rigid body dynamics - Proof that a force applied to the center of mass is the same as force applied off-center

There is a similar question that gives a bit of an explanation, but little mathematical proof here: force applied not on the center of mass

I would like mathematical proof that shows that the velocity of a rigid body when a force is applied to the center of mass is equal to the velocity of the same rigid body when the same force is applied to a point on the body other than the center of mass.

Thanks in advance!

Answer

Newton's second law $F=ma$ does not depend on the point of application of force because this law is valid only for point particles. Now to apply it to rigid bodies we must consider them as a system of particles.

Let a rigid body be made up of $N$ particles of mass $m_1,m_2,\cdots,m_N$. Now apply a force $f$ to some $i_{th}$ particle. All other particles will also exert internal forces on each other.

Therefore, the second law for all particles is
\begin{align}f_1^{int}&=\frac{dp_1}{dt}\\ f_2^{int}&=\frac{dp_2}{dt}\\ \cdots\\ f+f_i^{int}&=\frac{dp_i}{dt}\\ \cdots\\ f_N^{int}&=\frac{dp_N}{dt}\end{align}

Adding all these $$\sum_j f_j^{int}+f=\sum_j\frac{dp_j}{dt}$$

By the third law $$\sum_j f_j^{int}=0$$

Thus $$f=\frac{dP}{dt} \text{where } P=\sum_jp_j$$

Now if you apply the same force $f$ to the center of mass of the body, you get the same equation for total momentum. $$f=\frac{dP}{dt}$$

Therefore both situations will have identical solution for the total momentum and hence for the linear velocity of the center of mass of the rigid body.