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Question

(B) The moment of inertia of a square plate about an axis passing through its center of mass and parallel to an edge is $I_{CM} = \frac{M L^{2}}{12}$.

Vedclass · Accepted Answer

$\frac{11 M L^{2}}{24}$

Vedclass · Answer

(B) The moment of inertia of a square plate about an axis passing through its center of mass and parallel to an edge is $I_{CM} = \frac{M L^{2}}{12}$.Using the parallel axis theorem,the moment of inertia about an axis passing through the center of mass but rotated by an angle $	heta$ in the plane of the plate is $I_{	heta} = I_{CM} = \frac{M L^{2}}{12}$ (since for a square,the moment of inertia about any axis in the plane passing through the center of mass is the same).Now,we use the parallel axis theorem: $I = I_{CM} + M d^{2}$,where $d$ is the perpendicular distance from the center of mass to the given axis.The distance $d$ from the center of the square to the axis passing through a vertex at an angle of $15^{\circ}$ with the horizontal is $d = \frac{L}{\sqrt{2}} \sin(45^{\circ} + 15^{\circ}) = \frac{L}{\sqrt{2}} \sin(60^{\circ}) = \frac{L}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} L}{2 \sqrt{2}}$.Thus,$I = \frac{M L^{2}}{12} + M \left( \frac{\sqrt{3} L}{2 \sqrt{2}} ight)^{2} = \frac{M L^{2}}{12} + M \left( \frac{3 L^{2}}{8} ight) = \frac{M L^{2} + 4.5 M L^{2}}{12} = \frac{M L^{2}}{12} + \frac{9 M L^{2}}{24} = \frac{2 M L^{2} + 9 M L^{2}}{24} = \frac{11 M L^{2}}{24}$.

$A$ square plate of mass $M$ and edge $L$ is shown in the figure. The moment of inertia of the plate about the axis in the plane of the plate passing through one of its vertices making an angle $15^{\circ}$ with the horizontal is:

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