwing punches. The force they exert (static force) is similar to the effect of their acceleration when moving their fist quickly (dynamic). Both influences are relevant when calculating their overall strength and technique.

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