(B) The total moment of inertia $I$ of the system is the sum of the moments of inertia of the three discs about the axis $OO'$.For disc $D_1$,the axis

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(B) The total moment of inertia $I$ of the system is the sum of the moments of inertia of the three discs about the axis $OO'$.For disc $D_1$,the axis $OO'$ passes through its centre and is perpendicular to its plane. Thus,$I_1 = \frac{1}{2}MR^2$.For discs $D_2$ and $D_3$,the axis $OO'$ is parallel to their diameters and passes at a distance $R$ from their centres. Using the parallel axis theorem,$I_{2,3} = I_{cm} + Md^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.Since there are two such discs,the total moment of inertia is $I = I_1 + 2 \times I_{2,3} = \frac{1}{2}MR^2 + 2 \times (\frac{5}{4}MR^2) = \frac{1}{2}MR^2 + \frac{5}{2}MR^2 = 3MR^2$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia

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$A$ circular disc $D_1$ of mass $M$ and radius $R$ has two identical discs $D_2$ and $D_3$ of the same mass $M$ and radius $R$ attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis $OO'$,passing through the centre of $D_1$ as shown in the figure,will be:

JEE MAIN 2019 All JEE MAIN

A
$MR^2$
B
$3MR^2$
C
$\frac{4}{5}MR^2$
D
$\frac{2}{3}MR^2$

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System of Particles and Rotational Motion

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Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia

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