Question

Two uniform rods of mass m and length l form a cross moment of inertia of cross about an axis parallel to cd passing through a is

Answer 1

Mass moment of inertia of system about A is $\frac{7ML^{2}}{12}$.

Explanation:

Mass moment of inertia of rod AB about centre of gravity perpendicular to length  along CD ($I_{CD}^{AB}$)= $\frac{ML^{2}}{12}$  

Now from Parallel axis theorem mass Moment of inertia of rod AB  about A  $I^{AB}$ = $\frac{ML^{2}}{12} + M\times (\frac{L}{2})^{2} = \frac{ML^{2}}{3}$     ...1)

Mass moment of inertia of rod CD about its longitudinal axis $I_{CD}^{CD}$ = 0

Now from Parallel axis theorem mass Moment of inertia  of rod CD about A $I^{CD} = 0 + M \times (\frac{L}{2})^{2} = \frac{ML^{2}}{4}$    ...2)

Total Moment of inertia of system about $I^{A} = I^{AB} + I^{CD}$

$I^{A} = \frac{ML^{}{2}}{3}+ \frac{ML^{2}}{4} = \frac{7 ML^{2}}{12}$