Angular Momentum In Collisions

During the brief collision the pivot exerts large forces but no torque about itself, so angular momentum about the pivot is conserved even though linear momentum is not. Before impact, the clay's angular momentum about the centre is $L_i=Mv\left(\frac{L}{2}\right)$, since it strikes at distance $L/2$ from the axis. After impact the rod and the embedded clay rotate together with moment of inertia $I=\frac{M{L}^2}{12}+M{\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2}{3}$. Conservation gives $\frac{MvL}{2}=\frac{M{L}^2}{3}\omega$, so $\omega =\frac{MvL/2}{M{L}^2/3}=\frac{3v}{2L}$. The value v/L drops the geometric factors. The value 6v/L misplaces the factors by an order of magnitude. The value 2v/(3L) inverts the ratio of the conserved quantities. This is the NCERT angular-momentum approach to rotational collisions. As a dimensional check, $\frac{v}{L}$ carries units of s${}^{-1}$, the correct dimension of angular velocity. It is essential to recognise why linear momentum is not conserved here: the pivot supplies a large impulsive reaction force during the collision, so only angular momentum about that fixed axis survives. Being perfectly inelastic, the impact also dissipates part of the clay's initial kinetic energy as heat and deformation.