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Question: Four point masses, each of mass M are placed at corner of square ABCD of side L. The moment of inert...

Four point masses, each of mass M are placed at corner of square ABCD of side L. The moment of inertia of this system about an axis passing through A and parallel to BD is
(A) 3ML23M{L^2}
(B) ML2M{L^2}
(C) 2ML22M{L^2}
(D) 3ML2\sqrt 3 M{L^2}

Explanation

Solution

Since each corner of square ABCD has a mass and we need to find the inertia parallel to BD so we will be using the parallel axis theorem.
Formula Used:
I=Icm+Ma2I = {I_{cm}} + M{a^2}

Complete step by step answer
First, we will draw the diagram. Here, OP is the perpendicular drawn from O to AD. XX’ is drawn passing through A and parallel to BD. The length of AP and AD to be equal as L2\dfrac{L}{2}.

We know from properties of square that BAD=900\angle BAD = {90^0}, so OAP=450\angle OAP = {45^0}. To get OA, cos450=L/2AOAO=L2cos450=L2×1/2=L2\cos {45^0} = \dfrac{{L/2}}{{AO}} \Rightarrow AO = \dfrac{L}{{2\cos {{45}^0}}} = \dfrac{L}{{2 \times 1/2}} = \dfrac{L}{{\sqrt 2 }}.
To find moment of inertia at A, we will apply parallel axis theorem, I=Icm+Ma2I = {I_{cm}} + M{a^2}where
Icm{I_{cm}}is the moment of inertia at center of mass, M is the mass a be the distance between two axes.
IXX=Icm+M(AO)2{I_{XX'}} = {I_{cm}} + M{(AO)^2}
The moment of inertia at center will be because of all four points A, B, C, D.
So IXX=(IA+IB+IC+ID)+M(AO)2=M(L2)2+0+M(L2)2+0+4M(L2)2{I_{XX'}} = ({I_A} + {I_B} + {I_C} + {I_D}) + M{(AO)^2} = M{(\dfrac{L}{{\sqrt 2 }})^2} + 0 + M{(\dfrac{L}{{\sqrt 2 }})^2} + 0 + 4M{(\dfrac{L}{{\sqrt 2 }})^2}
IXX=ML22+ML22+4ML22=3ML2\Rightarrow {I_{XX'}} = \dfrac{{M{L^2}}}{2} + \dfrac{{M{L^2}}}{2} + \dfrac{{4M{L^2}}}{2} = 3M{L^2}

Correct answer is A. 3ML23M{L^2}

Additional information

There are basically two theorems used in calculating moment of inertia. They are –
Parallel Axis Theorem which states that a body’s moment of inertia about any axis is the sum of moment of inertia about a parallel axis which passes through the center of mass and the product of mass of body and perpendicular distance between axes square.
Perpendicular Axis Theorem which says that a body’s moment of inertia about a perpendicular to plane axis is the sum of moment of inertia of any two-perpendicular axis in the first axis intersecting plane.

Note
Moment of inertia at point B and D is zero as the perpendicular distance for both the points are zero.