$F = MA$ is what I’d call an empirical truth, something we know to be true but can’t be demonstrated without presupposing it.

But I’m wondering if $Fr = Iα$ can be demonstrated from $F = MA$ and the conservation of angular momentum—- or whatever other physics principle.

The proofs I’ve seen so far of $Fr = Iα$ only considers a single particle, and somehow just assumes that the proof will work for a multitude of particles too.

So, is it demonstrable?

If so, can you provide me a proof? And if it’s one of those proofs that proves $Fr = Iα$ for a single particle, can you tell me why it must also work for multiple particles too?


3 Answers 3


For a system of $n$ particles, it's easiest to directly prove the equation $\vec{\tau} = \dot{\vec{L}}$ in vector form; the form you've written it in is equivalent to this for 2-D motion of a rigid body. We will need to make three assumptions to show that this is true: Newton's Second Law, Newton's Third Law, and the assumption that all internal forces are central forces.

Suppose that we have a system of $n$ particles, all exerting forces on each other. Let $\vec{F}_{\alpha \beta}$ be the force exerted on particle $\alpha$ by particle $\beta$. Also, let $\vec{F}_{\alpha, \text{ext}}$ be the force exerted on particle $\alpha$ by any external forces (from outside the system of $n$ particles.

If we choose a particular origin, then the total angular momentum of the body will be $$ \vec{L} = \sum_\alpha m_\alpha \vec{r}_\alpha \times \dot{\vec{r}}_\alpha. $$ The rate of change of $\vec{L}$ will then be $$ \dot{\vec{L}} = \sum_\alpha m_\alpha \left[ \dot{\vec{r}}_\alpha \times \dot{\vec{r}}_\alpha + \vec{r}_\alpha \times \ddot{\vec{r}}_\alpha\right]. $$ Now, the first term vanishes (the cross product of any vector with itself is zero), while we can use Newton's Second Law to write the second term as $$ \sum_\alpha \vec{r}_\alpha \times (m_\alpha \ddot{\vec{r}}_\alpha) = \sum_{\alpha} \left[\vec{r}_\alpha \times \vec{F}_{\alpha, \text{total}} \right] = \sum_{\alpha} \left[\vec{r}_\alpha \times \left( \vec{F}_{\alpha, \text{ext}} + \sum_{\beta} \vec{F}_{\alpha \beta} \right) \right] $$ Now, the first term is $$ \sum_{\alpha} \vec{r}_\alpha \times \vec{F}_{\alpha, \text{ext}} = \vec{\tau}_\text{ext}, $$ i.e., the total torque exerted on the system by external forces. This is the $\vec{\tau}$ in the equation I wrote at the top. The second term, involving the double sum over $\alpha$ and $\beta$, $$ \sum_{\alpha, \beta} \vec{r}_\alpha \times \vec{F}_{\alpha \beta} $$ represents the net internal torque of the system. We wish to see when this latter term vanishes.

Let's look at all the terms in this double sum involving particles 1 and 2 (for example.) These will be $$ \sum_{\alpha, \beta} \vec{r}_\alpha \times \vec{F}_{\alpha \beta} = \vec{r}_1 \times \vec{F}_{12} + \vec{r}_2 \times \vec{F}_{21} + \dots $$ By Newton's Third Law, we have $ \vec{F}_{12} = - \vec{F}_{21}$ and so this becomes $$ \sum_{\alpha, \beta} \vec{r}_\alpha \times \vec{F}_{\alpha \beta} = \vec{r}_1 \times \vec{F}_{12} - \vec{r}_2 \times \vec{F}_{12} + \dots \\= (\vec{r}_1 - \vec{r}_2) \times \vec{F}_{12} + \dots $$ For completely arbitrary vectors, this will not vanish, and we are stuck.

However, if the internal forces are central—or in other words, if $\vec{F}_{12}$ acts along the line connecting particles 1 and 2—then $\vec{F}_{12}$ is parallel to $\vec{r}_1 - \vec{r}_2$ and so the cross product vanishes. We can see that all of the other terms in the double sum can be "paired off" in this way, and so under this additional assumption, the entire double sum will vanish. Thus, we will have $$ \dot{\vec{L}} = \vec{\tau}_\text{ext}, $$ as desired.

Note, finally, that the assumption that all internal forces are central is intimately related to the conservation of angular momentum. It's not hard to see that if we had an isolated system ($\vec{\tau}_\text{ext} = 0$) where the internal torques were not all central, then we could still have $\dot{\vec{L}} \neq 0$ for this isolated system; in other words, the angular momentum of an isolated system would change, which is the antithesis of a conservation law.

This answers the question as asked, but I would be remiss if I didn't also mention Noether's Theorem here. Conservation of angular momentum can be shown to be a consequence of the 3-D rotational symmetry of space, via Noether's Theorem. We can apply this to the case of inter-particle forces between two isolated particles. If we assume that the laws governing this force are rotationally symmetric, then the force between two isolated particles must be along the line connecting them; any other choice would break the symmetry between all of the directions orthogonal to the line connecting them. Thus, if you assume that the inter-particle forces are rotationally symmetric, it follows that angular momentum is conserved.

