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?
Best Answer
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 $F$.