Effect of Torque on Rigid Bodies

A rigid body with a nonzero net external torque $\tau$ acting about its axis of rotation experiences an angular acceleration $\alpha$ about that same axis, related to the torque by the scalar equation

where $I$ is the body's moment of inertia about that axis. This relation was already derived, from first principles, in §5.2.3 (for a point mass) and §5.2.6 (for a full rigid body, via $\tau=dL/dt$); here it is highlighted explicitly as the rotational-motion counterpart of Newton's second law in linear motion, $F=ma$ — torque takes the role that force plays, moment of inertia takes the role that mass plays, and angular acceleration takes the role that linear acceleration plays.

Worked application — a disc driven by a hanging mass. A disc of mass $m_1$ and radius $R$, free to rotate about a fixed axis, has a light, inextensible string wound around it several times, with a mass $m_2$ suspended from the free end of the string, so that the string makes the disc rotate without slipping over it. Analysing the disc and the hanging mass separately by their own free body diagrams: for the disc, the tension $T$ in the string, acting at the rim, produces the torque $RT=I\alpha$; for the hanging mass, gravity and tension give $m_2g-T=m_2a$ (Newton's second law along the vertical). Using $\alpha=a/R$ (since the string does not slip) and $I=m_1K^2$ (with $K$ the disc's radius of gyration), these two equations combine to give

$a=\frac{m_2}{m_1\left(\dfrac{K^2}{R^2}\right)+m_2}\,g.$ …