Angular velocity of a rod rotating around a vertical axis

by alaix
Tags: angular, axis, rotating, velocity, vertical
 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 $\omega$ = $\sqrt{3g/2L cos(θ)}$ 2. Relevant equations $\tau$ = r x F I = 1/2 m LČ $\tau$ = I$\alpha$ 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$\alpha$ Where I = 1/3 mLČ Therefore $\alpha$ = $\frac{3mgL sin(θ)}{2mL^{2}}$ = $\frac{3g sin(θ)}{2L}$ Since I'm looking for the angular VELOCITY, and since angular acceleration = d$\omega$/dθ $\alpha$ dθ = d$\omega$ By integrating both sides I find $\omega = -\frac{3g cos(θ)}{2L}$ Which is ALMOST the answer I'm looking for... what am I missing? Thanks! Attached Thumbnails
Emeritus
 Quote by 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 $\omega$ = $\sqrt{3g/2L cos(θ)}$
 Quote by alaix 2. Relevant equations Since I'm looking for the angular VELOCITY, and since angular acceleration = d$\omega$/dθ $\alpha$ dθ = d$\omega$
No, actually that's wrong: $\alpha \neq d\omega/d\theta$. Rather, $\alpha = d\omega / dt$.