- #1

- 68

- 0

So the inertia of a rod is I

_{Stick cm}= 1/12 *m*L

^{2}

Does L

^{2}= (2L/3)

^{2}= 4L

^{2}/9 -> 1/12 *m*[4L

^{2}/9] or is it something like

m*L

^{2}/12 + m* (2L/3)

^{2}?

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

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- B
- Thread starter Blockade
- Start date

- #1

- 68

- 0

So the inertia of a rod is I

Does L

m*L

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

Picture

- #2

- 350

- 81

- #3

- 68

- 0

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

- 350

- 81

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