Question

Derive the expression for moment of inertia of the thin rod about an axis through its center and perpendicular to its length.

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Solution

Consider a rod of length 'L' and mass 'M'.
Let 'λ' be the linear mass density.

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Let a small line element dx be at a distance x from O.
Moment of inertia of this small line element of mass λdx about O is,

To find the total moment of inertia of rod, integrate from x = -L/2 to x = L/2.
I =

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integral subscript negative straight L divided by 2 end subscript superscript straight L divided by 2 end superscript lambda space x squared space d x space

equals space straight lambda space open vertical bar straight x cubed over 3 close vertical bar subscript negative straight L divided by 2 end subscript superscript straight L divided by 2 end superscript space space equals space straight lambda over 3 open square brackets open parentheses straight L over 2 close parentheses cubed minus space open parentheses fraction numerator negative straight L over denominator 2 end fraction close parentheses cubed close square brackets space equals space fraction numerator straight M over denominator 3 straight L end fraction open square brackets straight L cubed over 8 space plus space straight L cubed over 8 close square brackets equals space 1 over 12 space ML squared

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Units And Measurement

Derive the expression for moment of inertia of the thin rod about an axis through its center and perpendicular to its length.

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