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Question

(B) The moment of inertia of a thin rod about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$

Vedclass · Accepted Answer

$\frac{ML^2}{9}$

Vedclass · Answer

(B) The moment of inertia of a thin rod about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the new axis.The center of mass of the rod is at $L/2$ from one end. The given axis is at $L/3$ from the same end.Therefore,the distance $d$ between the center of mass and the axis is $d = |L/2 - L/3| = L/6$.Substituting these values into the parallel axis theorem:$I = \frac{ML^2}{12} + M(L/6)^2$$I = \frac{ML^2}{12} + \frac{ML^2}{36}$$I = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.

Column-$I$	Column-$II$
$(1)$ Moment of inertia of a solid sphere about its diameter	$(a)$ $\frac{2}{3}MR^2$
$(2)$ Moment of inertia of a solid sphere about a tangent	$(b)$ $\frac{2}{5}MR^2$
	$(c)$ $\frac{7}{5}MR^2$

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

What is the moment of inertia of a thin rod of length $L$ and mass $M$ about an axis perpendicular to the rod and passing through a point at a distance of $L/3$ from one of its ends?

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