Rod rotating with moment of inertia at angles? HUH?

[JamesJames]

A slender rod of length l and mass m rotates with constant angular verlocity omega about an axis which is at an angle alpha to the rod and pases through its centre. Obtain an expression for the torque acting on the rod and get eth equation of motion of the rod once it is released from its constraint at the centre. T = (1/24)ml^2*omega^2*sin(2*alpha)

Here is what I have: Just replace theta by alpha. I chose the coordinate axes to be the principal axes so the cross terms in the moment of inertia vanish. I get T = (Iyy-Izz)*omega^2*sin(2*alpha) as you can see from the attachment.

Can someone help me to figure out the values of Izz and Iyy? I would really find it vey useful if I could solve this question. Thanks, James.

 

[JamesJames]

the axis on the left is x, on the right commin out of the page is y and along the bottle/rod is z. omega is shown at the angle aplha = theta as required.

 

[deda]

Mind if I ask what is moment of inertia?
Is it some ratio between the mass and its radius from the center of rotation?

 

[JamesJames]

it is m*l^2...but for various orientations it gets a bit tricky..that is where I am stuck..really anything to help me out would be great.

 

[JamesJames]

Come on guys something...I have done most of it and am confident of what I have ...I juse need help to finish the computation...anything!

 

[JamesJames]

Can anyone help?

 

[JamesJames]

Trust me you cannot go wrong with a suggestion

 

[JamesJames]

Just a hint maybe?

 

[e(ho0n3]

I'm going to reply to this since I found the problem interesting. Let O be the point where the rod and the rotation axis meet. Chop up the rod into little pieces of length dr of mass dm each. The angular momentum of each little piece is

\vec{l} = \vec{r} \times dm\vec{v}​

where

\vec{v} = \vec{r} \times \vec{\omega} \rightarrow v = r\omega\sin{\alpha}​

and so

l = rv dm = r^2\omega\sin{\alpha}\cdot{dm}.​

Taking the integral gives us the total angular momentum of the rod.

\int_m{r^2\omega\sin{\alpha}\cdot{dm} = \int_{-l/2}^{l/2}{mr^2/l\omega\sin{\alpha}\cdot{dr}}​

since dm = m/l\cdot{dr}. Simplifying gives L = ml^2\omega/12\sin{\alpha}.

At any given instance, \vec{L} is always changing direction (in fact, L is perpendicular to the rod and rotates about the axis of rotation of the rod) but will always make an angle of \pi/2 - \alpha with the rotation axis. By looking at the rotation from directly above the axis of rotation we see that dL = L\omega\cos{\alpha}\cdot{dt}. The net torque then is

\tau = \frac{dL}{dt} = L\omega\cos{\alpha} = \frac{ml^2\omega^2}{24}\sin{2\alpha}.​

As for the second part of the question involving the motion of the rod once it's free is still puzzling me. I guess since the center of mass of the rod isn't moving to begin with, the rod won't be going anywhere.