JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Rotational Mechanics

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 as 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 :

Answer: (B)

Solution

The axis

O O^{'}

passes through the centre of

D_{1}

perpendicular to its face, so

D_{1}

contributes its central value

\frac{M R ^{2}}{2}

. Each of

D_{2}

and

D_{3}

is attached at an opposite edge of

D_{1}

, with its centre a distance

R

from the axis, and contributes its moment about a diameter

\frac{M R ^{2}}{4}

plus the parallel-axis shift

M R^{2}

, that is

\frac{5 M R ^{2}}{4}

apiece. The total is

\frac{M R ^{2}}{2} + 2 \times \frac{5 M R ^{2}}{4} = \frac{M R ^{2}}{2} + \frac{5 M R ^{2}}{2} = 3 M R^{2}

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option