The moment of inertia of a rod of mass M and | vedclass

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(B) Let the rod be along the $x$-axis initially with its centre at the origin $(0,0)$. The rod consists of two halves,each of mass $m = M/2$ and lengt

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$\frac{1}{12} ML^2$

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(B) Let the rod be along the $x$-axis initially with its centre at the origin $(0,0)$. The rod consists of two halves,each of mass $m = M/2$ and length $l = L/2$.When the rod is bent at the centre,the centre of the rod remains at the origin. Let the axis of rotation be the $z$-axis passing through the origin.The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through one of its ends and making an angle $	heta$ with the rod is given by $I = \int r^2 dm$,where $r$ is the perpendicular distance from the axis.For each half of the rod,the distance $r$ of a point at a distance $x$ from the centre along the rod is $r = x \sin(	heta)$,where $	heta$ is the angle between the rod and the axis of rotation. Here,the axis is perpendicular to the original rod,so for the first half,$	heta = 90^o$,and for the second half,the angle changes.However,a simpler approach: The moment of inertia $I = \int r^2 dm$. Since the distance $r$ of every mass element $dm$ from the axis of rotation remains unchanged when the rod is bent at its centre (the centre point is on the axis),the total moment of inertia remains the same.Therefore,$I = \frac{1}{12} ML^2$.

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