So this is the question that's been bothering me:

Say you have a simple rigid body in space that is at rest or traveling with only translational motion at a constant speed. Say that the body is something like a rod and it's not rotating. So, at some point, an external force is acted on the rod, just for an instant and it is not acted along some axis that goes through the center of mass. What will happen??

My guess is that it will start rotating about the center of mass (because of torque) + it will get some translational motion towards the direction of the force. Is that correct? and if so, how much of the force becomes rotation and how much translational movement and why??

Please forgive my English and thnx in advance!!

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    $\begingroup$ Dear gkostra Welcome to Physics SE. Your english is impeccable, I shouldn't have know that it wasn't your mother tongue (it's mine) if you wouldn't have said so. $\endgroup$ Commented Aug 1, 2013 at 23:53

2 Answers 2


The answer to this question is as simple as you seem to think it is. Let's consider the case of two spatial dimensions for simplicity. Let's take our object to be moving inertially without rotation, and we choose a frame co-moving with the object. Additionally, lets choose our origin to be at the center of mass of the object.

Now suppose we apply an impulse $\mathbf{J}$. If the mass of the object is $m$, Newton's laws tell us the final center of mass velocity of the object must be $\mathbf{J}/m$. This tells us the translational part.

Next we consider the rotation. We know the angular impulse will depend on the place where the force is applied to the object. If the force is applied a position $\mathbf{r}$ and the object has moment of inertia $I$, then the angular impulse $K$ will be $\mathbf{r} \wedge \mathbf{J}$. (The wedge is basically a cross product). The resulting angular velocity about the center of mass is then $K/I$.

Your last question makes me think your intuition isn't that good. You say how much of the force becomes rotation and how much becomes translation. It doesn't work like that. All of the force goes into translation and all of the force goes into rotation. If you had a fixed amount of impulse you could generate and wanted to make a bigger rotational speed, you wouldn't have to sacrifice translational speed. In fact, you couldn't sacrifice translational speed---it will always be $J/m$. The only way you could get more rotational speed is by using a bigger lever arm, which doesn't affect the translational speed.

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    $\begingroup$ Many many thnx! your answer couldn't be more helpful!! $\endgroup$
    – gkosfra
    Commented Aug 2, 2013 at 8:44
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    $\begingroup$ "All of the force goes into translation and all of the force goes into rotation." This sentence solved my problem that I had been googling for weeks. Thanks. Maybe you should bold it. $\endgroup$
    – firelynx
    Commented Dec 26, 2017 at 13:52
  • $\begingroup$ @BrianMoths: Sorry for the downwote. I forgot the relationship between force and energy, and therefore mistakenly thought that the all of the force changing both translation and rotation broke energy conservation, when in fact the energy transfer is greater if the force act off-center rather than head-on. Now I cannot undo my mistake before the post is edited :( (Also, I deleted my erroneous comment ... guess I don't use StackExchange quite often enough) $\endgroup$
    – B. Bergtun
    Commented Apr 7, 2021 at 18:03

Here are the rules that govern rigid body motion:

  1. Sum of forces on body affect the acceleration of the center of mass only $$\sum \mathbf{F} = \frac{{\rm d}}{{\rm d}t} (m\,\mathbf{v}_C) = m\, \dot{\mathbf{v}}_C$$
  2. Sum of moments about the center of mass affect the rotational acceleration of the body $$ \sum \mathbf{M}_C =\frac{{\rm d}}{{\rm d}t} \left( \mathrm{I}_C \boldsymbol{\omega} \right) = {\rm I}_C \dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times {\rm I}_C \boldsymbol{\omega} $$

where point C designates the center of mass, $m$ is the mass, ${\rm I}_C$ is the 3×3 mass moment of inertia tensor about the center of mass, $\mathbf{v}_C$ and $\dot{\mathbf{v}}_C$ are the linear velocity and acceleration vectors of the center of mass, $\boldsymbol{\omega}$ and $\dot{\boldsymbol{\omega}}$ the rotational velocity and acceleration vector of the entire body.

For a full treatment of rigid bodies in space See this answer on the derivation of the equations of motion.

In your case there is a net force $\mathbf{F} \ne 0$ as well as a net moment $ \mathbf{M} = \mathbf{r} \times \mathbf{F} \ne 0$ causing both linear and rotational acceleration. Here $\mathbf{r}$ is the position vector of the force relative to the center of mass.

The combined motion is going to have an instantaneous rotation axis at every moment. More on this here. This axis describes a rotation about it, as well as a parallel translation along this axis as the general motion of a rigid body. This is described by Chasle's Theorem where the general motion of a rigid body is a screw.


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