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Grade: 12th pass
        
Find mi about an axis and posing through center of mass perpendicular to its plane
one month ago

Answers : (1)

Sarika Bishnoi
14 Points
							
Let d=mass per unit area
         =M/(2a)2-(a)= M/3a2
Now fill the empty space with a square of +d as well as -d .
We will have one square of +d with side 2a and one square of -d with side a.
Total moment of inertia= moment of inertia due to                                               square1+ due to square2
                                            = M1(L1)/6 +M2(L2)2/6
 
                                   =[d(2a)2](2a)2/6+[ -d(a)2](a)2/6
                                    =5da4/2
                                     5Ma2/6          
 
one month ago
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