How can I find the inertia of a rod with a pivot point?

  • #1
68
0

Main Question or Discussion Point

I have a question on finding the inertia of a rod with a pivot point somewhere along it's length.

So the inertia of a rod is IStick cm = 1/12 *m*L2
Does L2 = (2L/3)2 = 4L2/9 -> 1/12 *m*[4L2/9] or is it something like
m*L2/12 + m* (2L/3)2?

If it's the second one then may I ask why it's that?

Picture
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Answers and Replies

  • #2
348
80
Revise parallel axis theorem; I = Ic + (md^2) Ic is MI about CM and d is the distance between parallel axes passing through CM and pivot. It can be proved first and then can be used.
 
  • #3
68
0
Revise parallel axis theorem; I = Ic + (md^2) Ic is MI about CM and d is the distance between parallel axes passing through CM and pivot. It can be proved first and then can be used.
so m*L2/12 + m* (2L/3)2
where Ic = m*L2/12
and md^2 = m* (2L/3)2


when you said "MI about CM" you mean the mass around the center mass which is the entire rod? As for d is the "distance between parallel axes passing through CM and pivot" meaning the length between the end of the stick pass the center mass where it ends at the pivot?
 
  • #4
348
80
center of mass is at the mid point of rod at L/2 from one end and if pivot is at 2L/3 from one end then d = (2L/3) - (L/2) = L/6. Ic is the MI about the axis passing through the point at L/2 from one end.
 

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