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Newton's Second Law for Rotations

The rotational version of Newton's law $ f=ma$ 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 $ 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$:
\tau &=& \dot{L} \,\eqss \, I\dot{\omega}\,\isdefss \, I\alpha\\ [5pt]
f &=& \dot{p} \,\eqss \, m\dot{v}\,\isdefss \, ma
To show that Eq.$ \,$(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,

$\displaystyle f_t \eqsp ma_t

Multiplying both sides by $ R$ 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 $ I=mR^2$. 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 $ 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:

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

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Equations of Motion for Rigid Bodies
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