# Simple Harmonic Motion, rigid pendulum

sheepcountme

## 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.

## Answers and Replies

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.

sheepcountme
That's really weird because the book has an example with a meter stick and uses that formula for I.

Staff Emeritus