Two thin discs, each of mass \(M\) and radius \(r\) metre, are attached as shown in figure, to form a rigid…
Two thin discs, each of mass \(M\) and radius \(r\) metre, are attached as shown in figure, to form a rigid body. The rotational inertia of this body about an axis perpendicular to the plane of disc \(B\) passing through its centre is
\(2 M r^{2}\)
\(3 \mathrm{Mr}^{2}\)
\(4 \mathrm{Mr}^{2}\)
\(5 \mathrm{Mr}^{2}\)
Solution
The moment of inertia of the disc about the centroidal axis is \(\frac{\mathrm{Mr}^{2}}{2}\). The moment of inertia of the disc at A about the axis passing through B is given using parallel axis theorem as \(\frac{\mathrm{Mr}^{2}}{2}+\mathrm{M}(2 \mathrm{R})^{2}\)
Thus the total moment of inertia of the whole system is given as \(\frac{\mathrm{Mr}^{2}}{2}+\) \(\frac{\mathrm{Mr}^{2}}{2}+\mathrm{M}(2 \mathrm{R})^{2}=5 \mathrm{Mr}^{2}\)
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Asked in: JEE Mains - Rotational Motion - Chapter Test
