Expression for Angular Momentum in Terms of Moment of Inertia

To connect angular momentum to the moment of inertia built up in section 1.5, consider once again a rigid object made of N particles of masses $m_1,\ldots,m_N$ at respective perpendicular distances $r_1,\ldots,r_N$ from a fixed axis, all sharing the common angular speed $\omega$ but with individual linear speeds $v_i=r_i\omega$, directed tangentially. Particle i's linear momentum has magnitude $p_i=m_iv_i=m_ir_i\omega$, directed along its own velocity (tangentially); since its position vector $r_i$ is perpendicular to this tangential velocity, its angular momentum has magnitude $L_i=p_ir_i=m_ir_i^2\omega$ For a rigid body with a FIXED axis of rotation, every individual particle's angular momentum vector points along the SAME direction -- the axis itself, by the right-hand rule -- so, unlike a general vector sum, these magnitudes can simply be added algebraically: $L=\sum_i L_i=\sum_i m_ir_i^2\omega=\left(\sum_i m_ir_i^2\right)\omega=I\omega$ using the very definition of I from section 1.5. So the total angular momentum of a rigid body about a fixed axis is $L=I\omega$ directly analogous to …