There are a few questions here which are almost the same. But I still had a doubt I read that a force applied anywhere on a rigid body produces the same acceleration on the center of mass. How is that possible?

If I apply a force on the center of mass it just accelerates but if I apply a force off the center of mass it accelerates linearly and rotates. Since it rotates some of the force goes into rotating the body so the linear acceleration cannot be the same? Isn't it?

And is there a mathematical proof that a force applied any where on the rigid body produces the same acceleration? There was a proof in one the answers which used the action of force on discrete particles of the body. Somebody pls give a better proof

  • $\begingroup$ There is not better proof than the one about the action of the forces on the discrete particles. The key step is newtons third law which hows that the forces between the particles cancell out when you compute the motion of the CofM. By the way: understanding the bit about off-center force causing rotation is important. You need to appreciate that change on momentum is force $\times$ time, but change in energy is force $\times$ distance. $\endgroup$
    – mike stone
    Commented Feb 12, 2017 at 16:34
  • $\begingroup$ If you wrap string round a body and pull, you will need to move you hand further to get the same change in momentum as attatching it at the cofm would give. The extra work has gone into rotation. $\endgroup$
    – mike stone
    Commented Feb 12, 2017 at 16:42
  • $\begingroup$ @mikestone Is that also true if a force is applied at the COM and the same force is applied at some other point. If the body rotates and accelerates linearly will its linear acceleration be then equal to the acceleration that it would have when the force would have been applied to the COM . Or is it true only when the body somehow just accelerates and doesn't rotate even though the force is off center. i.e when there is no rotation only then the acceleration of the body due to the force off center equal to the acceleration of the body when force in on COM it's true when there is rotation $\endgroup$
    – E2n
    Commented Feb 12, 2017 at 18:00
  • $\begingroup$ The acceleration of the cofm is the same wherever the force is applied. $\endgroup$
    – mike stone
    Commented Feb 12, 2017 at 19:18
  • $\begingroup$ -1. What was wrong with the proof mentioned in your last paragraph? Please provide a link which identifies that answer. $\endgroup$ Commented Feb 12, 2017 at 19:56

1 Answer 1


This is a result of Newton's 2nd law. Force is the time derivative of linear momentum. And momentum of a collection of particles is defined as $${\bf p} = \sum_i m_i {\bf v}_i = \left( \sum_i m_i \right) {\bf v}_C $$ where $m_i$ is the individual mass, ${\bf v}_i$ the individual velocity and ${\bf v}_C$ the velocity of the center of mass. By taking the derivative of the above you get the relationship between force ${\bf F}$ and center of mass acceleration $\dot{{\bf v}}_C$

$${\bf F} = \left( \sum_i m_i \right) \dot{{\bf v}}_C $$

The center of mass location ${\bf r}_C$ is defined by

$$ \sum_i m_i {\bf r}_i = \left( \sum_i m_i \right) {\bf r}_C $$

and by direct differentiation of the above, you get

$$ \sum_i m_i {\bf v}_i = \left( \sum_i m_i \right) {\bf v}_C $$

where ${\bf v}_i = \dot{{\bf r}}_i$ and ${\bf v}_C = \dot{{\bf r}}_C$

So what happens when a force is applied away from the center of mass?

The point where the force is applied will accelerate at least as much as the center of mass. In general, it will accelerate more due to the rotation. The force will feel a reduced mass given by the relationship

$$ m_{eff} = \left( \frac{1}{m} + \frac{c^2}{I} \right)^{-1} $$ where $m$ is the mass, $I$ is the mass moment of inertia about the axis of rotation and $c$ is the moment arm of the force as seem by the center of mass.


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