# Simple Harmonic Motion, rigid pendulum

## 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 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$$\pi$$$$\sqrt{}$$((1/3)/gh)

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

vela
Staff Emeritus
Homework Helper
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.

vela
Staff Emeritus