The linear mass density of a thin rod AB of length L varies from A to B as λ(x) = λ₀ (1 + x/L), where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

Option 1 - <p>(5/12)ML²<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>(7/18)ML²<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>(2/5)ML²</p>
Option 4 - <p>(3/5)ML²</p>
6 Views|Posted 5 months ago
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1 Answer
A
5 months ago
Correct Option - 2
Detailed Solution:

dm = λdx = λ? (1 + x/L)dx
M = ∫? λ? (1 + x/L)dx = λ? [L + L²/2L] = 3λ? L/2
dI = dmx² = λ? (1 + x/L)dx * x²
I = λ? ∫? (x² + x³/L)dx = λ? [L³/3 + L? /4L]
I = (7λ? L³)/12 = (7/12) * (2M/3L) * L³ = (7/18)ML²

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