Why does Moment of Inertia have an R^2 dependence (logically)?

[Yug Ahuja]

I am trying to understand the physical and intuitive origin of the $𝑟^2$ factor in the moment of inertia (𝐼=𝑚𝑟2) from first principles, but I keep running into circular or unsatisfactory definitions in standard textbooks.

Usually, the [𝑟2] factor is justified using one of two methods:

1. The Kinetic Energy Derivation: Substituting (𝑣=𝜔𝑟) into (𝐾𝐸=$\frac{1}{2}𝑚𝑣^2$) yields ($\frac{1}{2}(𝑚𝑟^2)𝜔^2$). While mathematically elegant, this simply groups variables; it doesn't give a dynamic, force-based intuition as to why the resistance scales with the square of the distance.
2. The torque derivation: Using (𝜏=𝑟𝐹) and (𝑎=𝛼𝑟) to get ($\tau=mr^2\alpha$). I got a reasonable explanation for $𝑟^2$ explanation from somewhere but it require prior understanding of the torque. I have still stated the explanation below When you try to rotate an object (like spinning a ball on a string), the distance from the pivot point ([𝑟]) works against you twice:
   • First Factor of [r] (Leverage/Torque): To make something rotate, you apply a force. However, what actually causes rotation is torque ((𝜏=𝐹×𝑟)). If a mass is twice as far away ([2𝑟]), the force we apply has less "rotational leverage" over it. You have to push twice as hard just to apply the same rotational action to that distant mass.
   • Second Factor of [r] (Linear Speed): For a wheel to spin at a certain angular speed (RPM), the outer edge has to travel much faster than the inner part. If a mass is twice as far away ([2r]), it must travel along a circle that is twice as large. To keep up with the rotation, it has to move with twice the linear velocity ((𝑣=𝜔𝑟)). Accelerating a mass to twice the linear speed requires twice the effort. Because these two factors happen at the same time, they multiply together to give $𝑟^2$ dependence

My issue with the second method is that textbooks rarely define torque independently. They either call it a layman "rotational analogue of force" (which is just a metaphor), or they define it as ($\vec \tau =\frac{𝑑 \vec L}{ 𝑑𝑡}$), which requires defining angular momentum [𝐿⃗ ] first. If [𝐿⃗ ] is then defined as (𝐈𝜔⃗ ) or (𝑚𝑟⃗ ×𝑣⃗ ) by calling it a "rotational analogue of momentum," the entire argument becomes a circular tautology.

Finally, is there a way to understand 𝑟2 dependence without going circular, or by at least defining one of the things properly without using layman terms like rotational analogues and then throwing up formula. I mean a definition of something should be such that it is self explanatory and directly links to mathematical expression or it lays method by which we can derive the expression

 

[wrobel]

One just need to understand how the very concept of the moment of inertia arises.

Assume a rigid body rotates about a fixed point that coincides with the origin $O$ of an inertial coordinate frame. The rigid body consists of particles of masses $m_1,\ldots,m_N$ with position vectors $\boldsymbol r_1,\ldots, \boldsymbol r_N$. The angular momentum of the rigid body about its fixed point $O$ is$$\boldsymbol K_O=\sum_i m_i\boldsymbol r_i\times \boldsymbol v_i.$$Using the formula $\boldsymbol v_i= \boldsymbol \omega\times \boldsymbol r_i$, we have$$\boldsymbol K_O=J_O \boldsymbol \omega$$where
$$J_O \boldsymbol\omega=\sum_im_i \boldsymbol r_i\times( \boldsymbol \omega\times \boldsymbol r_i)$$
is the inertia operator about the point $O$.

The moment of inertia of the body about the axis passing through the point $O$ parallel to the unit vector $\boldsymbol e$ is defined as $I=(\boldsymbol e,J_O\boldsymbol e).$

 

[Yug Ahuja]

One just need to understand how the very concept of the moment of inertia arises.

Assume a rigid body rotates about a fixed point that coincides with the origin $O$ of an inertial coordinate frame. The rigid body consists of particles of masses $m_1,\ldots,m_N$ with position vectors $\boldsymbol r_1,\ldots, \boldsymbol r_N$. The angular momentum of the rigid body about its fixed point $O$ is$$\boldsymbol K_O=\sum_i m_i\boldsymbol r_i\times \boldsymbol v_i.$$Using the formula $\boldsymbol v_i= \boldsymbol \omega\times \boldsymbol r_i$, we have$$\boldsymbol K_O=J_O \boldsymbol \omega$$where
$$J_O \boldsymbol\omega=\sum_im_i \boldsymbol r_i\times( \boldsymbol \omega\times \boldsymbol r_i)$$
is the inertia operator about the point $O$.

The moment of inertia of the body about the axis passing through the point $O$ parallel to the unit vector $\boldsymbol e$ is defined as $I=(\boldsymbol e,J_O\boldsymbol e).$

but Sir, if we take angular momentum as starting axiom, how do we define it such that its formulation naturally come out to be $\vec p \times \vec v$

 

[wrobel]

I get it. That’s not really my area — I mostly just solve problems. But this forum has experts on the otherworldly essence of definitions. I’m sure they’ll be delighted to help.

 

[Dale]

I would use a Lagrangian approach for this. For a free body in cylindrical coordinates the Lagrangian is: $$L = \frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 + \dot z^2 \right)$$

Since $\theta$ doesn't show in the Lagrangian, it is a cyclic coordinate. So we immediately have the conservation law $$\frac{d}{dt} \frac{\partial L}{\partial \dot \theta} = \frac{d}{dt} m r^2 \dot \theta = 0$$

This is where the $r^2$ factor comes from.

