What is the moment acceleration of the center of mass for a rigid body under an external force

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

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?

• the center of mass is not always axis of rotation May 23, 2020 at 17:11
• @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?
– CPW
May 23, 2020 at 21:06

• 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.
• 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?