# Acceleration of The Centre of Mass of A Rod

The acceleration a of the center of mass of a system is equal to F/M, where F is the net external force and M is the total mass of all the particles of the system, or at least that is my current understanding of it

A uniform rod in space is subjected to a force F. Now, this force is the net external force and the acceleration a of the center of mass should be equal to F/M, regardless of the distance of application x from the centre of mass.

However, the net external torque, T does depend on x, and a greater T increases angular velocity, which in turn increases the total kinetic energy of the object.

My question essentially is, why is a, the translational acceleration of the center of mass of the object, dependent on the point of application of the force? And if it is not dependent on that, and a is in fact the same no matter what the x, how is total kinetic energy different for the same force applied over the same small distance?

The first point is that the acceleration of $$G$$ do not depend of the point of application of the force$$\overrightarrow{T}$$.
But there is no paradox : if the rod is rotating, the power associated to the force $$\overrightarrow{T}\cdot \overrightarrow{v}$$ is greater (or smaller) that if it is not rotating with $$\overrightarrow{T}$$ applied in $$G$$ : $$\overrightarrow{T}\cdot \overrightarrow{{{v}_{G}}}$$.