The moment of inertia of a rod about an axis | vedclass

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(A) Let the rod of length $L$ and mass $M$ be bent at the center. Each half has length $l = L/2$ and mass $m = M/2$.When bent at $90^{\circ}$,the dist

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

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(A) Let the rod of length $L$ and mass $M$ be bent at the center. Each half has length $l = L/2$ and mass $m = M/2$.When bent at $90^{\circ}$,the distance of the center of mass of each half-rod from the axis passing through the vertex (the bend) is $d = \frac{l}{2\sqrt{2}} = \frac{L}{4\sqrt{2}}$.However,the axis of rotation is the same as the original axis passing through the center of the original rod. In the bent configuration,the axis passes through the vertex (the midpoint of the original rod).The moment of inertia of each half-rod about an axis passing through its own center of mass is $I_{cm} = \frac{1}{12}m l^2 = \frac{1}{12} (M/2) (L/2)^2 = \frac{ML^2}{96}$.Using the parallel axis theorem for each half-rod,the distance $d$ from the center of each half-rod to the axis of rotation is $d = \frac{L}{4}$.$I_{half} = I_{cm} + m d^2 = \frac{ML^2}{96} + (M/2) (L/4)^2 = \frac{ML^2}{96} + \frac{ML^2}{32} = \frac{ML^2 + 3ML^2}{96} = \frac{4ML^2}{96} = \frac{ML^2}{24}$.Since there are two such halves,the total moment of inertia is $I = 2 	imes \frac{ML^2}{24} = \frac{ML^2}{12}$.

Column-$I$	Column-$II$
$(1)$ Perpendicular Axis Theorem	$(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem	$(b)$ $I_z = I_x + I_y$

Similar Questions

Match Column-$I$ with Column-$II$:
Column-$I$ Column-$II$
$(1)$ Perpendicular Axis Theorem $(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem $(b)$ $I_z = I_x + I_y$

Where, $d =$ distance between two parallel axes.

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