The Uneven Dumbbell

Figure 11.5: Definition sketch for the uneven dumbbell (i. e., with $M_2 > M_1$).

Suppose we wish to apply Newton's second law to two particles considered together as a single system. As we showed previously, only external forces act on the total momentum, $\mbox{\bf p}_{total} = \mbox{\bf p}_1 + \mbox{\bf p}_2$, of the two particles:

$$\mbox{\bf F}_{external} = \frac{d \mbox{\bf p}_{total}}{dt} .$$ (12.9)

If we invoke the non-relativistic definition of momentum in terms of mass and velocity, we find that

$$\mbox{\bf p}_{total} = M_1 \mbox{\bf v}_1 + M_2 \mbox{\bf v... ...v}_2}{M_{total}} \right) \equiv M_{total} \mbox{\bf V}_{cm} ,$$ (12.10)

where $M_{total} = M_1 + M_2$. The quantity $\mbox{\bf V}_{cm}$ is the velocity of the center of mass and can be expressed as the time derivative of the position of the center of mass, $\mbox{\bf R}_{cm}$,

$$\mbox{\bf V}_{cm} = \frac{d \mbox{\bf R}_{cm}}{dt} ,$$ (12.11)

where

$$\mbox{\bf R}_{cm} = \frac{M_1 \mbox{\bf r}_1 + M_2 \mbox{\bf r}_2}{M_{total}} .$$ (12.12)

We now see how the kinetic energy and the angular momentum of the dumbbell may be split into two parts, one having to do with the motion of the center of mass of the dumbbell, the other having to do with the motion of the dumbbell relative its center of mass. Figure 11.5 shows graphically how the vectors $\mbox{\bf r}'_1 = \mbox{\bf r}_1 - \mbox{\bf R}_{cm}$ and $\mbox{\bf r}'_2 = \mbox{\bf r}_2 - \mbox{\bf R}_{cm}$ are defined. These vectors represent the positions of the two masses relative to the center of mass. Substitution into equation (11.12) shows that $M_1 \mbox{\bf r}'_1 + M_2 \mbox{\bf r}'_2 = 0$. This leads to the conclusion that $M_1 d_1 = M_2 d_2$ in figure 11.5. We also define the velocity of each mass relative to the center of mass as $\mbox{\bf v}'_1 = d \mbox{\bf r}'_1 / dt$ and $\mbox{\bf v}'_2 = d \mbox{\bf r}'_2 / dt$, and we therefore have $M_1 \mbox{\bf v}'_1 + M_2 \mbox{\bf v}'_2 = 0$.

The kinetic energy of the dumbbell is just the sum of the kinetic energies of the two masses, $K = M_1 v_1^2 /2 + M_2 v_2^2 /2$, where $v_1$ and $v_2$ are the magnitudes of the corresponding vectors. Substitution of $\mbox{\bf v}_1 = \mbox{\bf V}_{cm} + \mbox{\bf v}'_1$ etc., into the kinetic energy formula and rearranging yields

$$K_{total} = K_{trans} + K_{intern} = [ M_{total} V_{cm}^2 /2 ] + [ M_1 v'^2_1 /2 + M_2 v'^2_2 /2 ] .$$ (12.13)

Terms like $\mbox{\bf V}_{cm} \cdot \mbox{\bf v}'_1$ cancel out because $M_1 \mbox{\bf v}'_1 + M_2 \mbox{\bf v}'_2 = 0$.

The first term on the right side of equation (11.13) in square brackets is the kinetic energy the dumbbell would have if both masses were concentrated at the center of mass. The second term is the kinetic energy it would have if it were observed from a reference frame in which the center of mass is stationary. The first is called the translational kinetic energy of the system while the second is called the internal kinetic energy.

The angular momentum of the dumbbell is just the sum of the angular momenta of the two masses: $\mbox{\bf L}_{total} = M_1 \mbox{\bf r}_1 \times \mbox{\bf v}_1 + M_2 \mbox{\bf r}_2 \times \mbox{\bf v}_2$. By reasoning similar to the case of kinetic energy, we can rewrite this as

$$\mbox{\bf L}_{total} = \mbox{\bf L}_{orb} + \mbox{\bf L}_{s... ...ox{\bf v}'_1 + M_2 \mbox{\bf r}'_2 \times \mbox{\bf v}'_2 ] .$$ (12.14)

The first term in square brackets on the right is called the orbital angular momentum while the second term is called the spin angular momentum. The former is the angular momentum the system would have if all the mass were concentrated at the center of mass, while the latter is the angular momentum of motion about the center of mass.

Interestingly, the idea of center of mass and the corresponding split of kinetic energy and angular momentum into orbital and spin parts has no useful relativistic generalization. This is due to the factor of $\gamma \equiv (1 - v^2 / c^2 )^{-1/2}$ in the relativistic definition of momentum, which means that

$$\frac{d m \mbox{\bf x}}{dt} \ne \mbox{\bf p} ~~~\mbox{(relativistic case)} .$$ (12.15)

Figure 11.6: Definition sketch for the rotating dumbbell attached to an axle labeled $\omega$. The axle attaches to the crossbar at the center of mass.

Figure 11.6 shows a dumbbell which is forced to rotate about an axle which is rigidly attached to the dumbbell at the center of mass. The crossbar of the dumbbell is oriented normal to the axle and rotates with angular frequency $\omega$. The axle is rigidly oriented in space. In this case the internal kinetic energy of the dumbbell is all in rotation about the center of mass and it is easy to show that this kinetic energy can be written

$$K_{intern} = I \omega^2 /2 ~~~ \mbox{(fixed axle)} ,$$ (12.16)

where $I = M_1 d_1^2 + M_2 d_2^2$ is called the moment of inertia. Similarly, the spin angular momentum, which is parallel to the axle, is

$$L_{spin} = I \omega ~~~ \mbox{(fixed axle)} .$$ (12.17)

Finally, Newton's second law for rotation becomes

$$\tau = \frac{d L_{spin}}{dt} = \frac{d I \omega}{dt} ~~~ \mbox{(fixed axle)} ,$$ (12.18)

where $\tau$ is the component of torque along the rotation axis.