Say I have a rigid body in space. I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.

What is the proof of this?

  • $\begingroup$ You mean like a mathematical proof on experimental proof? $\endgroup$ – Ataraxia Feb 9 '13 at 15:25
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    $\begingroup$ You could google for cars crashing on ice site:youtube.com. $\endgroup$ – dmckee Feb 9 '13 at 15:28
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    $\begingroup$ I could be crazy, but I don't think that statement is generally true. Consider a uniform, thin rod fixed at one endpoint, and suppose you apply a force on the other endpoint, then the rod will rotate about the fixed point, not the center of mass. $\endgroup$ – joshphysics Feb 10 '13 at 0:13
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    $\begingroup$ @joshphysics Yes, it isn't if the rod is fixed. That's why my object is in space. There is more than one force acting when it is fixed. $\endgroup$ – Alraxite Feb 10 '13 at 0:53
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    $\begingroup$ A tip: in physics, there aren't really 'proofs'. There are derivations, which start from a formula (like Newton's F=ma) and end at the statement you want. But proof is a touchy word in physics and they are not interchangeable. $\endgroup$ – user44430 Feb 12 '13 at 1:11

A very simple reason would be that if the body rotated about some point other than the center of mass, the center of mass in the ground frame would be in circular motion.

Now we know that the motion of the center of mass is governed by EXTERNAL FORCES ONLY, and in case of a force applied for a short time, there is no external force acting on the center of mass subsequently.

So we can say that the subsequent motion of the center of mass will be linear(and not circular , which it would be if the body rotated about some other point). And as the body has some angular momentum, it will rotate about the center of mass!

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    $\begingroup$ The COM does not need to be in circular motion in order for it to not be the center of rotation (or instant center of zero velocity, as it is otherwise known). It is merely necessary for its velocity in your chosen inertial reference frame to not be zero. The instant axis of rotation will pass through the point in the plane of rotation which has zero velocity in your chosen reference frame, due to a superposition of the COM linear velocity and the relative velocity imposed by angular rotation. $\endgroup$ – Asad Saeeduddin Oct 21 '15 at 11:54
  • $\begingroup$ @AsadSaeeduddin I guess my answer is incomplete. If we agree that the COM is in linear motion and the body is rigid, each point on the body will have to be at constant distance from the COM at all times. This is only possible if each point is in linear motion too with the same velocity as the COM(i.e. stationery w.r.t. COM) or in circular motion about the COM (w.r.t COM). The COM being in linear motion has nothing to do with it being the center of rotation. It is the other way around. $\endgroup$ – udiboy1209 Oct 22 '15 at 13:34
  • $\begingroup$ So if the points are in circular motion about the COM in the body frame, and the body is accelerating, that is proof positive that the points are not in circular motion about the COM in the inertial frame (except in the instantaneous special case of the COM velocity being zero). What you've just stated is a disproof of your own answer. $\endgroup$ – Asad Saeeduddin Oct 22 '15 at 14:59

I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.

To my understanding, your question is flawed. If a single force is applied to a rigid body under the influence of no other forces, either:

  1. The line of action of the force passes through the center of mass, causing a pure translation and no rotation
  2. The line of action of force does not pass through the center of mass, in which case you end up with a pure rotation about an axis which does not pass through the center of mass. In other words the instantaneous axis of zero velocity induced by a single force can never be the center of mass.

If you apply an eccentric force, the center of mass of the body will undergo a linear acceleration, and the body itself will undergo an angular acceleration. In a fixed reference frame, this can be viewed as a pure rotation about a certain point, but this point will never be the center of mass of the body.

