Expression for Torque in Terms of Moment of Inertia

Just as angular momentum turned out to be $I\omega$ (the rotational analogue of $p=mv$), the same style of argument gives the rotational analogue of Newton's second law, $F=ma$. Consider once more a rigid object of N particles at perpendicular distances $r_1,\ldots,r_N$ from a fixed axis, now rotating with a common ANGULAR ACCELERATION $\alpha$ (rather than a constant $\omega$). Every particle then has its own TANGENTIAL (linear) acceleration $a_i=r_i\alpha$, and hence experiences its own tangential force $f_i=m_ia_i=m_ir_i\alpha$. Since this force is tangential, its own perpendicular distance from the axis is simply $r_i$ (the same distance used for the acceleration), so the TORQUE this one particle's tangential force contributes is $\tau_i=f_ir_i=m_ir_i^2\alpha$ If the rotation is confined to a single plane, every particle's torque contribution points along the same direction (the axis), so -- exactly as for angular momentum -- these torque magnitudes simply add: $\tau=\sum_i\tau_i=\sum_i m_ir_i^2\alpha=\left(\sum_i m_ir_i^2\right)\alpha=I\alpha$ So the net torque on a rigid body equals $\tau=I\alpha$ the direct rotational analogue of $F=ma$, with I once again replacing mass -- confirming, from yet another direction, exactly why moment of inertia is the correct rotational stand-in for mass. Table 2 collects this and …