Would a force not cause acceleration $a=\frac{F}{m}$ if part of it is used to cause rotation?

Question

Let's say we have a massless rod of length $2\,\mathrm{m}$ that has a $1\, \mathrm{kg}$ ball at one end and another $1\,\mathrm{kg}$ ball at the other end.$$ \begin{array}{c} {\displaystyle{~\large{{\bigodot}\!\!\!\!\!\!\frac{\hspace{277px}}{2\,\mathrm{m}}\!\!\!\!\!\!\!\bigodot}}}\\ \begin{array}{lcr} 1 \, \mathrm{kg} & & 1 \, \mathrm{kg} \\ \hspace{249px} \end{array} \end{array} $$

If a force $F$ is applied at the center of mass, the center of mass will have an acceleration of $a=\frac{F}{m}.$

What is the effect on linear acceleration if the force is applied to the left ball on the rod? By the formula $F=ma$, the location of force shouldn't affect linear acceleration, but if the force causes the system to rotate, wouldn't the system not move as far linearly?

Essentially what I'm asking is in both cases where the force is applied either to the center of mass of the end of the rod, does the center of mass travel the same distance in a given time frame?

Answer 1

The center of mass will travel the same distance in the same time frame in both cases because a body will always move as though the external force were applied at it's center of mass.

In case 1, with the external force (for this scenario would we prefer to say a force over a very short time frame, e.g. an impulse?) applied at the CM, the body will only acquire a translational motion. In case 2, when the force is applied off-center, the body will acquire both translational and rotational motion.

The energy transferred to the object is different in the two cases (case 1: $KE = \frac{1}{2} m v^2$, and case 2: $KE=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2$), but the CM will have the same acceleration in both cases.

Answer 2

Let us suppose for simplicity that the external force acts instantly, like a push, transferring a momentum $\vec{P}$ to the system. Then the center of mass will move with this momentum in both cases, i.e., equally.