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$\frac{1}{12} {ML}^{2}$ is the moment of inertia of a rod about an axis through its centre and perpendicular to it. The rod is bent from the middle at an angle 60⁰. Find the moment of inertia of the bent rod about the same axis. (M is the mass and L is the length of the rod)

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$\frac{1}{48} {ML}^{2}$

$\frac{{ML}^{3}}{8 \sqrt{3}}$

$\frac{1}{24} {ML}^{2}$

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

answer is A.

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Detailed Solution

Given $I_{1} = \frac{1}{12} {ML}^{2}$

now as per the question

Both half rods are same plane.

now

$I = \frac{2 {ml}^{2}}{3} \sin^{2} θ$

where m is mass of each rod, l is its length and $θ$ is half of angle between them.

$I = \frac{2 (\frac{M}{2}) {(\frac{L}{2})}^{2}}{3} \times \sin^{2} 30^{\circ}$

$I = \frac{{ML}^{2}}{48}$

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