Deriving torque equation from Newton's 2nd Law I'm trying to understand the derivation of the torque equation $\vec{r} \times \vec{F} = I \alpha$. My textbook derives this easily enough from Newton's 2nd Law for a single point with mass $m$ and  radial distance $r$, with the force applied at the same distance $r$, as (if we drop the vector notation for simplicity) $F=ma=m(r\alpha)$, so $rF=mr^2 \alpha = I\alpha$. (Note, there are both 'ay's and 'alpha's there; they look similar.)
The textbook stops there and concludes that the equation holds for any rotating body. But then I tried this derivation for two points positioned along a rigid, massless rod (which points perpendicular to an axis about which it rotates). If there's a point-mass $m_1$ a distance $r_1$ from the axis of rotation and another point-mass $m_2$ at a distance $r_2$, and the force is applied at (for example) distance $r_2$, I get $F=m_1 a_1 + m_2 a_2 = m_1 r_1 \alpha + m_2 r_2 \alpha$, so  $r_2 F= (m_1 r_1 r_2 + m_2 {r_2}^2 )\alpha$. But $m_1 r_1 r_2 + m_2 {r_2}^2$ isn't the correct value for $I$ here. What am I missing?
 A: You cannot write $F=m_1 a_1 + m_2 a_2$, because $F$ is not the only force acting on the rod.
The rod is hinged at a point on the axis, and this hinge stops one end of the rod from moving. Thus this hing exerts a reaction force on the rod.
The torque equation is unchanged because this force passes through the axis, and so does not exert a torque.
Hint to find the reaction force: Newton's law states that $\sum F_{ext}=ma_{com}$. Try to find $a_{com}$(acc. of center of mass), and you'll get the reaction force in terms of $F$.
A: I believe this may be of use to you.
The Physical Origin of Torque and of the Rotational Second Law, Daniel J. Cross
Newton's laws by themselves are not enough to derive the effects of torque for anything other than a point mass. For more than one point mass, you need to add an additional constraint 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.

From here, you can see that the rigid body system you are using is not actually possible (if the forces they exert on each other can only lie on the line connecting them, then it is impossible for one mass to exert a tangential force on the other mass while remaining collinear, or rigid, with it and the axis) and that you must take into account deformations.
A: Ofcourse if your book stops there, then it needs to assert something like “the equation holds for every rotating body”, since it can't be so useful to multiply Newton's equation by $r$ (note also that $a_\theta =\alpha r$ only if the radius is constant, otherwise you would get an extra term $2\dot r \dot \theta $). What your book is proving is a very special case:

If a particle is at constant distance from the origin, and $F_\theta =ma_\theta $ holds, then $F_\theta r=ma_\theta r$ also holds.

The useful relationship is given by Euler's equations in the CM reference frame: $$\mathbf \Gamma = I\cdot \dot {\boldsymbol \omega}+\boldsymbol \omega \times I \cdot \boldsymbol \omega,$$
where $I$ is the matrix of inertia defined by the relation $\mathbf L = I\cdot \boldsymbol \omega $ and $\mathbf \Gamma$ is the momentum of the external forces. If $\ \omega \parallel \mathbf L$ (this happens for example if the body has cylindrical symmetry or the body is planar) this reduces to $$\mathbf \Gamma = I \cdot {\boldsymbol \alpha}.$$
Finally, if the rotation is around an axis of inertia $e_1$ and $\Gamma \parallel \omega $ (again, the case of planar bodies in planar motion), this can be written simply as $$\Gamma = I_1 \alpha ,$$
since $I$ is diagonal in this reference frame.
