In the diagram shown below,all three rods are | vedclass

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(A) The total moment of inertia of the system is the sum of the moments of inertia of the individual rods about the axis of rotation (rod $B$).$1$. Mo

Vedclass · Accepted Answer

$\frac{M L^2}{6}$

Vedclass · Answer

(A) The total moment of inertia of the system is the sum of the moments of inertia of the individual rods about the axis of rotation (rod $B$).$1$. Moment of inertia of rod $B$: Since rod $B$ itself is the axis of rotation,every mass element of rod $B$ lies on the axis. Therefore,its moment of inertia is $I_B = 0$.$2$. Moment of inertia of rod $A$: Rod $A$ is perpendicular to the axis $B$ and is attached at its center to the axis. The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I_A = \frac{1}{12} M L^2$.$3$. Moment of inertia of rod $C$: Similarly,rod $C$ is also perpendicular to the axis $B$ and attached at its center. Thus,$I_C = \frac{1}{12} M L^2$.Total moment of inertia $I = I_A + I_B + I_C = \frac{1}{12} M L^2 + 0 + \frac{1}{12} M L^2 = \frac{2}{12} M L^2 = \frac{1}{6} M L^2$.

In the diagram shown below,all three rods are of equal length $L$ and equal mass $M$. The system is rotated such that rod $B$ is the axis. What is the moment of inertia of the system?

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