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Moment Of Inertia Of A Rectangular Plate - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Moment of inertia for uniform rectangular lamina is considered one the most difficult concept.

  • 5 Questions around this concept.

Solve by difficulty

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:

Concepts Covered - 1

Moment of inertia for uniform rectangular lamina

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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Moment of inertia for uniform rectangular lamina

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