Finding angular speed of a thin rod.

AI Thread Summary
A thin 1.0-meter-long rod pivoted at one end falls from a horizontal position, and the goal is to find its angular speed when it becomes vertical. The moment of inertia for the rod is given as (1/3)ML^2. The solution involves using conservation of energy principles, where kinetic energy is expressed as (1/2)Iω^2 and potential energy as mgL. The change in potential energy is determined by the movement of the center of mass, which drops from a height of L to L/2, leading to the correct calculation of angular speed. The final answer for the angular speed when the rod is vertical is 5.4 rad/s.
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Homework Statement



A thin 1.0-meter-long rod pivoted at one end falls (rotates) from a horizantal position, starting from rest and with no friction. What is the angular speed of the rod when it is vertical?



Homework Equations



I (of the thin rod) = (1/3)ML^2



The Attempt at a Solution




I tried solving this problem using conservation of energy, but I'm not getting the right answer (5.4 rad/s).

For kinetic energy, I was using (1/2)I\omega^2; and for potential energy, I was using mgL. Then I solved for \omega. What's wrong with that?
 
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In figuring out the change in PE of the rod, consider what happens to its center of mass.
 
It accelerates?
 
How does its position change? That determines the change in PE of the rod.
 
The CoM moves from a height of L to a height of L/2 ?
 
Good. So what's the change in PE of the rod?
 
MgL - Mg(L/2)

I got the answer now, thanks!
 
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