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Homework Statement
A thin uniform rod is attached to an axis through its midpoint. The axis is standing on a disk rotating with constant angular speed [itex] \Omega [/itex] about its symmetry axis. The rod's midpoint is located directly above the rotational axis of the disk. Let [itex] \theta [/itex] denote the rod's angle with the horizontal plane. Determine how [itex] \theta [/itex] varies with time.
Pictures are attached below to make the above description clearer.
Homework Equations
Tensor of inertia, [itex] I_{\alpha \beta} = \int (r^2 \delta_{\alpha \beta} - r_{\alpha} r_{\beta}) dm [/itex]
Euler's equations for principal systems:
[itex] M_1 = I_{11} \dot{\omega}_1 - \omega_2 \omega_3 (I_{22} - I_{33}) [/itex]
[itex] M_2 = I_{22} \dot{\omega}_2 - \omega_3 \omega_1 (I_{33} - I_{11}) [/itex]
[itex] M_3 = I_{33} \dot{\omega}_3 - \omega_1 \omega_2 (I_{11} - I_{22}) [/itex]
where [itex] M_i [/itex] denotes the components of the net applied moment of force and [itex] \mathbf{\omega} = \omega_1 \mathbf{e}_1 + \omega_2 \mathbf{e}_2 + \omega_3 \mathbf{e}_3 [/itex] is the angular velocity vector, with [itex] \mathbf{e}_i [/itex] being the principal basis vectors.
The Attempt at a Solution
I have tried several things, none of which have yielded satisfactory results. This is amongst the first 3D rigid body dynamics problems I have tried to solve.
Assume an inertial Cartesian coordinate system with the xy-plane coinciding with the horizontal plane of the disk and the z-axis pointing upward perpendicular to this plane. Now introduce an accelerated reference frame constant in the rod. Let the x'-axis be perpendicular to the rod, such that it at a time t=0 is positioned directly above the x-axis with the angle between them being [itex] \theta [/itex]. In the same manner put the y'- and z'-axes, such that they make angles [itex] \theta [/itex] with the y- and z-axis respectively.
In this frame we can write [itex] \Omega \mathbf{e}_z = \Omega \sin \theta \mathbf{e}_{x'} + \Omega \cos \theta \mathbf{e}_{z'} [/itex].
I am not sure how the total angular velocity vector [itex] \omega [/itex] should look like now. I have tried both using the above and adding a term [itex] \dot{\theta} \mathbf{e}_{x'} [/itex] to it. Neither seems to give me a result I want.
Now, I could be wrong but I think this is a principal system, so the tensor of inertia is diagonal and its diagonal elements are given by
[itex] I_{x'x'} = \frac{1}{12} ml^2 [/itex]
[itex] I_{y'y'} = 0 [/itex] (I think, but the value shouldn't matter?)
[itex] I_{z'z'} = \frac{1}{12} ml^2 [/itex]
Euler's equations then gives
[itex] \frac{1}{12} ml^2 (\Omega \cos \theta \dot{\theta}+ \ddot{\theta}) = M_1 = 0. [/itex]
[itex] 0=0 [/itex]
[itex] \frac{1}{12} ml^2 (-\Omega \sin \theta \dot{\theta}) = M_3 = 0. [/itex]
Here I am not happy with what I have got, so I must have gone wrong somewhere in the above. Any hints to where I am messing up are much appreciated. Thank you.
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