# Disk let free to rotate

A rigid body moving with no constraints, in particular rotating, will rotate necessarily about a principal axis of inertia.

I thought that the reason of this is that otherwise, the angular momentum $\vec{L}$ would not be parallel to the angular velocity $\vec{\omega}$, hence it would follow a precession motion and that would imply the presence of a torque, which is not present if the body is let free.

But then I came up with this example.

A disk rotates about an axis $\hat{c}$, which is not parallel to a principal axis of inertia. The supports exert the torque necessary for this rotation. At $t_0$ the supports break. The angular momentum vector $\vec{L(t_0)}$ of the disk cannot change anymore, but how will the disk rotate? (In the picture there is the disk before and after supports break).

On textbook it is stated that disk starts rotating about its principal axis of inertia $\hat{u}$ passing through its center of mass with angular velocity $\vec{\omega_R}$ but it also follows a precession motion about the (constant) angular momentum vector, with angular velocity $\vec{\omega_P}$.

Why does this happen? A variation of $\vec{\omega}$ (in direction) means an angular acceleration, which in turns imply the presence of a torque. But there is no torque here, the disk is free.

On the other hand of course $\vec{L}$ cannot change, and it is not parallel to a principal axis of inertia.

So how can $\vec{\omega}$ change? And does a rigid body always rotate about a principal axis of inertia, no matter if the angular momentum is parallel to it or not?

• – David Hammen Apr 15 '16 at 15:15

A rigid body moving with no constraints, in particular rotating, will rotate necessarily about a principal axis of inertia.

I thought that the reason of this is that otherwise, the angular momentum $$\vec L$$ would not be parallel to the angular velocity $$\vec\omega$$, hence it would follow a precession motion and that would imply the presence of a torque, which is not present if the body is let free.

This is incorrect. A rigid body rotating in torque-free conditions can have any angular momentum. A torque-free precession results if this angular momentum happens to not be aligned with one of the principal axes.

The axis of rotation changing even when there is no applied torque is an effect which is possible in 3-dimensions but not in 2-dimensions.

The angular momentum about an axis $\vec L$ is given by $\vec L = I \vec \omega$ where $I$ is the moment of inertia about the axis and $\vec \omega$ is the angular velocity about the axis.

If you are observing an asymmetric body, like the disc, with total angular momentum $\vec L$ then if there are no forces (and hence no torques) acting on the body the total angular momentum of the body is constant.

In a frame of reference fixed relative to the centre of mass of the body the sum of the x, y and z components of the angular momentum will be constant.

$\vec L = \vec L_x + \vec L_y +\vec L_z$.

Because the body is asymmetric as it rotates relative to the frame of reference its moment of inertia about an axis will change eg from $I_x$ to $I_x^\prime$ and this in turn will change the angular velocity about that axis from $\omega_x$ to $\omega_x^\prime$ subject to the constraint that the angular momentum in the x-direction, $L_x = I_x \omega_x$ to $I_x^ \prime \omega_x^\prime$, stays constant.

This means that what you observe as the rotation of the object is the initial angular velocity, $\vec \omega = \vec \omega_x + \vec \omega_y +\vec \omega_z$, changing to a new angular velocity, $\vec \omega^\prime = \vec \omega_x^\prime + \vec \omega_y^\prime +\vec \omega_z^\prime$.
So you observe a change in the direction of the axis of rotation although the total angular momentum has not changed.

A full analysis requires one to keep the total kinetic energy of the body, $\frac{1}{2}I \omega^2$, constant.

See Poinsot's_ellipsoid and the references listed in the article although the wonderful "Poinsot construction in stereo 3D simulation" seems no longer to be available.