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# How to solve this question?

Figure shows an equilateral triangle OAB and axis of rotation passes through a corner O.

Moment of inertia I of system about the given axis of rotation is given as

I = IAB + IOA + IOB

Where IAB Is moment of inertia of rod AB . Similarly moment of inertia of other two rods are considered.

Since thin rod AB is symmetrically placed about axis of rotation, we have

IAB = M × [ ( √3 /2 ) L ]= (3/4) M L2

Moment of inertia IOA of rod OA is determined as follows

Let ρ be the linear density of rod , ρ = ( M / L ) .

Let us consider small element of length dl in the rod at a distaance l along the rod from O as shown in figure.

Moment of inertia dI of this small element , dI = dm (l cos30)2 .

Moment of inertia of rod of full length OA is determined as

Similarly we get , IOB = (1/4) M L2

I = IAB + IOA + IOB =  M L2 [ (3/4) + (1/4) + (1/4) ] = (5/4) M L2
Answered by Expert 7th May 2021, 6:18 PM
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