Calculate by direct integration the moment of inertia for a thin rod of mass M and length L about an axis located distance d from one end. Confirm that your answer agrees with Table 12.2 when d = 0 and when d = L /2.

Mass of rod = M and

length=L ,

rotating about an axis perpendicular to its length and passing at a distance d from one end,

Lets draw the diagram as below to understand and solve the problem

Consider a very small strip dx , will have mass dm

Moment of inertia is given by

$I=\int x^{2}dm..............................\left(1\right)$

So mass per unit length λ

$\lambda =\frac{M}{L}$

So we can get the mass of small strip dx as

$dm=\lambda dx=\frac{M}{L}dx........................\left(2\right)$

From equation(2) and (1), we get Inertia

$$\begin{gathered} I={\int }_{L-d}{x}^{2}dm \\ ={\int }_{-d}^L{x}^2\frac{M}{L}dx \end{gathered}$$

integrate

$I=\frac{M}{3}\left\{L^2-3Ld+3d^2\right\}..................................\left(3\right)$

Get inertia by substituting values

For end substitute d=0, we get

$I_{end}=\frac{M{L}^2}{3}$

For center substitute d=L/2

$$\begin{gathered} I_{\text{centre}}=\frac{M}{3}\left\{L^2-3L\times \frac{L}{2}+3{\left(\frac{L}{2}\right)}^2\right\} \\ {I}_{center}=\frac{M}{3}\left\{L^2-\frac{3}{2}{L}^2+\frac{3}{4}{L}^2\right\} \\ {I}_{\text{centre}}=\frac{M{L}^2}{12} \end{gathered}$$

Both results confirms the table 12.2