Rotational Motion

JEE Main level, single correct, medium

80 of 280

Question 80 of 280

A circular disc D1D_{1} of mass MM and radius RR has two identical discs⁡D2\operatorname{discs} D_{2} and D3D_{3} of the same mass MM 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 D1D_{1}, as shown in the figure, will be

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Answer: (B) 3MR23 \mathrm{MR}^{2}

Moment of inertia of disc D1D_{1} about OO′=I1=MR22O O^{\prime}=I_{1}=\frac{M R^{2}}{2}

M.O.I of D2\mathrm{D}_{2} about OO′\mathrm{OO}^{\prime} =I2=12(MR22)+MR2=MR24+MR2=I_{2}=\frac{1}{2}\left(\frac{M R^{2}}{2}\right)+M R^{2}=\frac{M R^{2}}{4}+M R^{2}

M.O.I of D3\mathrm{D}_{3} about OO′\mathrm{OO}^{\prime} =I3=12(MR22)+MR2=MR24+MR2=I_{3}=\frac{1}{2}\left(\frac{M R^{2}}{2}\right)+M R^{2}=\frac{M R^{2}}{4}+M R^{2}

so, resultant M.O.I about OO′\mathrm{OO}^{\prime} is I=I1+I2+I3\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}

⇒I=MR22+2(MR24+MR2)=MR22+MR22+2MR2=3MR2\begin{aligned} \Rightarrow & I=\frac{M R^{2}}{2}+2\left(\frac{M R^{2}}{4}+M R^{2}\right) \\ &=\frac{M R^{2}}{2}+\frac{M R^{2}}{2}+2 M R^{2}=3 M R^{2} \end{aligned}

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