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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

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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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-06-11 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University