Deriving Moment of Inertia


by turdferguson
Tags: deriving, inertia, moment
turdferguson
turdferguson is offline
#1
Dec7-06, 06:01 PM
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) ?
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radou
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#2
Dec7-06, 06:14 PM
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Quote Quote by turdferguson View Post
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.
turdferguson
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#3
Dec7-06, 06:24 PM
P: 312
Thanks a lot. So the period is 2pi*root[(1/12L^2 + x^2)/(gx)]


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