 #1
 11
 0
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 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]
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!
Attachments

2.2 KB Views: 415