Note that this is the Lagrangian of a free body. There is no force involved. The $r^2$ is not from the force but from the conserved angular momentum itself.

 

[PeroK]

but Sir, if we take angular momentum as starting axiom, how do we define it such that its formulation naturally come out to be $\vec p \times \vec v$

If momentum of a particle is conserved, then angular momentum about any point is conserved, where angular momentum is defined as $\vec r \times m\vec v$. And $\vec r$ is the vector from the point of interest to the particle.

The next step is to use Newton's second law to show how this new quantity changes if a particle is subject to a force.

Try working this out yourself.

 

[robphy]

You may be interested in https://en.wikipedia.org/wiki/Moment_(mathematics)

In mathematics, moments of a function are certain quantitative measures related to the shape of the function's graph.

For example, in physics, if the function represents mass density,
then the zeroth moment is the total mass,
the first moment is the center of mass (multiplied by total mass), and
the second moment is the moment of inertia.

In statistics, the function is a probability distribution,
then the first moment is the expected value,
the second central moment is the variance,
the third standardized moment is the skewness, and
the fourth standardized moment is the kurtosis.

In the spirit of @wrobel 's first post, one could uncover a way to group
a set of terms related by some property in terms of various mass-moments.
Such a grouping could be useful for thinking about what is going on physically
or developing ways to solve mathematical problems.

also useful:
https://en.wikipedia.org/wiki/Moment_of_inertia

 

[Matterwave]

I would use a Lagrangian approach for this. For a free body in cylindrical coordinates the Lagrangian is: $$L = \frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 + \dot z^2 \right)$$

Since $\theta$ doesn't show in the Lagrangian, it is a cyclic coordinate. So we immediately have the conservation law $$\frac{d}{dt} \frac{\partial L}{\partial \dot \theta} = \frac{d}{dt} m r^2 \dot \theta = 0$$

This is where the $r^2$ factor comes from.

One thing I'd like to emphasize here, just in case it gets lost in the math a little, is that the cylindrical coordinates are used here purely for convenience. One should not attribute physical consequences to coordinate choices. It's just that cylindrical coordinates allow you to see this conservation rule immediately with very little algebra.

The conservation of angular momentum is there due to the nature (isotropy) of the Euclidean space, the coordinates are bookkeeping.

 

[A.T.]

(logically)
...
if we take angular momentum as starting axiom ...
...
I keep running into circular or unsatisfactory definitions
...
the entire argument becomes a circular tautology.
...

You seem to have a general misconception about physics. Physics isn't math. Physics isn't linearly derived from axioms by "logic". Physics has no problem with combined definitions, which you call "circular".

For example:
- Charge is what gets affected by an electric field.
- An electric field is what affects charges.
Sounds circular? Who cares? As long it's self-consistent and useful to make predictions it's perfectly fine.

 

[A.T.]

.... force-based intuition ....

...it require prior understanding of the torque. ...

Note that you can derive the concept of torque from linear forces based on a simple static truss-structure, without relying on energy or angular momentum. Once you have torque from forces sorted out, you can proceed with the method 2 you mentioned, to justify the ~r² relationship for rotational inertia.

 

[Herman Trivilino]

I am trying to understand the physical and intuitive origin of the 𝑟2 factor in the moment of inertia (𝐼=𝑚𝑟2) from first principles

Are you okay with an introductory notion of torque as leverage? Think about using a pry bar. It's a lever. As you operate the lever the further from the pivot point you push or pull on the pry bar, the greater the effect. That is, the greater the torque. The torque increases in proportion to the distance from the pivot point.

Now consider a particle rotating about a point. Both the speed of the particle and the torque exerted on it scale with the distance $r$ from the axis of rotation.

That is, the further from the axis, the greater the linear speed; with the speed increasing in proportion to $r$. And as discussed above, the further from the axis, the greater the torque; with the torque increasing in direct proportion to $r$.

 

[Yug Ahuja]

Thank you everyone for the fascinating discussion. It is clear that explaining this factor rigorously requires advanced tools like Noether's Theorem and Lagrangian Mechanics, which are currently outside my scope as an incoming undergrad. Since I need to prioritize my upcoming exams, I won't be actively participating in the debate for now. However, I would love to keep this thread open for anyone else who might want to share a different perspective or alternative explanation. I will definitely look into new replies later once I have these mathematical tools under my belt. Thanks for your time!

 

[Herman Trivilino]

Thank you everyone for the fascinating discussion. It is clear that explaining this factor rigorously requires advanced tools

I don't think so.

 

[Dale]

explaining this factor rigorously requires advanced tools like Noether's Theorem and Lagrangian Mechanics

I don’t know that it actually “requires” it, but the nice thing about the Lagrangian approach is how easy it makes certain things. So I do encourage you to learn the Lagrangian and Hamiltonian approaches as soon as you can. They make many things much easier in physics.

In the meantime, don’t get too bogged down with such questions. It is good to have them but not good to get stuck on them. Circularity is not a real problem. Usually it appears because one part of physics takes $X$ as a definition and derives $Y$ while another takes $Y$ as a definition and derives $X$. There is nothing wrong with that, and other circularities are also similarly not problematic.

So definitely ask such questions, but don't wait for the answers before continuing your studies.

 

[Herman Trivilino]

This discussion of circularity caused me to dig out Arnold B. Aron's Teaching Introductory Physics to for a passage that I found, and still find, seminal:

Among the list of definitions at the beginning of [Newton's] Principia, we find the following Definition IV:

An impressed force is an action exerted upon a body in order, to change its state, either of rest, or of uniform motion in a right line.​

Then, as Law I of three Laws of Motion, we find:

Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.​

The circularity here is quite apparent, [...]