Question

In a rectangle ABCD, \[BC = 2AB\]. The moment of inertia is minimum along an axis through?

A. \[CD\]
B. \[BC\]
C. \[HF\]
D. \[EG\]

Answer 1

Hint:To calculate the moment of inertia about an axis through an object, two theorems are used. One is the parallel axis theorem and another is the perpendicular axis theorem. Both are used depending upon the situations given in the question. The moment of inertia due to rotation about an axis passing through one of the ends of a rod of length L and mass M and perpendicular to the rod is \[\dfrac{{M{L^2}}}{3}\].

Formula Used:
The moment of inertia, \[I\] due to rotation about an axis passing through one of the ends of a rod of length L and mass M and perpendicular to the length of the rod is,
\[I = \dfrac{{M{L^2}}}{3}\] ---- (1)
The moment of inertia due to rotation about an axis passing through the centre of the rod of length L and mass M and in a direction perpendicular to the length of the rod is,
\[{I_C} = \dfrac{{M{L^2}}}{{12}}\]--- (2)

Complete step by step solution:
Given: A rectangle ABCD where \[BC = 2AB\]
Let M be the mass of the rectangle ABCD and length of AB = L ---- (3) .
Then, \[AB = CD = L\]
Then, \[BC = 2AB = 2L\]--- (4).
Then, \[BC = AC = 2L\]
The moment of inertia about the axis CD is:
Using the condition from equation (1),
\[{I_{ < CD > }} = \dfrac{1}{3}M{(BC)^2}\]
\[{I_{ < CD > }} = \dfrac{1}{3}M{(2L)^2}\]
\[\Rightarrow {I_{ < CD > }} = \dfrac{8}{3}M{L^2}\]--- (5)

The moment of inertia about the axis BC is:
Using the condition from equation (1),
\[{I_{ < BC > }} = \dfrac{1}{3}M{(AB)^2}\]
\[\Rightarrow {I_{ < BC > }} = \dfrac{1}{3}M{(L)^2}\]---- (6)
The moment of inertia about the axis HF is:
As HF passes through the centre of edge AB, using the condition from equation (2),
\[{I_{ < HF > }} = \dfrac{1}{{12}}M{(AB)^2}\]
\[\Rightarrow {I_{ < HF > }} = \dfrac{1}{{12}}M{L^2}\]---- (7)

The moment of inertia about the axis EG is:
As EG passes through the centre of edge BC, using the condition from equation (2),
\[{I_{ < EG > }} = \dfrac{1}{{12}}M{(BC)^2}\]
\[\Rightarrow {I_{ < EG > }} = \dfrac{1}{{12}}M{(2L)^2}\]
\[\Rightarrow {I_{ < EG > }} = \dfrac{1}{3}M{L^2}\]---- (8)
From equations (5), (6), (7) and (8),
The moment of inertia about HF, \[{I_{ < HF > }} = \dfrac{1}{{12}}M{L^2}\] has the minimum value.

Hence, option C is the correct answer.

Note: If we have to calculate the moment of inertia about an axis parallel to the axis passing through the centre of the object, we use the theorem of the parallel axis. When the body has symmetry in about two out of three axes in a 3D plane, the perpendicular axis theorem is used. The complete rectangle is considered a single entity of mass M. If the axis of rotation is any of the diagonals, then the moment of inertia is \[\dfrac{1}{6}M{L^2}\].