Question

# Consider a thin uniform square sheet made of a rigid material. If its side is “a”, mass “m” and moment of inertia “I” about its diagonals, then:(A) I $=\dfrac{m{{a}^{2}}}{12}$ (B) I $=\dfrac{m{{a}^{2}}}{24}$ (C) I $>\dfrac{m{{a}^{2}}12}{{}}$ (D) $\dfrac{m{{a}^{2}}}{24}$$<I<$$\dfrac{m{{a}^{2}}}{12}$

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Hint: To calculate the moment of inertia of a thin uniform square sheet, we can divide the square into two triangular laminas.
Now the moment of inertia of triangular lamina is
$=\dfrac{1}{2}\text{M}{{\text{a}}^{2}}$ I
Here taking M $=\dfrac{\text{m}}{2}$ and a $=\dfrac{\text{a}}{\sqrt{2}}$ , we can find moment of inertia of triangular lamina and by multiplying it by 2, moment of inertia of uniform square sheet can be calculated.

Complete step by step solution
The moment of inertia of a triangular lamina is given by:
I $=\dfrac{1}{2}\text{M }{{\text{h}}^{2}}$
Here we will put
M $=\dfrac{\text{m}}{2}$ because mass of triangle is half of the total square
Also h $=\dfrac{\text{a}}{\sqrt{2}}$ the reason is that the diagonal of square is $\sqrt{{{\text{a}}^{2}}+{{a}^{2}}}=\sqrt{2{{\text{a}}^{2}}}=\sqrt{2}\text{a}$
Therefore, half of diagonal of square $=\dfrac{\sqrt{2}\text{a}}{2}$
h $=\dfrac{\text{a}}{\sqrt{2}}$
So
Moment of inertia of triangular lamina is
$\text{I=}\dfrac{1}{2}\times \left( \dfrac{\text{m}}{2} \right)\times {{\left( \dfrac{\text{a}}{\sqrt{2}} \right)}^{2}}$
$=\dfrac{1}{2}\times \dfrac{\text{m}}{2}\times \dfrac{{{\text{a}}^{2}}}{2}=\dfrac{1}{8}\text{m}{{\text{a}}^{2}}$
I $=\dfrac{1}{8}\text{m}{{\text{a}}^{2}}$
So
Moment of inertia of square lamina $=2\times$ moment of inertia of triangular lamina
I $=2\times \dfrac{1}{8}\text{m}{{\text{a}}^{2}}$
I $=\dfrac{1}{4}\text{m}{{\text{a}}^{2}}$ .

Note
Moment of inertia is a quantity expressing a body’s tendency to resist angular acceleration, which is the sum of the product of the mass of each particle in the body with the square of its distance from the axis of rotation.
Moment of inertia formulas for different shapes are:
Solid square $=\dfrac{1}{2}\text{M}{{\text{R}}^{2}}$
Uniform sphere (through center) $=\dfrac{2}{5}\text{M}{{\text{R}}^{2}}$
Thin loop (through center) $=\text{M}{{\text{R}}^{2}}$
Thin loop (through central diameter) $=\dfrac{1}{2}\text{M}{{\text{R}}^{2}}+\dfrac{1}{12}\text{M}{{\text{w}}^{2}}$