Two identical thin rods of mass M kg and len | vedclass

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(B) Let the vertical rod be $1$ and the horizontal rod be $2$.For rod $1$, the moment of inertia about an axis passing through $P$ (its end) and perpe

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$17$

Vedclass · Answer

(B) Let the vertical rod be $1$ and the horizontal rod be $2$.For rod $1$, the moment of inertia about an axis passing through $P$ (its end) and perpendicular to its length is $I_1 = \frac{ML^2}{3}$.For rod $2$, the moment of inertia about its center of mass is $I_{cm} = \frac{ML^2}{12}$. Using the parallel axis theorem, the moment of inertia about an axis passing through $P$ and perpendicular to the plane is $I_2 = I_{cm} + Md^2$, where $d = L$ is the distance from $P$ to the center of rod $2$.$I_2 = \frac{ML^2}{12} + M(L)^2 = \frac{13ML^2}{12}$.The total moment of inertia is $I = I_1 + I_2 = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}$.Comparing this with $\frac{x}{12} ML^2$, we get $x = 17$.

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

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