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JEE Main 2019 (11 Jan, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2019 (11 Jan, Shift 2)

Q162JEE Main 2019 · 11 JAN · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A circular disc D1 of mass M and radius R has two identical discs D2 and D3 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 D1 as shown in the figure, will be:

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Answer: (a)

$D_1$ about its own axis: $\frac{MR^2}{2}$ .
$D_2$ , $D_3$ : axis is parallel to their diameter at distance $R$ : $\frac{MR^2}{4} + MR^2 = \frac{5MR^2}{4}$ each.
$I = \frac{MR^2}{2} + 2 \times \frac{5MR^2}{4} = 3MR^2$
So the answer is (a).

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