Finding angular speed of a thin rod.

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Homework Help Overview

The problem involves a thin rod pivoted at one end, which falls from a horizontal position to a vertical position. The objective is to determine the angular speed of the rod when it reaches the vertical orientation, utilizing concepts from rotational dynamics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply conservation of energy to find the angular speed but encounters difficulties with their calculations. Some participants suggest examining the change in potential energy related to the center of mass of the rod.

Discussion Status

The discussion has progressed with participants providing guidance on how to analyze the potential energy change as the rod rotates. The original poster has indicated they arrived at an answer after considering the change in height of the center of mass.

Contextual Notes

Participants are discussing the implications of the rod's pivot point and the height change of its center of mass during the rotation. There is a focus on ensuring the correct application of energy conservation principles in the context of rotational motion.

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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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