  • $\begingroup$ +1 Asad your answer is more concise, and mine more derivation driven since the OP asked for a "proof". We both say the same thing actually. $\endgroup$ – ja72 Nov 27 '13 at 13:14
  • $\begingroup$ Could you pls explain briefly why the axis of pure rotation can never pass through the centre of mass ? Also then you would be disagreeing with the answer by udiboy1209 ? $\endgroup$ – Shubham Oct 21 '15 at 10:44
  • $\begingroup$ @Shubham If you agree that the COM will undergo linear acceleration due to an eccentric force, you've already conceded that it does not have zero velocity in your chosen inertial reference frame. How then could the instant axis of zero velocity pass through it? I've added my thoughts on udiboy's answer as a comment. $\endgroup$ – Asad Saeeduddin Oct 21 '15 at 12:05
  • $\begingroup$ What if I give an impulsive force to an end of a uniform rod ? is pure rotation about centre of mass still not possible, if my frame of reference is the centre of mass ? $\endgroup$ – Shubham Oct 22 '15 at 9:38
  • $\begingroup$ @Shubham The geometry of the body doesn't really matter. Does the line of action of the force pass through the center of mass? $\endgroup$ – Asad Saeeduddin Oct 22 '15 at 15:03

What you are talking about is called the instant center of percussion. To purely rotate a rigid body about an axis (the rotation axis) a force needs to be applied along the axis of percussion which is a) perpendicular to the rotation axis, b) on the far side of the center of gravity from the pivot and c) located a distance $ \ell =c + \frac{I}{m c}$ from the pivot ($m$ mass, $I$ mass moment of inertia about cm and $c$ distance between pivot and cm).


Consider a body with desired rotation $ \vec{\omega} = (0,0, \omega_z)$ about a point A aligned with a local $\hat k$ axis, and the center of gravity located along the local $\hat i$ axis, with coordinates $\vec{c} = (c_x,0,0)$.

An impulse with components $\vec{J}=(J_x,J_y,J_z)$ is applied at a location $\vec\ell = (l_x,l_y,l_z)$ relative to A with the equations of motion at the center of mass

$$ \vec{J} = m \left( \hat 0 + \vec{\omega} \times \vec{c} \right) \\ (\vec{\ell} -\vec{c} ) \times \vec{J} = I \vec{\omega} $$

in components the above is

$$ \begin{pmatrix} J_x \\ J_y \\ J_z \end{pmatrix} = m \begin{pmatrix} 0 \\ 0 \\ \omega_z \end{pmatrix} \times \begin{pmatrix} c_x \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ m c_x \omega_z \\ 0 \end{pmatrix} $$

So $J_x=J_z=0$ making $\vec{J}$ to be along the local $\hat{j}$ axis.

$$ \begin{pmatrix} \ell_x - c_x\\ \ell_y \\ \ell_z \end{pmatrix} \times \begin{pmatrix} 0 \\ J_y \\ 0 \end{pmatrix} = \begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \end{bmatrix} \begin{pmatrix} 0 \\ 0 \\ \omega_z \end{pmatrix} $$

$$\begin{pmatrix} -(m c_x \omega_z) \ell_z \\ 0 \\ (m c_x \omega_z) (\ell_x-c_x) \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ I_z \omega_z \\ \end{pmatrix}$$

with solution $\ell_z =0$ and $\boxed{\ell_x = c_x + \frac{I_z}{m c_x}}$. Note that the value of $\ell_y$ is irrelevant since it along the force axis $\vec{J}$.

Here are some reference posts:

See relevant answer to a similar question (https://physics.stackexchange.com/a/81078/392)

The full equations of motion about an arbitrary point are derived in (https://physics.stackexchange.com/a/80449/392)

