Two identical thin rods of mass $M \mathrm{~kg}$ and length $L \mathrm{~m}$ are connected as shown i

Question

Two identical thin rods of mass $M \mathrm{~kg}$ and length $L \mathrm{~m}$ are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point $P$ and perpendicular to the plane of the rods is $\frac{x}{12} \mathrm{ML}^2 \mathrm{~kg} \mathrm{~m}^2$. The value of $x$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

The system consists of two rods. For the vertical rod, the axis passes through one of its ends, so its moment of inertia is $I_1 = \frac{ML^2}{3} = \frac{4ML^2}{12}$. For the horizontal rod, its center of mass is at a distance $L$ from the axis. Using the parallel axis theorem, its moment of inertia is $I_2 = I_{CM} + Md^2 = \frac{ML^2}{12} + ML^2 = \frac{13ML^2}{12}$. The total moment of inertia is $I = I_1 + I_2 = \frac{4ML^2}{12} + \frac{13ML^2}{12} = \frac{17ML^2}{12}$. Comparing with $\frac{x}{12}ML^2$, we get $x = 17$.