91Ó°ÊÓ

A rectangular plate with length and width = L

Axis of rotation passing through center and perpendicular to plane

Lets assume a strip of small width say dx has mass of dm as in figure below

dm = σ x L x dx ............................(1)

where σ is mass per unit area and L is length and dx is width of trip

$\mathrm{σ}=\frac{M}{L^2}$

Substitute in equation(1) we get

$$\begin{gathered} dm=\frac{M}{L^2}×L×dx \\ dm=\frac{M}{L}dx..........................\left(2\right) \end{gathered}$$

So the moment of inertia of this strip about its central axis perpendicular length,

$d{I}_C=\frac{dm×l^2}{12}$

Substitute the value of dm from equation (2), we get

$$\begin{gathered} d{I}_C=\frac{\left({\displaystyle \frac{M}{L}}dx\right)×L^2}{12} \\ d{I}_C=\frac{ML}{12}dx \end{gathered}$$

Using the parallel axis theorem, find the moment of inertia of this strip about the given axis of rotation

$$\begin{gathered} dI=d{I}_c+dm×x^2 \\ =\frac{ML}{12}dx+\frac{M}{L}{x}^{2}dx \\ =\frac{ML}{12}\left(1+\frac{12}{L^2}{x}^2\right)dx \end{gathered}$$

Integrate this to get the inertia of the plate

localid="1649147922795" $$\begin{gathered} I=∫dI \\ =\frac{ML}{12}{∫}_{-\frac{L}{2}}^{+L_2}\left(1+\frac{12}{L_2}{x}^2\right)dx \\ I=\frac{M{L}^2}{6} \end{gathered}$$