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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!

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    $\begingroup$ The statement is false. I can have a baseball moving at 10 miles per hour with a force applied to its center of mass, and a baseball moving at 20 miles per hour with a force applied to a point other than the center of mass. $\endgroup$ – Brian Moths Jun 9 '14 at 3:33
  • $\begingroup$ Incorrect. See my link, Edwin's answer, Wikipedia, physics lectures.... $\endgroup$ – Josh Jun 9 '14 at 5:27
  • $\begingroup$ I guess it's a rigid body so it ought to have the saMe velocity where ever you apply the force $\endgroup$ – Shashaank Feb 23 '17 at 18:13
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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.

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Well,, its not sufficiently mathematically sound, But

On any body, wherever you apply a force as long as the force is on that body,,, we can apply F=ma (newtons second law) which is independent of the position of application of force as long as the body as rigid (or else mass may change(breaking off, etc)

However,, when it comes to torque the equation being |T|=mar sin x ,, varies with r

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  • $\begingroup$ Isn't this just a restatement of the problem at hand? $\endgroup$ – lionelbrits Jan 8 '15 at 12:36
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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

I disagree with the above statement that you made. For instance, if I have a long, narrow block and apply a force at its center of mass all the force I applied will get turned into translational motion of the object, i.e., the object will start moving to another location. However, if I take the same object and apply a force near one of the ends of the block perpendicular to the length of the block some to all of the force will produce a rotational motion in the block; thus even with the same force the block will not have the same final translational velocity as in the first case because it will also have some rotational velocity (usually called angular velocity).

However (and perhaps this is what you mean in your question) the final kinetic energies of the objects in both cases will be the same (as long as the same force is applied over the same displacement). In the first case, all the work from applying the force will become translational kinetic energy while in the second case some of the work from the applied force will become translational kinetic energy while some becomes rotational kinetic energy. However, as I said before, for the same force applied over the same displacement the final total kinetic energies in both cases will be the same.

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    $\begingroup$ Nope. The translational velocity WILL be the same. See my link for just one place backing me up. $\endgroup$ – Josh Jun 9 '14 at 5:25
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    $\begingroup$ I agree with Josh, assuming you don't change the direction of the force by making it follow the rotation (thus making it a different force), a lack of counter force will necessitate that the linear acceleration in the force's direction will be the same in both cases. This results in the same translational velocity $\endgroup$ – Jim Jun 9 '14 at 12:46
  • $\begingroup$ I think the issue here is that you are using your intuition that something is being conserved, and since you are causing a rotational motion, less motion should be translational. The thing that is being conserved is energy, so you will have to apply your force over a greater distance to affect the same translational motion. But that is in line with the fact that your contact point is rotating. $\endgroup$ – lionelbrits Jan 8 '15 at 12:40

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