Free Books

Two Masses Connected by a Rod

Figure B.5: Two ideal point-masses $ m$ connected by an ideal, rigid, massless rod of length $ 2r$.
\includegraphics[width=1.5in]{eps/massrodmass}


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 densityB.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 impedance7.1) ``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.

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}
$


Striking One of the Masses

Now let $ f(t,x) = \delta(t)\delta(x-r)$. That is, we apply an impulse of vertical momentum to the mass on the right at time 0. In this case, the unit of vertical momentum is transferred entirely to the mass on the right, so that

$\displaystyle p \eqsp 1 \eqsp m v_r \,\,\Rightarrow\,\,v_r \eqsp \frac{1}{m},
$

which is twice as fast as before. Just after time zero, we have $ v_r=1/m$, $ v_{-r}=0$, and, because the massless rod remains rigid, $ v_0=1/(2m)$. Note that the velocity of the center-of-mass $ 1/(2m)$ is the same as it was when we hit the midpoint of the rod. This is an important general equivalence: The sum of all external force vectors acting on a rigid body can be applied as a single resultant force vector to the total mass concentrated at the center of mass to find the linear (translational) motion produced. (Recall from §B.4.1 that such a sum is the same as the sum of all radially acting external force components, since the tangential components contribute only to rotation and not to translation.) All of the kinetic energy is in the mass on the right just after time zero:

$\displaystyle E_K \eqsp \frac{1}{2}mv_r^2 \eqsp \frac{1}{2}m\left(\frac{1}{m}\right)^2 \eqsp \frac{1}{2m} \protect$ (B.13)

However, after time zero, things get more complicated, because the mass on the left gets dragged into a rotation about the center of mass. To simplify ongoing analysis, we can define a body-fixed frame of referenceB.16 having its origin at the center of mass. Let $ v'$ denote a velocity in this frame. Since the velocity of the center of mass is $ v_0=1/(2m)$, we can convert any velocity $ v'$ in the body-fixed frame to a velocity $ v$ in the original frame by adding $ v_0$ to it, viz.,

$\displaystyle v \eqsp v' + v_0 \eqsp v' + \frac{1}{2m}.
$

The mass velocities in the body-fixed frame are now

$\displaystyle v'_{-r} \eqsp -\frac{1}{2m},\qquad\qquad v'_r \eqsp \frac{1}{2m},
$

and of course $ v'_0=0$. In the body-fixed frame, all kinetic energy is rotational about the origin. Recall (Eq.$ \,$(B.9)) that the moment of inertia for this system, with respect to the center of mass at $ x=0$, is

$\displaystyle I \eqsp m(-r)^2 + m r^2 \eqsp 2mr^2.
$

Thus, the rotational kinetic energyB.4.3) is found to be

$\displaystyle E'_R \eqsp \frac{1}{2}I\omega^2 \eqsp
\frac{1}{2}(2mr^2)\left(\frac{v'_r}{r}\right)^2
\eqsp mr^2\left(\frac{1}{2mr}\right)^2
\eqsp \frac{1}{4m}.
$

This is half of the kinetic energy we computed in the original ``space-fixed'' frame (Eq.$ \,$(B.13) above). The other half is in the translational kinetic energy not seen in the body-fixed frame. As we saw in §B.4.2 above, we can easily calculate the translational kinetic energy as that of the total mass $ M=2m$ traveling at the center-of-mass velocity $ 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}
$

Adding this translational kinetic energy to the rotational kinetic energy in the body-fixed frame yields the total kinetic energy, as it must. In summary, we defined a moving body-fixed frame having its origin at the center-of-mass, and the total kinetic energy was computed to be

$\displaystyle E_K \eqsp E'_K + E'_R
\eqsp \frac{1}{4m} + \frac{1}{4m}
\eqsp \frac{1}{2m}
$

in agreement with the more complicated (after time zero) space-fixed analysis in Eq.$ \,$(B.13). It is important to note that, after time zero, both the linear momentum of the center-of-mass ( $ p_0=Mv_0=(2m)(v_r/2) =
mv_r=m\cdot(1/(2m))=1/2$), and the angular momentum in the body-fixed frame ( $ L'=I\omega= (2mr^2)(v'_r/r)=(2mr^2)(1/(2mr))=r/2$) remain constant over time.B.17 In the original space-fixed frame, on the other hand, there is a complex transfer of momentum back and forth between the masses after time zero. Similarly, the translational kinetic energy of the total mass, treated as being concentrated at its center-of-mass, and the rotational kinetic energy in the body-fixed frame, are both constant after time zero, while in the space-fixed frame, kinetic energy transfers back and forth between the two masses. At all times, however, the total kinetic energy is the same in both formulations.
Next Section:
Angular Velocity Vector
Previous Section:
Radius of Gyration