Angular velocity of a rod rotating around a vertical axis

alaix

1. The problem statement, all variables and given/known data

An homogeneous rod is fixed to an extremity and is rotating around a vertical axis, doing an angle of theta with the vertical. (in the scheme I made I wrote alpha but it's theta! (θ))

If the lenght of the rod is L, show that the angular velocity needed to make it turn is

[itex]\omega[/itex] = [itex]\sqrt{3g/2L cos(θ)}[/itex]


2. Relevant equations

[itex]\tau[/itex] = r x F
I = 1/2 m L²

[itex]\tau[/itex] = I[itex]\alpha[/itex]

3. The attempt at a solution

Here is what I tried

I considered that all the exterior forces (ie. gravity) was acting on the center of mass of the rod, which is situated in the middle, at L/2.

Therefore

Torque = r x F = 1/2 L mg sin(θ)

Torque = I[itex]\alpha[/itex]

Where I = 1/3 mL²

Therefore

[itex]\alpha[/itex] = [itex]\frac{3mgL sin(θ)}{2mL^{2}}[/itex] = [itex]\frac{3g sin(θ)}{2L}[/itex]

Since I'm looking for the angular VELOCITY, and since angular acceleration = d[itex]\omega[/itex]/dθ

[itex]\alpha[/itex] dθ = d[itex]\omega[/itex]

By integrating both sides I find

[itex]\omega = -\frac{3g cos(θ)}{2L}[/itex]


Which is ALMOST the answer I'm looking for... what am I missing?

Thanks!

Attached Thumbnails

 

cepheid

There is a lot of ambiguity here as to what the final answer is supposed to be. You should include some extra parentheses under the square root sign to make it clear what's in the numerator and what's in the denominator.

No, actually that's wrong: [itex] \alpha \neq d\omega/d\theta [/itex]. Rather, [itex] \alpha = d\omega / dt [/itex].