angular momentumangular velocitymoment of inertiarotational-kinematics
What are the ways to create a mathematical model for the moment of inertia of a football? Can the moment of inertia of the football be simplified to two cones stack against each other?
I'm trying to find the angular momentum of the football by using $L=I \omega$
Is the angular momentum of football actually conserved?
What are the ways one can obtain the angular velocity on a rotating football?
The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as: $$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$ We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain constant. But the angular velocity is not, and that is what which changes with change in angular momentum, because: $$\vec{L\,} = I\vec{\omega\,}$$ For example, ice skaters when have their arms outstretched, their moment of inertia is high and so angular velocity is low, but if they draw in their arms, their moment of inertia decreases and correspondingly, without any external torque, their angular speed increases!
The revision to your question has made it further interesting. Imagine that the rod is connected to a motor. Now, once the insect starts crawling towards the end, the moment of inertia of the entire system increases. According to our equations, the angular speed of the system should correspondingly decrease. But, we are told that it remains a constant $\omega$. This means that the motor has continuously apply a torque to keep the angular velocity constant! This is the external torque that we have find in the question, and it is responsible for the increasing angular momentum as well as kinetic energy of the system.
In the first case, there was no constraint keeping the angular velocity constant, unlike your second question.
The 3×3 mass moment of inertia represents a tensor that expresses a single radius of gyration for each plane passing through the center of of mass.
What is a radius of gyration?
The radius of gyration (RGYR) expresses the distribution of mass around the rotation axis (perpendicular to said plane) as an equivalent ring or cylinder with the entire mass on a single radius from the axis.
But it does more. It defines also where the percussion axis is for a given rotation away from the center of mass.
What is a percussion axis?
The percussion axis, is commonly referred to as the sweet spot in sports is the axis in space which when impacted induces a particular rotation.
How?
In 2D it is kind of magical. Suppose you have a rigid body with radius of gyration $r_G$ and you want to rotate it about a pivot located a distance $c$ from the center of mass. Here is the sketch on the plane perpendicular to the rotation
I have drawn the radius of gyration from the center of mass.
Now follow these steps:
You have now the percussion axis defined
$$ p = \frac{I_{\rm plane}}{m c} = \frac{r_G^2}{c} $$
Notice that the percussion axis is purely a geometrical construct once the radius of gyration is known. The above is related to a pole-polar mapping in geometry.
Lemma
The radius of gyration on a plane can map each point on the plane (rotation center) to a unique line on the plane (percussion axis) and vise versa. If the rotation point is at infinity (a pure translation) then the percussion axis passes through the center of mass (a force through CM translates a body). Additionally, if the rotation point is at the center of mass then the percussion axis is at infinity which represents a pure torque on the body. Hence a pure torque will always rotate a body about its center of mass.
What About 3D?
In 3D the 3×3 mass moment tensor represents three radiii of gyration and a mass.
$$ \boldsymbol{\rm I} = m \begin{vmatrix} r_y^2+r_z^2 & -r_x r_y & -r_x r_z \\ -r_x r_y & r_x^2+r_z^2 &-r_y r_z \\ -r_x r_z & -r_y r_z & r_x^2+r_y^2 \end{vmatrix} $$
The above is reduced into three principal radii of gyration about some rotated coordinate system which eliminates the cross terms (non diagonal terms).
Proof of the percussion geometry
Consider the triangle ABC from the rotation to the tangent point and the center of mass.
The angle $\theta$ is found by $\sin \theta = \frac{r_G}{c}$
Now consider the small triangle BDC which has a side $p = r_G \sin \theta$ because it is similar to ABC.
The percussion axis on the mirror point B' is thus
$$ \boxed{ p = r_G \sin\theta = \frac{r_G^2}{c} }$$
I will leave it up to the reader to prove that a force through the percussion axis on a resting body will cause a rotation about the pivot without any reaction forces.
Best Answer
Generally, moment of inertia isn't scalar $I$ but tensor $\tilde{I}$. The exact equation is
$$\vec{L} = \tilde{I} \vec{\omega}$$
which means that generally vectors $\vec{I}$ and $\vec{\omega}$ do not have the same direction! The expression you wrote
$$\vec{L} = I \vec{\omega}$$
where $I$ is simply a scalar is that it is valid only if the body rotates only around one of the three principal axes of moment of inertia.
For all other rotations things become increasingly complex and you have to use Euler's equations. Angular momentum $\vec{L}$ is of course always conserved if there are no external moments, and $\vec{L}$ is therefore constant, but taking Euler's equations angular velocity $\vec{\omega}$ generally is not constant!
Since football is highly symmetric body, it is easy to identify its principal axes of moment of inertia and it is obvious that $I_{xx} = I_{yy} > I_{zz}$.