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Plz solve it step by step making me understand the formula being applied here
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Asked by vishakhachandan026 | 23 Jul, 2019, 08:59: AM
answered-by-expert Expert Answer
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
 
begin mathsize 14px style I space equals space rho integral subscript 0 superscript l l apostrophe squared d l apostrophe space space equals space rho l cubed over 3 space equals space left parenthesis space rho space l space right parenthesis space l squared over 3 space equals m l squared over 3 end style......................(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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