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Angular velocity of a rod rotating around a vertical axis 
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#1
Nov1811, 07:11 PM

P: 11

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 LČ [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 mLČ 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! 


#2
Nov1811, 08:48 PM

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