As shown in the following, a rigid body experiences an external force $\overrightarrow{F}$ away from its center of mass C: enter image description here

The rigid body is assumed to be free in space, experiencing no other force.

  • According to the theorem for the center of mass, the center of mass should have an moment acceleration $\frac{\overrightarrow{F}}{M}$, in which $M$ is the mass of the rigid body.

  • But since the center of mass C is the rotation center of the rigid body, $\overrightarrow{F}$ can be decomposed according to line $\overrightarrow{CP}$ into $\overrightarrow{Fr}$ to cause the rotation, and $\overrightarrow{Fm}$ to cause the momentum acceleration.

To reconciliate these two pictures, I have to assume that in additional to causing rotation, $\overrightarrow{Fr}$ has to be as efficient as $\overrightarrow{Fm}$ to cause the motion of the center of mass. It is hard for me to understand this. Anyone can explain?

  • $\begingroup$ the center of mass is not always axis of rotation $\endgroup$
    – maverick
    Commented May 23, 2020 at 17:11
  • $\begingroup$ @maverick: without external force, the center of mass has to move in a constant speed (including 0), so the center of mass for a free rigid body has to be the rotation center, right? $\endgroup$
    – CPW
    Commented May 23, 2020 at 21:06

1 Answer 1


According to the theorem, Fr acting alone would give the CM a linear acceleration in the direction of Fr and proportional to Fr. Combine this with the acceleration produced by Fm and you get back to the resultant caused by F. The torque relative to the CM produced by Fr is equal to that produced by F. With a freely moving mass, the resultant force accounts for both the linear and the angular acceleration of the mass.

  • $\begingroup$ Thanks. Still, it is not easy to understand why $\overrightarrow{Fr}$ is as efficient as $\overrightarrow{Fm}$ in cause linear motion. 1) $\overrightarrow{Fm}$ directly passes the center of mass, while $\overrightarrow{Fr}$ does not. 2) $\overrightarrow{Fm}$ cause the change of linear momentum, while $\overrightarrow{Fr}$ cause the change of linear momentum and the angular momentum. $\endgroup$
    – CPW
    Commented May 23, 2020 at 20:19
  • $\begingroup$ For a same burst of force, $\overrightarrow{Fm}$ only results in linear kinetic energy, while $\overrightarrow{Fr}$ results in both linear and angular kinetic energy. If I have a directional dynamite, why the energy effect are so different? $\endgroup$
    – CPW
    Commented May 23, 2020 at 21:04
  • $\begingroup$ Note that the magnitudes (lengths) of $Fm$ and $Fr$ are not necessarily the same. You could also decompose this into a force $F$ acting on $C$ in the same direction as $F$ is in the drawing above but passing through $C$, and a torgue $T$ around C with the magnitude of $F*l$, where $l$ is the perpendicular length between the point $F$ acts on and $C$. $\endgroup$
    – 18th Shard
    Commented Jun 27 at 0:41

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