  • $\begingroup$ This is not a proof from first principles. $\endgroup$ – Asad Saeeduddin Nov 27 '13 at 1:52
  • $\begingroup$ It follows naturally from the definition of linear and angular momentum (and conservation thereof). What part you do not understand and I will try to clarify? Oh, and the provided link has a derivation from first principles. $\endgroup$ – ja72 Nov 27 '13 at 2:39
  • $\begingroup$ "To purely rotate a rigid body about an axis" (this is not what the OP asked. The OP's question asked about a single force causing a rotation about the center of mass, which is impossible) $\endgroup$ – Asad Saeeduddin Nov 27 '13 at 2:55
  • $\begingroup$ Fundamentally any instantaneous motion of a rigid body can be described as a rotation about an axis plus a translation along the axis (screw motion). I start generally with what it takes to rotate about a specific axis and then try to apply it to the situation of the OP. With my answer the result is a rotation about any arbitrary point, including the c.m. if one chooses so ($c_x=0$). $\endgroup$ – ja72 Nov 27 '13 at 13:06

Assume a very small particle embedded in the Rigid body of mass $m$. Let us find out its Torque or moment of force $\vec{\tau}$ about an arbitrary point $p$.

$\vec{\tau} = \vec{f} \times \vec{r}$

where $\vec{r}$ is a displacement of this particle from point $p$.

The total Torque on the rigid body will be some of $\tau$ of all the particles. If this $\tau$ has a non-zero value then the body will be rotating.

Lets find out the total Torque, $\Gamma$

$\Gamma = \Sigma{\tau}$

$=> \Gamma = \Sigma{ \vec{f} \times \vec{r}}$

$=> \Gamma = \Sigma{ m \, \vec{a} \times \vec{r}}$

As The body is said to be rigid, therefore all the points on this body will be having same accelerations at ever instance. Also, Cross product is distributive ref, therefore, we can take $\vec{a}$ out of summation.

$=> \Gamma = \vec{a} \times \Sigma{ m \, \vec{r}}$

now, if point $p$ is center of mass then, $\Sigma{ m \, \vec{r}}$ is zero. ref

Therefore, $\Gamma$ is zero and rigid body will not rotate at all.

NOTE: $\times$ is the vector cross product operator.

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    $\begingroup$ "As The body is said to be rigid, therefore all the points on this body will be having same accelerations at ever instance." I don't think that's quite true. $\endgroup$ – Keep these mind Mar 5 '13 at 17:34
  • $\begingroup$ It seems that you have put the conclusion to your answer in as a premise. $\endgroup$ – Keep these mind Mar 5 '13 at 18:32
  • $\begingroup$ I agree with @Gugg. If the body is rotating, every point will have an acceleration of $\omega^2r$ towards the instantaneous axis of rotation. It is truly not constant for all points on the rigid body. $\endgroup$ – udiboy1209 Jul 12 '13 at 14:27
  • $\begingroup$ Assuming the notation "$\times$" refers to the vector cross-product, torque is properly defined as $\vec{\tau}=\vec{r}\times\vec{F}\neq\vec{F}\times\vec{r}$. This does not change the qualitative aspect of your answer because we are dealing with zeros, but it does change the some of the specifics... $\endgroup$ – Bryson S. Nov 13 '14 at 19:22

I think the point is that in free space linear and angular momentum are both separately conserved quantities. (This is implied so long as your space has translational and rotational symmetry.) If the total linear momentum of the rigid body is always constant after the push, then the center of mass must be moving in a straight line, at constant velocity. From this it follows that the rotation must be about the center of mass.


One can make reasonable assumptions to investigate the problem in a simple manner. Here is my reasoning about this question.

For the sake of simplicity let us assume we have a spherical object of radius R in the outer space. Let there be a hook at the surface of the sphere from which we can attach a string. Imagine we are equipped with a rocket system that can give us momentum to move about.

Now, we hold one end of the string and move away from the sphere in a direction that the string, when it becomes taut, is not parallel to the radius of the sphere. The force we exert on the sphere in that direction can be analysed into the tangent and the perpendicular to the surface of the sphere. If $\theta$ is the angle between the string and the normal to the sphere we have:

Tangent component: $F_T=F\sin(\theta)$

Normal component: $F_N=F\cos(\theta)$.

