I Trying to rotate a disc about two perpendicular axes

[Kashmir]

I've a disc which can rotate freely about two perpendicular axis (fixed to the body)
If I simultaneous try to rotate it about the two axis, what will happen?

 

[ergospherical]

What you have drawn is rotation about an axis parallel to $\mathbf{i} \omega_x + \mathbf{j} \omega_y$.

 

[Lnewqban]

Please, see:
https://en.m.wikipedia.org/wiki/Precession

 

[kuruman]

I've a disc which can rotate freely about two perpendicular axis.
If I simultaneous try to rotate it about the two axis, what will happen?View attachment 285546

Nothing will happen that is not already familiar to you. If the y-axis is perpendicular to the plane of the ecliptic and the x-axis in that plane, then, ignoring nutation, the Earth spins about its axis with angular velocity $\vec \omega=\omega_0\{\sin(23.5^o),~\cos(23.5^o)\}$ where $\omega_0=\frac{2\pi}{86400}~\text{rad/s}.$

On edit: This assumes that the axes are not attached to the body.

 

[Dale]

I've a disc which can rotate freely about two perpendicular axis

Are the axes attached to the body or are the axes fixed in an inertial frame?

 

[Kashmir]

Are the axes attached to the body or are the axes fixed in an inertial frame?

They are attached to the body

 

[Dale]

They are attached to the body

I think that all of the answers so far assumed that they were fixed in space. So I don't think that the previous answers apply. The question was unclear. Unfortunately, I don't know the answer to your actual question which is quite a bit more complicated. You still have a torque about a single axis at 45 degrees from the orthogonal axes, but as that axis moves with the body the resulting motion is complicated.

 

[ergospherical]

Mine's still fine! It's also actually more common in rigid body dynamics to refer the angular velocity to body fixed axes.

 

[Dale]

Excellent, maybe I was the only one confused then!

 

[Kashmir]

Mine's still fine! It's also actually more common in rigid body dynamics to refer the angular velocity to body fixed axes.

Thank you. So you mean that the body will rotate around the tilted axis?

 

[ergospherical]

Well not really, the angular velocity is $\boldsymbol{\omega} = \mathbf{i} \omega_x + \mathbf{j} \omega_y$ but $\mathbf{i}(t)$ and $\mathbf{j}(t)$ are functions of time so the motion looks different in the space frame.

The usual procedure is to choose as $\{ \mathbf{i}, \mathbf{j}, \mathbf{k} \}$ the eigenvectors of the inertia tensor $I$ at a convenient point (the centre of mass, say). Then, manipulation of $\dfrac{d\mathbf{L}}{dt} = \mathbf{M}$ gets you to the Euler equation $I \dfrac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega} \times (I \boldsymbol{\omega}) = \mathbf{M}$ which you can solve in the eigenvector basis and then finally convert to a space fixed basis.

 

[Kashmir]

Well not really, the angular velocity is $\boldsymbol{\omega} = \mathbf{i} \omega_x + \mathbf{j} \omega_y$ but $\mathbf{i}(t)$ and $\mathbf{j}(t)$ are functions of time so the motion looks different in the space frame.

The usual procedure is to choose as $\{ \mathbf{i}, \mathbf{j}, \mathbf{k} \}$ the eigenvectors of the inertia tensor $I$ at a convenient point (the centre of mass, say). Then, manipulation of $\dfrac{d\mathbf{L}}{dt} = \mathbf{M}$ gets you to the Euler equation $I \dfrac{d\boldsymbol{\omega}}{dt} + \boldsymbol{\omega} \times (I \boldsymbol{\omega}) = \mathbf{M}$ which you can solve in the body fixed basis and then finally convert to a space fixed basis.

Thank you. I've not done Euler equation and will be doing them soon. Suppose instead we focus on the instantaneous movement of the body at the instant the torques are applied as I've drawn, then at that instant can I say the body is instantaneously rotating about the tilted axis?

 

[ergospherical]

Yeah, that's fine!

By the way, if you haven't learned about the Euler equations then you can still make progress. In the eigenvector basis $I$ is diagonal so $\mathbf{L} = I \boldsymbol{\omega} = a \mathbf{i} \times \dfrac{d\mathbf{i}}{dt} + b(\boldsymbol{\omega} \cdot \mathbf{i}) \mathbf{i}$ for $a, b \in \mathbb{R}$ being the diagonal components, so\begin{align*}
a \mathbf{i} \times \dfrac{d^2\mathbf{i}}{dt^2} + b \left( \dfrac{d(\boldsymbol{\omega} \cdot \mathbf{i})}{dt} \mathbf{i} + (\boldsymbol{\omega} \cdot \mathbf{i}) \dfrac{d\mathbf{i}}{dt} \right) = \mathbf{M}
\end{align*}If you say what is $\mathbf{M}$ then you can go further.

 

[Kashmir]

Yeah, that's fine!

By the way, if you haven't learned about the Euler equations then you can still make progress. In the eigenvector basis $I$ is diagonal so $\mathbf{L} = I \boldsymbol{\omega} = a \mathbf{i} \times \dfrac{d\mathbf{i}}{dt} + b(\boldsymbol{\omega} \cdot \mathbf{i}) \mathbf{i}$ for $a, b \in \mathbb{R}$ being the diagonal components, so\begin{align*}
a \mathbf{i} \times \dfrac{d^2\mathbf{i}}{dt^2} + b \left( \dfrac{d(\boldsymbol{\omega} \cdot \mathbf{i})}{dt} \mathbf{i} + (\boldsymbol{\omega} \cdot \mathbf{i}) \dfrac{d\mathbf{i}}{dt} \right) = \mathbf{M}
\end{align*}If you say what is $\mathbf{M}$ then you can go further.

What's M ? And is there any way to upvote etc to show thanks?

 

[ergospherical]

$\mathbf{M}$ is the moment applied the disc, taken about the same point as the inertia tensor (for your example that's definitely most conveniently chosen as the centre).

 

[Kashmir]

$\mathbf{M}$ is the moment applied the disc, taken about the same point as the inertia tensor (for your example that's definitely most conveniently chosen as the centre).

Thank you million times dear. I'll go through all of your equations and try to derive them. :)