# Which is the axis of rotation?

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.)

-
–  Qmechanic Jan 28 '13 at 19:54
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. –  ja72 Jan 28 '13 at 20:25

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.

-
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? –  yarnamc Jan 28 '13 at 19:30
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. –  queueoverflow Jan 28 '13 at 19:33
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). –  yarnamc Jan 28 '13 at 19:45
Only forces can accelerate the center of gravity, and thus lack of forces (pure torque) results in steady motion for the c.g. –  ja72 Jan 28 '13 at 20:28

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.

Cheers!

-
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 –  Jaime Jan 29 '13 at 4:34
@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. –  joshphysics Jan 29 '13 at 4:42
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. –  Jaime Jan 29 '13 at 4:53