This should be simple, but it keeps bothering me. If a rigid body has no fixed axis, and a torque (defined relative to a point $A$) is applied, it will rotate around $A$. But often I can also calculate the torque relative to another point $B$ (which often seems to be non-zero too). So does this mean that the rigid body will have an angular acceleration about both axises? This seems a bit strange to me.

(For a fixed axis I assume that a rotation around any axis (other than the fixed axis) is impossible, because there will always be zero torque around those axes.)

  • $\begingroup$ Related: Two axes for rotational motion and links therein. $\endgroup$
    – Qmechanic
    Jan 28, 2013 at 19:54
  • $\begingroup$ A rigidbody under the influence of a pure torque will rotate about the center of gravity, always. So point $A$ has to be your c.g. in the situation above. $\endgroup$ Jan 28, 2013 at 20:25

4 Answers 4


Given a rigid body, a basic result in mechanics is that there exists a family of time-dependent rotations $R(t)$ such that for a given a fixed point $\vec x_0(t)$ in the body and for each point $\vec x_\alpha(t)$ in the body, one has $$ \vec x_\alpha(t) = \vec x_0(t) + R(t)(\vec x_\alpha(0) - \vec x_0(0)) $$ In words, this equation says that the motion of a rigid body can be described by a translation of any chosen fixed point, plus a rotation about that point. For each time $t$, the axis of rotation of $R(t)$ defines the axis of rotation of the rigid body. Notice that this rotation does not depend on the reference point fixed in the body that we choose. We could pick either the center of mass, or any other point in the body, but the description will remain the same. So, as far as I can tell, the question of "which point the body is rotating about" does not have a unique answer; the answer depends on how you choose to describe its motion, namely which fixed point $\vec x_0(t)$ you have chosen.


  • $\begingroup$ While your anlysys is correct, the canonical answer to "which point the body is rotating about" would be that it is rotating around the instant centre of rotation: en.wikipedia.org/wiki/Instant_centre_of_rotation $\endgroup$
    – Jaime
    Jan 29, 2013 at 4:34
  • $\begingroup$ @Jaime If you read this article carefully, you'll notice that the instant centre of rotation refers to a point in a "body undergoing planar movement." As far as I am aware, the generalization for a rigid body moving arbitrarily in three dimensions is the instantaneous axis of rotation which is determined uniquely by the rotation $R(t)$ and points along the direction of the angular velocity vector $\vec\omega$. Thanks for the comment; I was unaware of this terminology. $\endgroup$ Jan 29, 2013 at 4:42
  • $\begingroup$ Of course in three dimensions things don't rotate around a point, but around an axis. Still there always is a well defined, instantaneously stationary line, around which the object is rotating at any given moment. $\endgroup$
    – Jaime
    Jan 29, 2013 at 4:53

It will not rotate around A, since the centrifugal forces will make the body rotate about its center of mass. You can move the torque, it is not fixed to a given point.

  • $\begingroup$ So a body without fixed axis can only rotate around his center of mass? But when there is no net torque around the center of mass, but there is torque around another point, won't it rotate around that point? $\endgroup$
    – yarnamc
    Jan 28, 2013 at 19:30
  • $\begingroup$ How do you apply torque around a point? You fix some point and apply a tangential force somewhere else. But then, you fixed an axis! If you do not fix anything, you will need to apply some tangential force on the left side (say up) and some down force on the right side. Then it will rotate about its center of mass. $\endgroup$ Jan 28, 2013 at 19:33
  • 2
    $\begingroup$ Aha, I think that i start to understand; so to summarize, if a body has no fixed axis, it can only rotate around its center of mass. So if the torque about the center of mass is zero, it won't rotate (and will be in equilibrium if the forces in x, y and z direction sum to zero). $\endgroup$
    – yarnamc
    Jan 28, 2013 at 19:45
  • $\begingroup$ Only forces can accelerate the center of gravity, and thus lack of forces (pure torque) results in steady motion for the c.g. $\endgroup$ Jan 28, 2013 at 20:28
  • $\begingroup$ One can observe the motion of a rigid body from any point on it. And from the frame of any such point, the body would appear to be performing a purely rotational motion about that point itself. So 'it will not roatate about A' seems to be an incorrect statement to me. Please point out the mistake in my arguement. $\endgroup$ May 12, 2020 at 10:14

There is another issue that must be clarified before making any definite conclusion. The question is:"Is rigid body in free state or is it constrained"? This question must be asked when dealing with rotors constrained by gravity in the bearings, constituting a rotor/ bearing system. Then next question arises:"Does this system has a resonance (natural motion) at particular angular velocity"? The experiments showed that rotor will rotate around the axis(natural motion), at which torque is applied, and mass axis (of the eccentric rotor) will precess around torque (forced centroidal axis). Above the resonance velocity, natural tendency of rotating body is to rotate around its mass axis, and torque applied axis will precess now around rotor mass axis. When this happens on rotating machines, it generates "vibrations" at constraints , i.e. the bearings.

  • $\begingroup$ Welcome to Physics.SE! I suggest the following: 1) Take the tour! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. $\endgroup$
    – auden
    Jun 26, 2016 at 15:02

In an inertial frame, Rotation is always about the Centre of Mass or Barycentre, If no constraints such a fulcrum's equivalent is present

If the choice of frame is not a matter of concern, You can take a different axis and you will get the exact same analysis if you account for pseudo forces and moment caused by that pseudo force.

Take C to be the centre of mass. B being another point and describe angular kinematic variables $\omega$ and $\alpha$

Now B goes with linear acceleration $\alpha r_B$.

Take B to be FIXED, ie apply $-\alpha r dm$ Everywhere in the body

this will give an effective torque which if you divide by I at B, will give you same $\alpha$ as before.

Now you can take any axis, but not forget pseudo force. At non-inertial frame rotation happens around Centre of Mass though.


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