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Question

(A) The moment of inertia of a square plate of mass $m$ and side $a$ about an axis passing through its center and perpendicular to its plane is $I_{cm

Vedclass · Accepted Answer

$\frac{2}{3}ma^2$

Vedclass · Answer

(A) The moment of inertia of a square plate of mass $m$ and side $a$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{ma^2}{6}$.Using the perpendicular axis theorem,the moment of inertia about an axis passing through the center and perpendicular to the plane is the sum of moments of inertia about two perpendicular axes in the plane: $I_{cm} = I_x + I_y$. Since $I_x = I_y = \frac{ma^2}{12}$,we have $I_{cm} = \frac{ma^2}{12} + \frac{ma^2}{12} = \frac{ma^2}{6}$.Now,we use the parallel axis theorem to find the moment of inertia about an axis passing through a vertex and perpendicular to the plane. The distance $d$ from the center to the vertex is $d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}$.Thus,$d^2 = \frac{a^2}{2}$.The moment of inertia about the vertex is $I = I_{cm} + md^2 = \frac{ma^2}{6} + m(\frac{a^2}{2}) = \frac{ma^2 + 3ma^2}{6} = \frac{4ma^2}{6} = \frac{2}{3}ma^2$.

Consider a uniform square plate of mass $m$ and side length $a$. What is the moment of inertia of this plate about an axis passing through one of its vertices and perpendicular to its plane?

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