Newton's Second Law for Rotations

The rotational version of Newton's law $f=ma$ is

$$\tau \eqsp I\alpha, \protect$$ (B.28)

where $\alpha\isdeftext \dot{\omega}$ denotes the angular acceleration. As in the previous section, $\tau$ is torque (tangential force $f_t$ times a moment arm $R$ ), and $I$ is the mass moment of inertia. Thus, the net applied torque $\tau$ equals the time derivative of angular momentum $L=I\omega$ , just as force $f$ equals the time-derivative of linear momentum $p$ :

$$\begin{eqnarray*} \tau &=& \dot{L} \,\eqss \, I\dot{\omega}\,\isdefss \, I\alpha\\ [5pt] f &=& \dot{p} \,\eqss \, m\dot{v}\,\isdefss \, ma \end{eqnarray*}$$

To show that Eq.(B.28) results from Newton's second law $f=ma$ , consider again a mass $m$ rotating at a distance $R$ from an axis of rotation, as in §B.4.3 above, and let $f_t$ denote a tangential force on the mass, and $a_t$ the corresponding tangential acceleration. Then we have, by Newton's second law,

$$f_t \eqsp ma_t$$

Multiplying both sides by $R$ gives

$$f_tR \eqsp ma_tR \isdefs m\dot{v}_tR \isdefs m\dot{\omega}R^2 \eqsp I\dot{\omega} \eqsp I\alpha.$$

where we used the definitions $\omega=v_tR$ and $I=mR^2$ . Furthermore, the left-hand side is the definition of torque $\tau=f_tR$ . Thus, we have derived

$$\tau\eqsp I\alpha$$

from Newton's second law $f_t=ma_t$ applied to the tangential force $f_t$ and acceleration $a_t$ of the mass $m$ .

In summary, force equals the time-derivative of linear momentum, and torque equals the time-derivative of angular momentum. By Newton's laws, the time-derivative of linear momentum is mass times acceleration, and the time-derivative of angular momentum is the mass moment of inertia times angular acceleration:

$$\dot{p_t}=ma_t\;\;\;\Leftrightarrow\;\;\; \dot{L}=I\alpha$$