Simple Harmonic Motion, rigid pendulum

  • #1
sheepcountme
80
1

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[tex]\pi[/tex][tex]\sqrt{}[/tex]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 cancelled 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[tex]\pi[/tex][tex]\sqrt{}[/tex]((1/3)/gh)

And then I solved for h getting .493 meters but my book says this is incorrect.
 

Answers and Replies

  • #2
vela
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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.
 
  • #3
sheepcountme
80
1
That's really weird because the book has an example with a meter stick and uses that formula for I.
 
  • #4
vela
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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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