Simple Harmonic Motion, rigid pendulum

AI Thread Summary
To determine the pivot point for a meter stick acting as a physical pendulum with a period of 1.65 seconds, the moment of inertia must be accurately calculated. The initial attempt used the incorrect formula for moment of inertia, which applies to a rod pivoted at one end, rather than at a point along its length. The correct approach involves using the parallel-axis theorem to find the moment of inertia relative to the new pivot point. This adjustment leads to a different calculation for the distance from the center of mass. Accurate application of these principles is essential for solving the problem correctly.
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Homework Statement



A small hole is drilled in a meter stick is to act as a pivot. The meter stick swings in a short arc as a physical pendulum. How far from the center of mass should the pivot point be for a period of 1.65 seconds?

Homework Equations



Moment of inertia as given in book: I=(1/3)ml^2
T=2\pi\sqrt{}I/mgh
(where t is period, I is moment of inertia, m is mass g is gravity of course, and h is the distance from the pivot oint to the center of mass)

The Attempt at a Solution


I substituted the I in the second equation. The masses canceled each other out in the square root, l is 1 since it is a meter stick and I used the given T so I end up with the equation...

1.65=2\pi\sqrt{}((1/3)/gh)

And then I solved for h getting .493 meters but my book says this is incorrect.
 
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The formula you used for the moment of inertia is wrong. The expression you used is for a long rod pivoted at the end, but that's not the case here. You'll want to use the parallel-axis theorem to get the correct expression.
 
That's really weird because the book has an example with a meter stick and uses that formula for I.
 
In the example, the meter stick is probably pivoted at its end. That's not what's happening in this problem, so you can't use that expression for I.
 
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