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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:

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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)$:

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


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Striking One of the Masses
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Circular Cross-Section