Rotational Motion

JEE Main level, single correct, medium

13 of 280

Question 13 of 280

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius RR and length L(R<L)L(\mathrm{R}\lt \mathrm{L}) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as MM) :

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Answer: (B) 38MR2+16ML2\frac{3}{8} M R^{2}+\frac{1}{6} M L^{2}

Each cylinder has mass M/4M/4, radius RR, length LL.
Axis through midpoints of top and bottom sides (along y-direction).
Top/bottom cylinders (perpendicular to axis, axis through midpoint):
I=McR24+McL212I = \frac{M_c R^2}{4} + \frac{M_c L^2}{12} each.
Left/right cylinders (parallel to axis, distance L/2L/2 from it):
I=McR22+Mc(L/2)2I = \frac{M_c R^2}{2} + M_c(L/2)^2 each.
Total =2(McR24+McL212)+2(McR22+McL24)= 2\left(\frac{M_c R^2}{4}+\frac{M_c L^2}{12}\right) + 2\left(\frac{M_c R^2}{2}+\frac{M_c L^2}{4}\right)
=M2(3R24+L23)=3MR28+ML26= \frac{M}{2}\left(\frac{3R^2}{4}+\frac{L^2}{3}\right) = \frac{3MR^2}{8}+\frac{ML^2}{6}.

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