# Deriving Moment of Inertia

by turdferguson
Tags: deriving, inertia, moment
 P: 312 1. The problem statement, all variables and given/known data A uniform rod of mass M and length L is free to rotate about a horizontal axis perpendicular to the rod and through one end. A) Find the period of oscillation for small angular displacements. B) Find the period if the axis is a distance x from the center of mass 2. Relevant equations I = summation(Mx^2) T restoring = k*theta = mgtheta*x period = 2pi*root(I/k) 3. The attempt at a solution The first part is no problem. I = 1/3ML^2 and the restoring constant is LMg/2. T = 2pi*root(2L/3G) For the second part, I know that mgtheta acts at the distance x, but how do I derive moment of inertia for an axis x distance from the com? I know it changes with the axis and it has to fall inbetween 1/3ML^2 and 1/12ML^2. But Im not familliar with the integration that goes into deriving I. Should I leave it as 2pi*root(newI/xMg) ?
HW Helper
P: 3,224
 Quote by turdferguson For the second part, I know that mgtheta acts at the distance x, but how do I derive moment of inertia for an axis x distance from the com?
This should help: http://en.wikipedia.org/wiki/Parallel_axis_theorem.
 P: 312 Thanks a lot. So the period is 2pi*root[(1/12L^2 + x^2)/(gx)]

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