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Asked by anshadali111 | 07 May, 2021, 04:08: PM
Expert Answer
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 ]2 = (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
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