First slide

Moment of intertia

Question

Two uniform, thin identical rods each of mass M and length L are joined at middle so as to form a cross as shown. The moment of inertia of the cross about a bisector line EF (in the plane of cross) is

Moderate

A

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

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

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

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

Solution

We have to find $I_{1}$ ?
$I_{1} = I_{2}$ ----------(1)
$I_{0} = \frac{{ML}^{2}}{12} + \frac{{ML}^{2}}{12} = \frac{{ML}^{2}}{6}$
now $I_{1} + I_{2} = I_{0}$ --------(2)
From (1) and (2), $I_{1} = \frac{I_{0}}{2} = \frac{{ML}^{2}}{12}$

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