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Moment of inertia for uniform rectangular lamina is considered one the most difficult concept.
5 Questions around this concept.
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point ) is:
Let $I_{y y}$=Moment of inertia for uniform rectangular lamina about y- axis passing through its centre .
To calculate $I_{y y}$
Consider a uniform rectangular lamina of length l, and breadth b and mass M
mass per unit Area of rectangular lamina $=\sigma=\frac{M}{A}=\frac{M}{l * b}$
Take a small element of mass dm with length dx at a distance x from the y-axis as shown in figure.
$
\begin{aligned}
& d m=\sigma d A=\sigma(b d x) \\
& \Rightarrow d I=x^2 d m
\end{aligned}
$
Now integrate this dl between the limits $\frac{-l}{2}$ to $\frac{l}{2}$
$
I_{y y}=\int d I=\int x^2 d m=\int_{\frac{-l}{2}}^{\frac{l}{2}} \frac{M}{l b} x^2 *(b) d x=\frac{M}{l} \int_{\frac{-l}{2}}^{\frac{l}{2}} x^2 d x=\frac{M l^2}{12}
$
Similarly
Let $I_{x x}=$ Moment of inertia for uniform rectangular lamina about x - axis passing through its centre
To calculate $I_{x x}$
Take a small element of mass dm with length dx at a distance x from the x-axis as shown in figure.
mass per unit Area of rectangular lamina $=\sigma=\frac{M}{A}=\frac{M}{l * b}$
$
\begin{aligned}
& d m=\sigma d A=\sigma(l d x) \\
& \Rightarrow d I=x^2 d m
\end{aligned}
$
Now integrate this dl between the limits $\frac{-b}{2}$ to $\frac{b}{2}$
$
I_{x x}=\int d I=\int x^2 d m=\int_{\frac{-b}{2}}^{\frac{b}{2}} \frac{M}{l b} x^2 *(l) d x=\frac{M}{b} \int_{\frac{-b}{2}}^{\frac{b}{2}} x^2 d x=\frac{M b^2}{12}
$
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