Deriving Moment of Inertia

  • #1
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Homework Statement


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


Homework Equations


I = summation(Mx^2)
T restoring = k*theta = mgtheta*x
period = 2pi*root(I/k)

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) ?
 

Answers and Replies

  • #2
radou
Homework Helper
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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" [Broken].
 
Last edited by a moderator:
  • #3
312
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Thanks a lot. So the period is 2pi*root[(1/12L^2 + x^2)/(gx)]
 

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