What is the derivation for the moment of inertia of a point mass?

Question

Why is the moment of inertia of a point mass defined as $mr^2$? I've tried looking for a derivation online but to no avail. I'm thinking it is derived from the conservation of angular momentum somehow but can't figure out how.

Answer 1

Why is the moment of inertia of a point mass defined as $mr^2$?

This is a good question.

If we wanted to understand this quantity $I = mr^2$ which has some definition, the first thing we could do is think about what the definition means.

If taken literally, saying that $I = mr^2 = r (mr)$ is the 'moment of inertia' of a particle actually implies (see below) that $mr$ is the 'inertia' of a particle, which nobody interprets it as (as the tendency of an object to resist changes in it's motion does not even depend on $r$).

To appreciate why you would even end up with something like $I = mr^2$ in Newtonian mechanics, it's useful to go back to the meaning of the word 'moment'.

The etymology of moment and momentum relate to motion/movement apparently via the latin verb 'moveo' meaning 'to move'.

Apparently the first English use of the word moment meant it in the sense of 'importance' on a lever, i.e. equilibrium means the block on a level have equal importance, non-equilibrium means one will find rotational motion of the level in one direction over the other making one of them more 'important' then the other.

Thus simply due to the 'importance' of Archimedes, historically talking about other circular motions in a way that allows one to easily compare to Archimedes makes sense, so if we're going to use one word related to the Latin 'moveo' to relate to what is called momentum, we can use another word when talking about specifically rotational motion the way Archimedes set it up. We could for example agree to call motion along the direction of one specific sheet of a hyperbolic paraboloid the 'motator'.

The moment of a vector quantity $\vec{A}$ in three dimensions is defined as $$\vec{r} \times \vec{A}.$$ The reason one would even care to do this, and how it relates to the above discussion, is, if we think of the plane/three-dimensional space as a space in which we can move vectors about as is usual in mechanics, i.e. we don't treat them all as fixed at the origin, then we think of $\vec{A}$ as an arrow starting at the tip of $\vec{r}$ and pointing in some direction. If you decompose the vector $\vec{A}$ into a basis where radial motion is one of the orthogonal directions, say a spherical polar coordinate basis, so that $$\vec{A} = A_r \hat{r} + A_{\theta} \hat{\theta} + A_{\phi} \hat{\phi}$$ (can always choose the orientation so that one can ignore the last component and treat it as a planar rotation at a given instant if you prefer, so lets set $A_{\phi}=0$.) then clearly $$\vec{r} \times \vec{A} = 0 + A_{\theta} \vec{r} \times \hat{\theta} + 0 = A_{\theta} \vec{r} \times \hat{\theta}$$ thus the moment of a vector tells us how much $\vec{A}$ wants to move in a direction (in a plane through the origin) orthogonal to $\hat{r}$ at a given instant i.e. to rotate about the vector $\vec{r}$. Thus the moment of force $$\vec{r} \times \vec{F}$$ is just a convenient way to isolate the motion of the 'rotational components' of a force, in general it neglects the radial behavior of that force which obviously matters a lot. In special cases such as where $\vec{F}$ is always purely rotational this is obviously the only motion. Another special case is when $\vec{F}$ is purely radial, then this can obviously be used (as in the Kepler problem) to derive a conservation law for the rotational motion of the system. Force is obviously so important and is interpreted as acting on a mass situated at $\vec{r}$, so it's given the extra name 'torque' as this is the component that 'twists' the particle around the origin we see things from. It's common to loosely refer to the torque of a particle as the scalar quantity $r F_{\theta}$ or suppressing indices as $rF$.

In components the moment from above is $$r F_{\theta} .$$ So, any scalar quantity which is of the form $r B$, with $r$ the length of the position vector, could conceivably be interpreted as a 'moment' of a vector whose component in the rotational direction is $B$.

(Said backwards, the scalar quantity $rB$ encodes the idea of a 'moment', it represents the area of a parallelogram with sides $r$ and $B$. Half this, $\frac{1}{2} rB$ is the area of a triangle with base $B$ and height $r$, which more naturally relates to the vector picture of a motion in the direction the base of the triangle points in, situated at the place where the height ends, and becomes non-trivial thinking of the importance Kepler's law of equal areas which can be derived from this interpretation. To turn it into a vector quantity we treat $B$ as the component of $\vec{B}$ orthogonal to $\vec{r}$ in a plane containing $\vec{r}$ and $\vec{B}$.)

Thus at this stage we could interpret a scalar quantity like $mr^2 = r (mr)$ (pulled out of thin air at this stage) as the moment of some new vector whose motion in the direction orthogonal to $\vec{r}$ has magnitude $mr$. The vector is obviously not the position vector $\vec{r}$ times mass, i.e. $m \vec{r} = m r \hat{r}$, since $\vec{r} \times m \vec{r} = 0$. We can arbitrarily be ready to work with a vector $m r \hat{\theta}$ whose magnitude is $mr$.

Indeed, from this perspective, $$r (mr)$$ could be interpreted as the 'moment of mass/matter at the point $\vec{r}$', and apparently (Ref 1., Pages 7-9) [assuming the translation is correct, an assumption that seems to cause problems in these discussions e.g. with Archimedes and the origin of 'moment'] this is more or less what Euler first called the quantity now called 'moment of inertia', a term he coined over a decade after first framing things in terms of "the moment of the matter".