  • $\begingroup$ I’m a bit confused as to how the sum of all the mini-torques is the actual torque. For example, let’s say there’s two particles of equal mass and a fixed point about which they rotate. And let’s say the internal force between them is central, and they are aligned with each other. Furthermore, let’s assume one particle is 2 meters from the fixed point, and the other 1 meter. Why is it so that when I push the particle 2 meters away with a force of 2.5 newtons, the result would be the same as if I pushed the particle 2 meters away with 2 Newtons and the particle 1 meter away with 1 newton? $\endgroup$ Commented Jul 12, 2020 at 2:33
  • $\begingroup$ Or another variation of my question is: let’s say there’s a seesaw. The seesaw has a length of 4 meters, and each side is 2 meters. If I place a 1 kg object on one side 2 meters away from the center, and a 2 kg object on the other side 1 meter away from the center, why doesn’t the seesaw rotate? I’ve heard an answer to this in terms of energy, but I’m not really satisfied with it because energy is not intuitive. $\endgroup$ Commented Jul 12, 2020 at 2:38
  • $\begingroup$ I’ll grant that the sum of all forces creates a net force, but that doesn’t imply in itself that the sum of all torques creates a net torque which just so happens to be the actual torque or applied force a certain distance away from the axis of rotation $\endgroup$ Commented Jul 12, 2020 at 16:52

Yes it can. Well the actual law which is stated in vector form commonly as:

$$ \boldsymbol{\tau}_{\rm com} = \mathbf{I}_{\rm com} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathbf{I}_{\rm com} \boldsymbol{\omega} \tag{1} $$

where the subscript $_{\rm com}$ designates the center of mass, $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$ to the torque due to a force, $\mathbf{I}$ is the mass moment of inertia 3×3 tensor, $\boldsymbol{\omega}$ the rotational velocity vector and $\boldsymbol{\alpha}$ the rotational acceleration vector. Caution must be taken, because the MMOI is defined along the body coordinates as $\mathbf{I}_{\rm body}$ and it needs to be transformed to the world coordinates using the congruent transformation $\mathbf{I} = \boldsymbol{R}\, \mathbf{I}_{\rm body} \boldsymbol{R}^\top$ and the 3×3 rotation matrix $\boldsymbol{R}$.

Only along a single axis, or on a plane the above becomes $r F = I \alpha$.

First, see the derivation of linear and rotational momentum from a system of particles that move together. $$ \begin{aligned} \boldsymbol{p} & = m \boldsymbol{v}_{\rm com} \\ \boldsymbol{L}_{\rm com} & = \mathbf{I}_{\rm com} \boldsymbol{\omega} \end{aligned} \tag{2} $$

Then Newton's second law states that

$$ \begin{aligned} \boldsymbol{F} &= \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = m\,\boldsymbol{a}_{\rm com} \\ \boldsymbol{\tau}_{\rm com} &= \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_{\rm com} = \mathbf{I}_{\rm com} \tfrac{\rm d}{{\rm d}t} \boldsymbol{\omega} + (\tfrac{\rm d}{{\rm d}t} \mathbf{I}_{\rm com} ) \boldsymbol{\omega} \end{aligned} \tag{3} $$

Note that the time derivative of a vector riding on a rigid body is $\tfrac{\rm d}{{\rm d}t} \boldsymbol{A} = \boldsymbol{\omega} \times \boldsymbol{A}$, and so this can be applied to each column of $\mathbf{I}$ to give $ (\tfrac{\rm d}{{\rm d}t} \mathbf{I}) \boldsymbol{\omega} = \boldsymbol{\omega} \times \mathbf{I} \boldsymbol{\omega} $. So (3) becomes

$$ \begin{aligned} \boldsymbol{F} &= m\,\boldsymbol{a}_{\rm com} \\ \boldsymbol{\tau}_{\rm com} &= \mathbf{I}_{\rm com} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathbf{I}_{\rm com} \boldsymbol{\omega} \end{aligned} \tag{4} $$



The concept that relates $F = ma$ and its rotational counterpart $Fr = Ia$ is rotational symmetry.

More generally, rotational symmetry is the general relation between linear mechanics and rotational mechanics.

$F = ma \quad $ for linear acceleration

$Fr = I \alpha \quad $ for angular acceleration

Obviously, the minimum number of spatial dimensions for rotation to be possible at all is 2.

Our space has three spatial dimensions, so a general treatment should cover three spatial dimensions, but a discussion of the simplified case of two spatial dimensions is sufficient to establish the relation between linear mechanics and rotational mechanics.

So: I simplify the case to two spatial dimensions.

For the purpose of this discussion the simplest object is a (hula) hoop. A hoop with radius $r$

We can think of that hoop as divided in subsections, we can make those subsections as small as we want. We can think of a force that causes angular acceleration of that hoop as distributed along the entire perimeter of that hoop. That is, we can think of each subsection of the hoop as being subjected to a force that acts on that particular subsection. We set up this demonstration in such a way that that at every point along the perimeter the force is always tangential to the local perimeter. We can think of an accumulated version of the force as a summation over all the subsections.

The point of this setup is to capitalize on rotational symmetry.

For the purpose of calculation you can treat the hoop as a hoop with radius $r$.

Or: you can with the same result treat this case as if all of the mass is in a single point, moving along at a fixed distance $r$ to a center of rotation (in this case the center of mass of the hoop), subject to the accumulated force.

I will of course express the angular acceleration in radians per second squared.

The relation between linear acceleration and angular acceleration is of course the same as the relation between linear velocity $v$ and angular velocity $\omega$.

$ a = \alpha r \qquad (1)$

The local acceleration tangential to the circle can be expressed as a linear acceleration.

$F = ma \qquad (2) $

Replacing linear acceleration $a$ with angular acceleration $\alpha$

$F = m\alpha r \qquad (3) $

Adding a factor $r$ for 'radius' on both sides so that the left hand side of the equation expresses torque:

$Fr = m\alpha r r \qquad (4) $


$Fr = mr^2\alpha \qquad (5) $

(5) contains a factor $mr^2$. At this point it is a natural step to define a concept of moment of inertia: $I$, defining its magnitude as $mr^2$.

$Fr = I\alpha \qquad (6) $

General discussion

The necessary condition for this reasoning to hold good is that the force you are dealing with behaves the same in all spatial orientations. In other words, the necessary condition is rotational symmetry.


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