Two Masses Connected by a Rod

Figure B.5: Two ideal point-masses $m$ connected by an ideal, rigid, massless rod of length $2r$ .

As an introduction to the decomposition of rigid-body motion into translational and rotational components, consider the simple system shown in Fig.B.5. The excitation force density^B.15 $f(t,x)$ can be applied anywhere between $x=-r$ and $x=r$ along the connecting rod. We will deliver a vertical impulse of momentum to the mass on the right, and show, among other observations, that the total kinetic energy is split equally into (1) the rotational kinetic energy about the center of mass, and (2) the translational kinetic energy of the total mass, treated as being located at the center of mass. This is accomplished by defining a new frame of reference (i.e., a moving coordinate system) that has its origin at the center of mass.

First, note that the driving-point impedance (§7.1) ``seen'' by the driving force $f(t,x)dx$ varies as a function of $x$ . At $x=0$ , The excitation $f(t,0)dx$ sees a ``point mass'' $2m$ , and no rotation is excited by the force (by symmetry). At $x=\pm R$ , on the other hand, the excitation $f(t,\pm R)dx$ only sees mass $m$ at time 0, because the vertical motion of either point-mass initially only rotates the other point-mass via the massless connecting rod. Thus, an observation we can make right away is that the driving point impedance seen by $f(t,x)$ depends on the striking point $x$ and, away from $x=0$ , it depends on time $t$ as well.

To avoid dealing with a time-varying driving-point impedance, we will use an impulsive force input at time $t=0$ . Since momentum is the time-integral of force ( $f=ma=m\dot{v}\,\,\Rightarrow\,\,mv=\int f\,dt$ ), our excitation will be a unit momentum transferred to the two-mass system at time 0 .