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## Homework Statement

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 length 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]

## Homework Equations

[itex]\tau[/itex] = r x F

I = 1/2 m L²

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

## 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!