Can torque relations actually be *derived* from Newton's Laws, or is it something extra?

For a long time I have wondered if there is a way to show that the rotational analogs of Newton's Laws are a direct consequence of just those laws, or are we adding more to them?

I understand mathematically that we take Newton's second law and do "r cross both sides," but that has always struck me as using more than just the 2nd law.

Here it matters where forces on the body are applied, but I don't see where Newton's Laws talk about points where forces are applied to a body. For translational motion it doesn't matter. So are we adding a sort of "extra law" when we do this for rotation?

The only thing I can think of would be to "disassemble" an extended body into differential point masses and work with internal constraint forces that make the body rigid. If every mass is just a point mass it would get around the issue I'm having. I've never seen any discussion along those lines, though.

Short answer is "yes". Newton Laws are all that is required to derive relations such as "Net external torque = moment of inertia x angular acceleration".

And yes, to derive these relations it's necessary to "disassemble" the extended body into many infinitesimally small bodies. Each of these small bodies interacts with others and it's important that the force some body A acts on body B is opposite to the force the body B acts on body A. It's also important, that the direction of these forces is the same as direction of AB.

In these circumstances the total and total torque of all the internal forces is summed up to zero and these forces get eliminated from the final result.

• Note that the assumption that "the direction of these forces is the same as direction of AB" is not part of Newton's Laws. In other words, Newton's Laws on their own do not guarantee that the the torque relations hold for anything other than a point mass. – Michael Seifert Dec 6 '18 at 17:06
• Agreed with @MichaelSeifert that Newton's Three Laws by themselves do not guarantee "that the direction of these forces is the same as the direction of AB." That's an additional postulate, so it is incorrect to say that "Newton Laws are all that is required to derive..." – Luke Pritchett Dec 6 '20 at 15:05

Also note that angular momentum is not a quantity that stands on its own. It is a manifestation of linear momentum at a distance, just like linear velocity is a manifestation of a rotation at a distance.

For me linear momentum is incorrectly characterized as $\vec{p} = m \vec{v}_C$ (where C indicates the center of mass). At any instant a rigid body is rotating about an axis, and if you use $\vec{c}$ the location of the center of mass relative to the rotation axis then the equations of motion are

\begin{align} \vec{p} & = m \vec{\omega} \times \vec{c} & \vec{F} &= \frac{\rm d}{{\rm d}t} \vec{p} \\ \vec{L}_C &= I_C \vec{\omega} & \vec{\tau}_C & = \frac{\rm d}{{\rm d}t} \vec{L}_C \\ \vec{L} &= \vec{L}_C + \vec{c} \times \vec{p} & \vec{\tau} &= \vec{\tau}_C + \vec{c} \times \vec{F} \end{align}

So your question is simply if you have a clump of particles, each obeying the first row of the equations above, can you derive second row?

Yes. this is how. Take an infinitesimal mass ${\rm d}m$ and find it's contribution to linear momentum (from first equation above).

$${\rm d}\vec{p} = (-\vec{c}\times \vec{\omega})\, {\rm d}m$$

Now use the transformation rule (third row of equations) for the infinitesimal particle

$${\rm d} \vec{L} = \vec{c} \times {\rm d} \vec{p} = (-\vec{c}\times \vec{c}\times \vec{\omega})\, {\rm d}m$$

A simple integration yields

$$\vec{L} = \int (-\vec{c}\times \vec{c}\times \vec{\omega})\, {\rm d}m = \left[ \int (-\vec{c}\times \vec{c}\times )\, {\rm d}m \right] \vec{\omega}$$

But what is inside the integral? Each $[\vec{c}\times]$ is actually a 3×3 skew-symmetric matrix representing the cross product operator $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}\times = \begin{vmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{vmatrix}$$

Using the position vector $\vec{c} = (c_x,c_y,c_z)$ the contents of the integral are

$$I = \int \begin{vmatrix} c_y^2+c_z^2 & -c_x c_y & -c_x c_z \\ -c_x c_y & c_x^2 + c_z^2 & -c_y c_z \\ -c_x c_z & -c_y c_z & c_x^2 + c_y^2 \end{vmatrix}\,{\rm d}m$$

You will recognize the above as the definition of the mass moment of inertia (about the rotation axis) if you look at any dynamics book. Se we have shown that $$\vec{L} = I \vec{\omega}$$. The reason that we do the equations of motion about the center of mass and not the center of rotation is that in general we know where the COM location is, but not where the COR is. Also in general the mass moment of inertia matrix is defined about the COM. If $I=I_C$ and $\vec{L} = \vec{L}_C$ which gives us the second row.

The torque equation is derived from the transformation $$\vec{\tau} = \vec{c} \times \vec{F} = \vec{c} \times \frac{\rm d}{{\rm d}t} \vec{p} = \frac{\rm d}{{\rm d}t} \vec{L}$$

I believe this may be of use to you.

The Physical Origin of Torque and of the Rotational Second Law, Daniel J. Cross

According to him, no.

Basically, He explains how an additional "constraint" is required on Newton's original third law such that the idealized rigid body model is no longer feasible, stating that

when two masses interact, the forces they exert on each other, in addition to being equal and opposite, both lie along the line joining the masses.

Obviously, if the forces they exert on each other only lie along the line connecting them, then it would be impossible for a mass to tangentially accelerate another mass about an axis while remaining collinear (or rigid) with it and the axis.

He then goes on to demonstrate that, if the body is NOT idealized as a rigid body, then Newton's laws (coupled with his additional constraint) perfectly predict the torque effect of a force applied to it.

(Please comment if you think I misinterpreted the text of if you feel my summary is not sufficient)