Striking the Rod in the Middle

First, consider $f(t,x)=\delta(t)\delta(x)$ . That is, we apply an upward unit-force impulse at time 0 in the middle of the rod. The total momentum delivered in the neighborhood of $x=0$ and $t=0$ is obtained by integrating the applied force density with respect to time and position:

$$p \eqsp \iint f(t,x)\,dt\,dx \eqsp \iint \delta(t)\delta(x)\,dt\,dx \eqsp 1$$

This unit momentum is transferred to the two masses $m$ . By symmetry, we have $v_{-r} = v_r = v_0$ . We can also refer to $v_0$ as the velocity of the center of mass, again obvious by symmetry. Continuing to refer to Fig.B.5, we have

$$p \eqsp 1 \eqsp mv_{-r} + mv_r \eqsp (2m)v_0 \,\,\Rightarrow\,\,v_0 \eqsp \frac{1}{2m}.$$

Thus, after time zero, each mass is traveling upward at speed $v_0=1/(2m)$ , and there is no rotation about the center of mass at $x=0$ .

The kinetic energy of the system after time zero is

$$E_K \eqsp \frac{1}{2} mv_{-r}^2 + \frac{1}{2}mv_r^2 \eqsp m\left(\frac{1}{2m}\right)^2 \eqsp \frac{1}{4m}.$$

Note that we can also compute $E_K$ in terms of the total mass $M=2m$ and the velocity of the center of mass $v_0=1/(2m)$ :

$$E_K \eqsp \frac{1}{2} Mv_0^2 \eqsp \frac{1}{2} (2m)\left(\frac{1}{2m}\right)^2 \eqsp \frac{1}{4m}$$