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Question

(A) Let the side length of the equilateral triangle $ABC$ be $L$. The area of the triangle is $A = \frac{\sqrt{3}}{4}L^2$. The moment of inertia of a

Vedclass · Accepted Answer

$I = \frac{15}{16}I_0$

Vedclass · Answer

(A) Let the side length of the equilateral triangle $ABC$ be $L$. The area of the triangle is $A = \frac{\sqrt{3}}{4}L^2$. The moment of inertia of a thin uniform equilateral triangle about an axis passing through its centroid and perpendicular to its plane is given by $I = \frac{1}{6} M L^2$,where $M$ is the mass of the triangle. Since the sheet is uniform,the mass $M$ is proportional to the area $A$,so $M = \sigma A$,where $\sigma$ is the surface mass density. Thus,$I \propto A \cdot L^2 \propto L^2 \cdot L^2 = L^4$.Let $I_0$ be the moment of inertia of the original triangle $ABC$ with side length $L$. So,$I_0 = k L^4$ for some constant $k$.The smaller triangle $DEF$ has a side length of $L/2$. Its mass $m$ is $1/4$ of the mass $M$ of the original triangle because its area is $1/4$ of the original area. The moment of inertia of the smaller triangle $DEF$ about its own centroid (which is also $G$) is $I_{DEF} = k (L/2)^4 = k \frac{L^4}{16} = \frac{I_0}{16}$.The moment of inertia of the remaining figure is the difference between the moment of inertia of the original triangle and the removed triangle: $I = I_0 - I_{DEF} = I_0 - \frac{I_0}{16} = \frac{15}{16}I_0$.

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