torque needed to precess the angle of rotation If a rigid body with the inertia matrix $I_B$, has the angular velocity $\overrightarrow{\omega_1}$ what torque is needed to rotate it around another axis say $\overrightarrow{\omega_2}$ while keeping it's original rotation ... Sort of the torque needed to change the position of the rigid body while it rotates. A know that it's angular momentum $\overrightarrow{L_1} = I_B\overrightarrow{\omega_1}$.  I don't know how to proceed ... There will be another angular momentum while rotating about $\overrightarrow{\omega_2}$ that is $\overrightarrow{L_2} = I_B\overrightarrow{\omega_2}$ so the body will feel $\overrightarrow{L} = \overrightarrow{L_1} + \overrightarrow{L_2}$ ?! The torque needed is $\overrightarrow{\tau} = \frac{d\overrightarrow{L}}{dt}$ so a sort of $\overrightarrow{\tau} \approx \frac{\overrightarrow{L_2}}{T}$ where $T$ is some time ... ?? Or the body is rotating around it's first axis of rotation hence having the angular momentum $L = I\omega$ in it's coordinate system. Now if the coordinate system is rotated around another axis $
\omega_2$, what angular momentum will the body feel? It's inertia matrix in the rotating frame will be $I_r = R^{T}IR$ and from here the angular momentum will be $L_2 = I_r\omega_2$ ?
 A: You can think about this in two different ways. 
One way is to look at the initial and final angular momentum. If you go from $L\cdot(0,1, 0)$ to $L\cdot(1,0,0)$ you need to remove the $Y$ component and add the $X$ component. If you just calculate the difference in the angular momentum, then you get 
$$\Delta L = L\cdot (1,-1,1)$$
which would immediately imply that a torque about the $(1,-1,0)$ axis will provide the desired result.
Interestingly, the same thing can be achieved by applying a torque that is always perpendicular to the current direction of motion: this is what happens during the precession of a horizonally mounted gyroscope, for example. In that case, you might start with the axis of rotation pointing along the Y direction; and the torque generated by the force of gravity on the center of mass of the gyroscope will result in precession; after a certain time, this can result in the gyroscope pointing along the $X$ direction. Since the torque in this case keeps changing direction, you would have to integrated over all directions - and find once again that the average torque integrated over time is the same as you would have calculated before.
In both cases, the equation of motion is
$$\frac{d\vec{L}}{dt} = \vec{\Gamma}$$
Going with the first approach, the angular momentum $\vec{L}=I\vec{\omega}$, from which it follows that you need a torque $\Gamma$ for a time $t$ such that
$$\Delta L = \Gamma t$$
Which in you case means
$$\vec{\Gamma} = \frac{I\omega(1,-1,0)}{ t}$$
Can you figure it out from here?
A: I shall present my conclusion. Given the following: the world coordinate system $(i_w, j_w, k_w)$, the body's inertia matrix in this coordinate system $I_w$ and the rotation axis $\omega$ in the world coordinate system, one can obtain the angular momentum $L$ as $L = I_w \cdot \omega$. If the rotation axis is precessed in $\omega_2 = R(t)^T\cdot \omega$, with $R(t)$ a rotation matrix then the inertia matrix of the rigid body in the world coordinate system is now $I_{w2} = R^T\cdot I_w \cdot R$ and the new angular momentum is $L_2 = I_{w2}\cdot \omega_2 = R^T\cdot I_w \cdot R \cdot R^T \cdot \omega = R^T \cdot L$. Hence, $$ \tau = \lim_{t \to 0}\frac{L2 - L}{t} = \lim_{t \to 0}\frac{(R(t)^T - I_3)}{t}\cdot L = \dot{R^T}\cdot L$$ where $I_3$ is the identity matrix.
