I'm doing some calculations on the motion of a rigid body as part of a project, and (as a tangent) I've come across something that I can't quite explain.

Case 1: If I apply a force $F$ though the centre of mass of a rigid body, I can make it move in a straight line with some velocity.

Case 2: If, however, I apply the same force $F$ so it is off-set from the centre of mass by some perpendicular distance $d$, I can make it move in a straight line, but also make it rotate.

It's a fairly straight forward proof to show that the off-set force in Case 2 can be replaced by a force $F$ at the centre of mass (i.e. identical to Case 1) plus a force couple resulting in the moment $M=F\times d$ (see wikipedia).

Now the sticking point - in Case 1 the force (let's say it's an instantaneous impulse) will create some linear motion. If the body has mass $m$ and resultant velocity $v$, then the energy of the system is $E=\frac{1}{2} m v^2$. In Case 2, I have the same force applied at the COM, so I should get the same resultant velocity and the energy would be $E_{\text{linear}}=\frac{1}{2} m v^2$, but I also get a rotation, and that energy will be $E_{\text{rotational}} = \frac{1}{2} I \omega^2$, where $\omega$ is whatever angular velocity results.

Now I know there are some details I'm missing here but from this (albeit lacking in some detail) perspective, it looks like there is suddenly more energy in the system, just by pushing at a different point. That just doesn't sound right to me.

What's going on here? Is there actually more energy in Case 2, or is there some detail I'm missing/neglecting/glazing over?

  • $\begingroup$ Check out @jkej's answer here: physics.stackexchange.com/questions/66960/… similar to the answer below. $\endgroup$ Commented Nov 12, 2020 at 9:19
  • $\begingroup$ How do you resolve "…the same force… is off-set from the centre of mass…" and "…the same force is applied at the COM…", please? $\endgroup$ Commented Nov 13, 2020 at 2:51

1 Answer 1


The problem here is that no force can be perfectly impulsive because then the distance moved while the force was being applied would be zero and therefore the work done would be zero. If we assume some finite impulse $I$ then:

$$ I = F t $$

where we are applying the force $F$ for a time $t$. The perfectly impulsive force would be the limit $t \to 0$, and that requires $F \to \infty$. We end up with the impulse being given by the product of infinity times zero, which is undefined.

So let's consider a non-infinite force $F$ applied for a short but non-zero time $t$. When you apply the force in line with the centre of mass the object moves some distance $s$ and does work:

$$ W = Fs $$

The increase in the linear kinetic energy is equal to this work.

If you now apply the force offset from the centre of mass by a distance $r$ we still get the same linear acceleration, but now we get an angular acceleration as well. So in addition to moving a distance $s$ the object rotates by some angle $\theta$, and the rotation by a (small) angle $\theta$ means the point of application of the force moves an additional distance $s' = r\theta$. This means the total work done is now:

$$ W = F(s + r\theta) $$

It is the extra term $Fr\theta$ that goes into the rotational kinetic energy.

So the solution to the paradox is that the force does more work when applied off centre because the point of application of the force moves an increased distance.

  • 1
    $\begingroup$ Right, I suspected the devil might be in the detail. So to summarise, in a true physical system, there actually is extra energy because of the extra distance covered due to the rotation. $\endgroup$
    – Bamboo
    Commented Nov 12, 2020 at 9:18
  • $\begingroup$ @Phill yes, that's correct. $\endgroup$ Commented Nov 12, 2020 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.