Rotational Motion Question 14

Question: Four point masses, each of value m, are placed at the corners of a square ABCD of side 1. The moment of inertia of this system about an axis passing through A and parallel to BD is

[AIEEE 2006]

Options:

A) $ 2ml^{2} $

B) $ \sqrt{3}ml^{2} $

C) $ 3ml^{2} $

D) $ ml^{2} $

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Answer:

Correct Answer: C

Solution:

  • [c] $I_A$ = 0

$I_B=I_D$ = $mr^2$

=$ m \frac {l}{√2}^2 $

=$\frac {ml{^2}}{2}$

$IC=m(√2l)^2$

$IC=2ml^2$

$I=I_A+I_B+I_C+I_D$

$ I = 0 + \frac{ml^2}{2}+\frac {ml^2}{2}+2ml^2 $

$I=3ml^2$



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