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Consider a uniform square plate of side $a$ and mass $M$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its comers is
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Verified Answer
The correct answer is:
$\frac{2}{3} M a^2$
Step 1: Given information
A uniform square plate,
Side is "a" and mass "m".
According to the question, the diagram will be:
Step 2: Calculation
The moment of inertia of a square plate around an axis perpendicular to the plane going throughout its center of mass is determined by:
$\Rightarrow \mathrm{I}_{\mathrm{CM}}=\frac{\mathrm{ma}^2}{6}$
The moment of inertia around a parallel axis through one of its corners is determined by:
$\Rightarrow \mathrm{I}=\mathrm{I}_{\mathrm{CM}}+m\left(\frac{\mathrm{a}}{\sqrt{2}}\right)^2$
By substituting the values of $\mathrm{I}_{\mathrm{CM}}$,
$\begin{aligned}
= & \frac{\mathrm{ma}^2}{6}+\frac{\mathrm{ma}^2}{2} \\
& =\frac{\mathrm{ma}^2+3 \mathrm{ma}^2}{6} \\
& =\frac{4}{6} \mathrm{ma}^2 \\
& =\frac{2}{3} \mathrm{ma}^2
\end{aligned}$
Therefore, the moment of inertia is $\frac{2}{3} \mathrm{ma}^2$.
A uniform square plate,
Side is "a" and mass "m".
According to the question, the diagram will be:
Step 2: Calculation
The moment of inertia of a square plate around an axis perpendicular to the plane going throughout its center of mass is determined by:
$\Rightarrow \mathrm{I}_{\mathrm{CM}}=\frac{\mathrm{ma}^2}{6}$
The moment of inertia around a parallel axis through one of its corners is determined by:
$\Rightarrow \mathrm{I}=\mathrm{I}_{\mathrm{CM}}+m\left(\frac{\mathrm{a}}{\sqrt{2}}\right)^2$
By substituting the values of $\mathrm{I}_{\mathrm{CM}}$,
$\begin{aligned}
= & \frac{\mathrm{ma}^2}{6}+\frac{\mathrm{ma}^2}{2} \\
& =\frac{\mathrm{ma}^2+3 \mathrm{ma}^2}{6} \\
& =\frac{4}{6} \mathrm{ma}^2 \\
& =\frac{2}{3} \mathrm{ma}^2
\end{aligned}$
Therefore, the moment of inertia is $\frac{2}{3} \mathrm{ma}^2$.
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