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Question : 91 of 160
Marks:
+1 ,
-0
Solution:
(*) Mass per unit length,
‌ = λ | x | ⇒ d M = λ | x | d x
Moment of inertia about axis
A A I A A = ∫ x 2 d M Put the value of
d M I A A = x 2 × λ | x | d x = x 2 × 16 | x | d x [ ∵ λ = 16 k g ∕ m 2 ‌ (given)] ‌ = 16 ‌ x 2 | x | d x = 16 [ x 2 ( − x ) d x + x 2 ( + x ) d x ] [ ∵ | x | = { − x ‌ ‌ ‌ if ‌ x < 0 + x ‌ ‌ ‌ if ‌ x ≥ 0
] ‌ So, ‌ I A A ‌ ‌ = 16 [ ( − x 3 ) d x + ( x 3 ) d x ] ‌ ‌ = 16 [ − [ ‌ ] − L ∕ 2 0 + [ ‌ ] 0 + L ∕ 2 ] ‌ ‌ = 4 [ − [ x 4 ] − L ∕ 2 0 + [ x 4 ] 0 + L ∕ 2 ] ‌ ‌ = 4 [ − { ( 0 ) 4 − ( − ‌ ) 4 } + { ( + ‌ ) 4 − ( 0 ) 4 } ] ‌ ‌ = 4 [ { 0 − ‌ } + { ‌ − 0 } ] ‌ ‌ = 4 [ ‌ + ‌ ] = 4 [ ‌ ] = ‌ Now by the theorem of parallel axes, the moment of inertia about axis
B B ,
I B B = I C O M + ∫ ( ‌ ) 2 d M = I C O M + ‌ ‌ ∫ d M Put the value of
d M and
I ‌ COM ‌ I B B = I M + ‌ ‌ λ | x | d x = ‌ + ‌ λ | x | d x = ‌ ‌ ‌ + ‌ × 16 ‌ ( + x ) d x [ ∵ λ = 16 k g ∕ m 2 ‌ (given)] ‌ = ‌ + 4 L 2 [ ‌ ] 0 L ∕ 2 = ‌ + 2 L 2 [ x 2 ] 0 l ∕ 2 = ‌ + 2 L 2 [ ( ‌ ) 2 − ( 0 ) 2 ] = ‌ + 2 L 2 [ ‌ − 0 ] = ‌ + ‌ = L 4 = 1 4 = 1 k g − m 2 © examsiri.com
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