Four thin rods of same mass M and same length | vedclass

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(A) The moment of inertia of a single rod $AB$ about an axis passing through its centre $P$ and perpendicular to its length is $I_{cm} = \frac{1}{12}M

Vedclass · Accepted Answer

$\frac{4}{3}M{l^2}$

Vedclass · Answer

(A) The moment of inertia of a single rod $AB$ about an axis passing through its centre $P$ and perpendicular to its length is $I_{cm} = \frac{1}{12}Ml^2$.To find the moment of inertia of rod $AB$ about an axis passing through the centre $O$ of the square and perpendicular to its plane,we use the parallel axis theorem: $I = I_{cm} + Md^2$.Here,the distance $d$ between the centre of the rod $P$ and the centre of the square $O$ is $d = \frac{l}{2}$.So,the moment of inertia of one rod about the axis through $O$ is $I_{rod} = \frac{1}{12}Ml^2 + M(\frac{l}{2})^2 = \frac{1}{12}Ml^2 + \frac{1}{4}Ml^2 = \frac{1+3}{12}Ml^2 = \frac{4}{12}Ml^2 = \frac{1}{3}Ml^2$.Since the system consists of $4$ identical rods,the total moment of inertia of the system is $I_{system} = 4 	imes I_{rod} = 4 	imes \frac{1}{3}Ml^2 = \frac{4}{3}Ml^2$.

Four thin rods of same mass $M$ and same length $l$ form a square as shown in the figure. The moment of inertia of this system about an axis passing through the centre $O$ and perpendicular to its plane is

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