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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 $L_2 = I_r\omega_2$ ?

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  • $\begingroup$ $T$ is transpose... You may want to check Euler equation for rigid body at Wikipedia. This is a simple application of Euler equation $\endgroup$ – RoderickLee Jul 13 '16 at 21:33
  • $\begingroup$ T measures seconds if correct ... You mean Newton-Euler equations ? I think I need some guidelines thou ... just a little, to see how it applies. $\endgroup$ – C Marius Jul 13 '16 at 21:37
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    $\begingroup$ You are using the symbol $T$ twice: once as "transpose", and once to indicate "time during which torque is applied". That is mighty confusing. $\endgroup$ – Floris Jul 13 '16 at 23:08
  • $\begingroup$ The conclusion from your last equation should be that the amount of torque required depends on how quickly you want to change the angular momentum, as Floris' answer shows. First decide how quickly (ie choose a value for T), then you can calculate the torque $\vec \tau$. Same as deciding what force is required to change linear momentum. $\endgroup$ – sammy gerbil Jul 14 '16 at 11:48
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    $\begingroup$ I think that the correct physics words are "torque needed to precess the angle of rotation" as @Floris figured out :). I modified the title too $\endgroup$ – C Marius Jul 15 '16 at 8:37
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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?

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    $\begingroup$ Your last eqn does not seem to be correct. It implies the longer the time you apply the torque, the larger the torque which is required to cause the change in angular momentum. $\endgroup$ – sammy gerbil Jul 14 '16 at 0:49
  • $\begingroup$ @sammygerbil you are right - I think I fixed it. $\endgroup$ – Floris Jul 14 '16 at 2:18
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    $\begingroup$ What is the final angular momentum? 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 $L_2 = I_r\omega_2$ ?? $R$ is a rotation matrix which depends on $\omega$ $\endgroup$ – C Marius Jul 14 '16 at 16:15
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    $\begingroup$ Maybe I am misunderstanding your question. Do you want the object rotate about a different axis (in a wheel, you you want to end up rotating about a spoke, for example?) or do you want to precess the angle of rotation? $\endgroup$ – Floris Jul 14 '16 at 16:18
  • $\begingroup$ I want to precess the angle of rotation. Sorry for the poor words $\endgroup$ – C Marius Jul 15 '16 at 7:53
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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.

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