The normal component is parallel to the radius of the sphere and passes through the centre (CM) and has no moment. This component will pull the sphere in the normal direction.

The tangent component has a moment with respect to the centre


This component would rotate the sphere, should the axis of the sphere be pivoted, but it is not! However, I believe that, due to the inertia of the mass of the sphere, it would be sufficient to give pivotal leverage for the tangent force to rotate the sphere. The law of conservation of energy must be written, for a short time interval of application of the force, in the form

${\bf {F.x}} = {\frac {1}{2}}mv^2+ {\frac {1}{2}}I{\omega}^2 $

where:$\bf x$ is the displacement of the sphere, while the first term on the RHS is the kinetic energy due to the linear motion, and the second is the kinetic energy due to the rotational motion. Note that, as the sphere has no fixed axis, it will rotate about the axis which is peprepndicular to the great circle passing through the point of the hook, and the $F_T$ is tangent to it. Hence the axis will be perpendicular to $F_T$ and $F_N$ and so it is perpendicular to the force $\bf F$. This will be the case for any direction of $\bf F$.

Why should the axis of rotation pass through the CM? The poitn here is that the object is rotating freely. Is not constrained to rotate about an arbitrary axis. Without going into mathematics, a quick argument from physics point of view is that, if the axis passed through another point, the rotational motion would be unstable. I mean that for a freely rotating object, there is a minimum state of energy, and this is when the axis of rotation passes through the CM. If it passed through some other point, then according to the parallel axis theorem, the inertia of the object would be higher, hence higher energy of the system. It is like you bring an object at a certain height near the surface of the earth and then you set it free. It will fall to the lowest energy state, and that is when it is on the ground.

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    $\begingroup$ I appreciate you writing the answer but consider removing it. This does not answer the question. You just gave an analysis of a force acting on a sphere and then you gave the expression for the kinetic and rotational energy for the sphere. That is all. $\endgroup$ – Alraxite Feb 12 '13 at 1:39
  • $\begingroup$ You didn't give any arguments for why the axis should go through the center of mass even in the special case of a sphere. $\endgroup$ – Alraxite Feb 12 '13 at 1:39
  • $\begingroup$ @Alraxite Sorry I did not have the chance to repond earlyer. Thanks for bringing to my attention the CM part of the question. I have edited my answer to include an argument about that. Please read it. $\endgroup$ – JKL Feb 12 '13 at 11:32

You can find the change in angular momentum of a rigid body by simply evaluating it: $\frac{d\mathbf{L}_{total}}{dt} = \sum_p m_p\left(\mathbf{R}+\mathbf{r}_p\right) \times \frac{d}{dt}\left(\mathbf{V}+\mathbf{v}_p\right)\,$

$\frac{d\mathbf{L}_{total}}{dt}= M \mathbf{R} \times \frac{d\mathbf{V}}{dt} + \sum_p m_p \mathbf{r}_p\times \frac{d\mathbf{v}_p}{dt}$ Here I've broken up the position of the $p^{th}$ component into a center of mass part and a relative part.

Note that $m_p \frac{d\mathbf{v}_p}{dt}$ is precisely the force acting on one part of the body. You can show that internal forces (forces between particles) don't contribute to the torque (basically due to them being equal and opposite, so they cancel when you sum), so only external forces are important.

Only the component of that force that is perpendicular to $\vec{r}_p$ survives the cross product, and sets the body rotating. In other words, the statement " it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass" is a property of the cross product in that equation. Why about the center of mass? Well, you can evaluate the angular momentum about any line (more precisely, in any plane), and it neatly factors into a part that is the motion of the COM about that axis, and a part that is the motion of the body about the COM. If you choose the axis to go through the COM then the first part vanishes by the cross product. Anyway, the calculation above factors in the same way, as you can see.

You can check out the nitty gritty here which I typed up a long time ago. Hopefully it isn't too confusing.



protected by Qmechanic Oct 26 '13 at 21:58

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