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Plz solve it step by step making me understand the formula being applied here
Asked by vishakhachandan026 | 23 Jul, 2019, 08:59: AM
Fig.1 shows  two rods which are inclined to vertical axis by 45°. Let length of each rod be l and mass of each rod be m.

Let us consider moment of inertial of small mass elements dm from both the rods,
which is of length dl' and at a distance l' from point O.

Moment of inertia dI about the axis passing through centre as shown in fig.1 is given by

dI = 2×dm×(l'/√2)2  = ( ρ×dl' )×(l')2
where ρ is mass of rod per unit length.

Moment of inertia I of system of two rods about the axis passing through centre as shown in fig. 1 is given by

......................(1)
Moment of inertia I of system of four rods about the axis passing through centre as shown in fig. 2 is given by

I = m×(2/3)×l2 ...........................(2)

Moment of inertia I of system of four rods about the axis passing through corner as shown in fig. 3 is obtained
by applying parallel axis theorem.

Hence, noment of inertia I = m×(2/3)×l2 + 4m×(l/√2)2 =  (8/3)m×l2
Answered by Thiyagarajan K | 23 Jul, 2019, 02:30: PM

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