Shortest period for a physical pendulum

AI Thread Summary
The discussion centers on determining the shortest period of oscillation for a physical pendulum, specifically a meter stick pivoted at various points. The equation T = 2π√(I/mgh) is referenced to analyze the relationship between the pivot point and the period. Participants suggest that a shorter distance from the pivot should yield a shorter period, yet the conclusion points to a pivot at 0.3m as providing the shortest period. The conversation highlights confusion around why this specific distance results in the shortest oscillation time. Understanding the moment of inertia and the effect of pivot location is crucial for solving the problem accurately.
drunknfox
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Homework Statement


meter stick pivoted at point a from its center and swings as a physical pendulum. At which values of a gives you the shortest period of oscillation...1m, .2m, .3m, .4m. .5m


Homework Equations


T = 2pi*Sqrt(I/mgh)


The Attempt at a Solution


the shortest a should give you the shortest period. The longer the distance = longer period but the answer is .3...why?
 
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welcome to pf!

hi drunknfox! welcome to pf! :smile:

(have a square-root: √ :wink:)
drunknfox said:
the shortest a should give you the shortest period. The longer the distance = longer period but the answer is .3...why?

erm :redface: … you tell us!

start by writing out the relevant equations :smile:​
 
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