First slide

Moment of intertia

Question

Two uniform thin identical rods. AB and CD each of mass M and length I are joined so as to form a cross as shown in fig. The moment of inertia of the cross
about the bisector line EF is

Moderate

A

$\frac{{ML}^{2}}{6}$
B

$\frac{{ML}^{2}}{4}$
C

$\frac{{ML}^{2}}{12}$
D

$\frac{{ML}^{2}}{3}$

Solution

The line EF makes an angle of $45^{0}$ with each rod .
Therefore I = $I_{1} + I_{2} = (\frac{1}{12} M L^{2} \sin^{2} 45^{0}) \times 2 = \frac{1}{12} M L^{2}$

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