Moment Of Inertia Of A Rectangular Plate MCQ - Practice Questions with Answers

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 the 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 $\frac{l}{2}$

$$
I_{y y}=\int d I=\int x^2 d m=\int_{\frac{-I}{2}}^{\frac{1}{2}} \frac{M}{l b} x^2 *(b) d x=\frac{M}{l} \int_{\frac{-I}{2}}^{\frac{1}{2}} x^2 d x=\frac{M l^2}{12}
$$

Similarly
Let $I_{x x}=$ Moment of inertia for uniform rectangular lamina about $x$ - axis passing through its center.
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 the 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 $\frac{b}{2}$

$$
I_{x x}=\int d I=\int x^2 d m=\int_{\frac{-b}{2}}^{\frac{b}{2}} \frac{M}{l b} x^2 *(l) d x=\frac{M}{b} \int_{\frac{-b}{2}}^{\frac{b}{2}} x^2 d x=\frac{M b^2}{12}
$$