Torque and Angular Acceleration

Consider a rigid body constrained to rotate about a fixed axis, and within it a single point mass $m$ executing circular motion about that axis at distance $r$. A tangential force $\vec F$, perpendicular to the position vector $\vec r$ of the point mass, is what actually drives this rotation.

The torque produced by this force about the axis is (since $\vec r\perp\vec F$, so $\theta=90°$ and $\sin90°=1$):

Since the force is entirely tangential, it produces a tangential acceleration $a=r\alpha$ (where $\alpha$ is the point mass's angular acceleration about the axis), so by Newton's second law $F=ma=mr\alpha$. Substituting:

So for this single point mass, torque and angular acceleration are directly proportional, via the quantity $mr^2$ — this quantity is precisely the moment of inertia $I$ of the point mass about the axis (studied fully in §5.4), so this last equation reads $\tau=I\alpha$ for the single point mass. In vector form, $\vec\tau=(mr^2)\vec\alpha$, with both $\vec\tau$ and $\vec\alpha$ directed along the axis of rotation.

Generalising to a full rigid body. A rigid body is made up of a great many such point masses, so its total moment of inertia is the sum over all of them, $I=\sum_i m_ir_i^2$, and the torque-angular-acceleration relation for the whole body becomes

This relation — the rotational form of Newton's second law — is developed fully, together with the significance and calculation of moment of inertia $I$ itself, in §5.4 and §5.5. …