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A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^\circ$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is -
A
$\dfrac{ML^2}{6}$
B
$\dfrac{\sqrt{2}\,ML^2}{24}$
C
$\dfrac{ML^2}{24}$
D
$\dfrac{ML^2}{12}$
Detailed Solution
After bending, each half is a thin rod of mass $\dfrac{M}{2}$ and length $\dfrac{L}{2}$.
The axis passes through the bending point, which is one end of each half, and is perpendicular to both halves.
Moment of inertia of a thin rod of mass $m$ and length $l$ about a perpendicular axis through its end $= \dfrac{ml^2}{3}$
For one half: $I_1 = \dfrac{(M/2)(L/2)^2}{3} = \dfrac{ML^2}{24}$
Moment of inertia of the system: $I = \dfrac{(M/2)(L/2)^2}{3} + \dfrac{(M/2)(L/2)^2}{3}$
$I = \dfrac{ML^2}{24} + \dfrac{ML^2}{24}$
$I = \dfrac{ML^2}{12}$
The angle between the halves does not matter, because every element stays at the same distance from the axis.
The axis passes through the bending point, which is one end of each half, and is perpendicular to both halves.
Moment of inertia of a thin rod of mass $m$ and length $l$ about a perpendicular axis through its end $= \dfrac{ml^2}{3}$
For one half: $I_1 = \dfrac{(M/2)(L/2)^2}{3} = \dfrac{ML^2}{24}$
Moment of inertia of the system: $I = \dfrac{(M/2)(L/2)^2}{3} + \dfrac{(M/2)(L/2)^2}{3}$
$I = \dfrac{ML^2}{24} + \dfrac{ML^2}{24}$
$I = \dfrac{ML^2}{12}$
The angle between the halves does not matter, because every element stays at the same distance from the axis.