Clarification on force applied off COM

Question

I was reviewing some dynamics for rigid bodies and system of particles.

In this link the author writes that the change in velocity of the COM is independent of where a force is applied to a system of particles:

https://physics.stackexchange.com/a/118468/147291

While I can follow his derivation there is something I must be missing because for example, if you apply a pure torque on a rigid body you only induce spin and not a change in linear velocity. Therefore the linear velocity of the COM will be different depending upon where you apply the force.

Would appreciate clarification.

Answer 1

The acceleration of the center of mass $\vec a$ of a system is determined by $\vec F = m \vec a$ where $\vec F $ is the net external force and $m$ is the mass. If the vector sum of all the external forces is zero, the CM is not accelerated regardless of where the forces are applied.

For evaluations about the CM, $\vec N ={d \over dt} (\bf I \vec \omega)$ where $\vec N$ is the total external torque, $\bf I$ is the inertia tensor, and $\vec \omega$ is the angular velocity. For rotation in a plane this simplifies to $\vec N = I \vec \alpha$ where $I$ is the moment of inertia, and $\alpha$ the angular acceleration. Note the torque (and angular momentum) depend on the point about which they are evaluated.

By pure torque I assume you mean what is also called a couple where the total force is zero. For example, two forces equal in magnitude but opposite in direction applied to each end of a rod on a pivot at its center. Look up questions regarding "couple" on this site.

You can have no net external force, but external torque about a selected point, and vice versa.