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Units And Measurement

Question
CBSEENPH11017511

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

Solution
Consider a rod of length L and mass M.
Let λ be the linear mass density.
therefore space space space space space straight lambda space equals space straight M over straight L

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,

dl = λdx x2 

To find the total moment of inertia of rod, integrate from x = 0 to x = L
straight I space equals integral dI equals integral from 0 to straight I of λx squared dx
space space equals straight lambda open vertical bar straight x cubed over 3 close vertical bar subscript 0 superscript straight I space

space space equals fraction numerator straight M over denominator 3 straight L end fraction open square brackets straight L cubed minus 0 cubed close square brackets space

space space space equals 1 third ML squared