Newton's Second Law for Rotations

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The rotational version of Newton's law is

$\displaystyle \tau \eqsp I\alpha, \protect$

(B.28)

where $\alpha\isdeftext \dot{\omega}$ denotes the angular acceleration. As in the previous section, $\tau$ is torque (tangential force

times a moment arm

), and

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

equals the time-derivative of linear momentum

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

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

$\displaystyle f_t \eqsp ma_t$

Multiplying both sides by

gives

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

where we used the definitions $\omega=v_tR$ and

. Furthermore, the left-hand side is the definition of torque $\tau=f_tR$ . Thus, we have derived

$\displaystyle \tau\eqsp I\alpha$

from Newton's second law

applied to the tangential force

and acceleration

of the mass

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:

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