If you've never seen how to arrive at this quantity $I = mr^2$ out of thin air from some other concept in Newtonian mechanics like $F = ma$, one way to end up with it is to consider purely rotational motion about a fixed axis, applying $\vec{r} \times$ to $\vec{F} = m \vec{a}$ for $\vec{F}$ as above results in \begin{align} \vec{r} \times \vec{F} &= \vec{\tau} \\ &= \vec{r} \times \frac{d \vec{p}}{dt} = \vec{r} \times \frac{d^2 }{dt^2} (m r\hat{r} ) = m r \vec{r} \times \frac{d }{dt} ( \dot{\theta} \hat{\theta} ) = m r \vec{r} \times ( \ddot{\theta} \hat{\theta} - \dot{\theta}^2 \hat{r}) \\ &= m r \ddot{\theta} \vec{r} \times \hat{\theta} = m r \alpha r \hat{r} \times \hat{\theta} = m r^2 \alpha \hat{r} \times \hat{\theta} \\ &= I \vec{\alpha} \end{align} we see from $\vec{\tau} = I \vec{\alpha}$ with $I = mr^2$ that $I$ also behaves like mass does in $F = ma$, it acts as a rotational analogue of inertia i.e. resistance to change.

Indeed, in the book (chapter 3, section 363 of reference 2 below) in which Euler introduces the term (again depending on the accuracy of the translation) it seems he motivates the use of the term 'moment of inertia' from the fact that $\tau = I \alpha$ is similar to $F = ma$ except that force $F$ is replaced by the moment of force, and $m$ is replaced by some messy integral (in our case, i.e. for a single point, it's $mr^2$) so we'll call it moment of mas... no, moment of inertia. The only question is why he didn't choose the more obvious 'moment of mass/matter' by this logic.

Given the history in which even the coiner of the term interpreted $I = mr^2 = r(mr)$ as a moment of something other than 'inertia' initially ("the moment of the matter"), the fact that it acts as 'rotational inertia' in $\tau = I \alpha$ way mass does in $F = ma$, and the fact that the coiner of the term only re-framed it as 'moment of inertia' years later by thinking of forming the moment $\vec{r} \times$ as affecting the left-hand side of $\vec{F} = m \vec{a}$ in the natural way, while on the right-hand side it's natural to replace $m$ by some scalar quantity, it's no wonder the term can be a bit confusing.

Note that in applying this operation $\vec{r} \times$ to an equality like $\vec{F} = m \vec{a}$ (which we do because we expect rotational dynamics to fall out of it because the cross product knocks out forces and acceleration in the radial direction, when talking about fixed-axis motion), it's completely natural to expect that the dimensional quantity $m$ multiplying the dimensional quantity $\vec{r}$ (ignoring $d^2/dt^2$ dimensions) in $F = ma$ is going to change from $m$ to some new quantity with new dimensions, not only from the $r$ in the $\vec{r} \times$ operation, but also because of the fact that angles i.e. arguments of trigonometric functions, are dimensionless (or measured in radians etc) so if you want an angular analogue of $\vec{F} = m \vec{a}$, the dimensions of the multiplier of the acceleration term are going to have to change when acceleration gets turned into angular acceleration (which only has dimensions $[T]^{-2}$ from the second time derivative).

If you wanted to push the literal definition, you could do something like define the vector $m r \hat{\theta}$ as the 'matter at the point $\vec{r}$' and call $\vec{r} \times (m r \hat{\theta}) = m r^2 \hat{r} \times \hat{\theta}$ the 'moment of the matter at the point $\vec{r}$', but all one cares about is the component orthogonal to $\vec{r}$ so $m r \hat{\theta}$ is superficial.

References.

1. 'Euler, Newton, and Foundations for Mechanics', Marius Stan.
2. 'Theoria Motus Corporum Solidorum seu Rigidorum', Euler - translated and annotated by Ian Bruce.

Answer 2

Inertia is the constant in:

$$ F = ma = m\dot v$$

so, the moment of inertia is the constant in:

$$ \tau = I\dot \omega $$

So if I apply a force, $F$, to a point mass at the origin, is accelerates at:

$$ \dot v = \frac F m $$

(This is all occurring in the $x$-direction)

In this case, torque and angular velocity are all zero.

Now if I move the origin in the $y$-direction over to $(0, r)$,

The force is now a torque:

$$ \tau = Fr $$

and the velocity is an angular velocity:

$$ \omega = \frac v r $$

If I plug those into my form of Newton's Law, you get:

$$ \frac d {dt}(r\omega) = \frac 1 m \frac{\tau}r $$

since $\dot r = 0$ in this configuration:

$$ \dot \omega r = \frac 1 m \frac{\tau}r $$

$$ mr^2 = \frac{\tau}{\dot\omega} = I$$

So you get one $m$ because it is inertia, one power of $r$ because torque increases linearly with $r$, and another power of $r$ because angular velocity decreases as $1/r$.

Answer 3

A point mass is orbiting about the origin with rotational velocity $\vec{\omega}$. If the particle location is at $\vec{r}$, then its speed is $\vec{v} = \vec{\omega}\times\vec{r}$, where $\times$ is the vector cross product.

The momentum of the particle is $\vec{p} = m \vec{v} = m (\vec{\omega} \times \vec{r})$

The moment of momentum (angular momentum) is $\vec{L} = \vec{r} \times \vec{p}$, or

$$ \vec{L} = \vec{r} \times \left( m (\vec{\omega} \times \vec{r}) \right) $$

the above is re-grouped as

$$ \vec{L} = \underbrace{(-m \vec{r}\times \vec{r}\times ) }_{\mathrm{I}} \vec{\omega} $$

where the mass moment of inertia tensor $\rm I$ is

$$ \mathrm{I} = m \begin{bmatrix} y^2+z^2 & - x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & - y z & x^2+y^2 \end{bmatrix} $$

given the location $\vec{r} = \pmatrix{x \\y \\z}$

Project it down to 2D with $r^2 =x^2+y^2$ and you have $I_z = m r^2$.