Question
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) 3ML2
(B) ML2
(C) 2ML2
(D) 3ML2
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+Ma2
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 2L.
We know from properties of square that ∠BAD=900, so ∠OAP=450. To get OA, cos450=AOL/2⇒AO=2cos450L=2×1/2L=2L.
To find moment of inertia at A, we will apply parallel axis theorem, I=Icm+Ma2where
Icmis the moment of inertia at center of mass, M is the mass a be the distance between two axes.
IXX′=Icm+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(2L)2+0+M(2L)2+0+4M(2L)2
⇒IXX′=2ML2+2ML2+24ML2=3ML2
Correct answer is A. 3ML2
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.