Chapter 11 Problem 51

Problem

A thin rod of mass M and length rests on a frictionless table and is struck at a point from its CM by a clay ball of mass m moving at speed. The ball sticks to the rod. Determine the translational and rotational motion of the rod after the collision.

Solution

Linear momentum of the center of mass is conserved in the totally inelastic collision.

$p_{\text{initial}}=p_{\text{final}}$ $\rightarrow mv=\left(m+M\right)v_{\text{CM,final}}$ $\rightarrow v_{\text{CM,final}}={\frac {m}{m+M}}v$

Angular momentum about the center of mass of the system is conserved. First we find the center of mass, relative to the center of mass of the rod, taking up as the positive direction.

$y_{\text{CM}}={\frac {m\left({\frac {1}{4}}l\right)+M\left(0\right)}{m+M}}={\frac {ml}{4(m+M)}}$

The distance of the stuck clay ball from the system's center of mass is:

$y_{\text{clay}}={\frac {1}{4}}l-y_{\text{CM}}={\frac {1}{4}}l-{\frac {ml}{4(m+M)}}={\frac {Ml}{4(m+M)}}$

Use parallel axis theorem to find the moment of inertia of the rod about the center of mass of the system.

$I_{\text{rod}}={\frac {1}{12}}Ml^{2}+M\left[{\frac {ml}{4(m+M)}}\right]^{2}$

Now the conservation of angular momentum.

$L_{\text{initial}}=L_{\text{final}}$ $\rightarrow mvy_{\text{clay}}=\left(I_{\text{rod}}+I_{\text{clay}}\right)\omega _{\text{final}}$ $\rightarrow \omega _{\text{final}}={\frac {mvy_{\text{clay}}}{\left(I_{\text{rod}}+I_{\text{clay}}\right)}}$ $={\frac {mvy_{\text{clay}}}{\left({\frac {1}{12}}Ml^{2}+M\left[{\frac {ml}{4(m+M)}}\right]^{2}+my_{\text{clay}}^{2}\right)}}$ $={\frac {mv{\frac {Ml}{4(m+M)}}}{\left({\frac {1}{12}}Ml^{2}+M\left[{\frac {ml}{4(m+M)}}\right]^{2}+m\left[{\frac {Ml}{4(m+M)}}\right]^{2}\right)}}$ $={\frac {12mv(m+M)}{l(7m^{2}+11mM+4M^{2})}}$