91影视

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

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

So mass per unit length 位

$位=\frac{M}{L}$

So we can get the mass of small strip dx as

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

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

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

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

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

Both results confirms the table 12.2