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Question:

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, and F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :

31Io32

1516Io

78Io

3Io4

Solution:

Let the moment of inertia of the equilateral triangular lamina ABC about the axis passing through its center O and perpendicular to its plane be Io.
The lamina ABC is divided into four smaller equilateral triangles, namely, ADE, BDF, CEF, and DEF. The triangles ADE, BDF, and CEF are congruent and have the same moment of inertia.
Let I be the moment of inertia of each of the smaller triangles ADE, BDF, and CEF about the axis passing through O and perpendicular to the plane. Then, by the parallel axes theorem, the moment of inertia of each smaller triangle about the axis through its own centroid and parallel to the given axis is I - m(a/2)^2, where m is the mass of the smaller triangle and a is the side length of the original triangle.
Since the triangles ADE, BDF, and CEF are congruent, they have equal masses and moments of inertia. Thus, the moment of inertia of triangle ABC is:
Io = 3I + I_DEF
The moment of inertia of a triangle about its centroid is given by I = (ma^2)/12. Therefore, the moment of inertia of each of the small triangles is:
I_small = (m/4)(a/2)^2 / 12 = ma^2 / 192
Applying the parallel axis theorem:
I = I_small + (m/4)(a/(2√3))^2 = ma^2/192 + ma^2/48 = ma^2/32
Io = 3(ma^2/32) + I_DEF = 3ma^2/32 + (m/4)(a^2/12) = (3/32 + 1/48)ma^2 = 15/96 ma^2 = (5/32) ma^2
The mass of the larger triangle is m, so the moment of inertia of the larger triangle is Io = (1/6)ma^2.
The moment of inertia of the remaining part of the lamina is:
Io - I_DEF = Io - (1/16)Io = (15/16)Io
Therefore, the moment of inertia of the remaining part of the lamina about the same axis is (15/16)Io