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 $
$I=3ml^2$
BETA
