question_answer

Given a uniform disc of mass M and radius R. A small disc of radius \[R/2\] is cut from this disc in such a way that the distance between the centres of the two discs is\[R/2\]. Find the moment of inertia of the remaining disc about a diameter of the original disc perpendicular to the line connecting the centres of the two discs

A) \[\frac{3M{{R}^{2}}}{32}\]                

B) \[\frac{5M{{R}^{2}}}{16}\]

C) \[\frac{11M{{R}^{2}}}{64}\]              

D) none of these

Correct Answer: C

Solution :

Mass of cut disc: \[{{m}_{1}} = M/4\] Moment of Inertia of original disc about axis 1: \[I=\frac{1}{4}\,M{{R}^{2}}\] Moment of Inertia of small disc about axis 1: \[I'=\frac{1}{4}{{m}_{1}}\,{{\left( \frac{R}{2} \right)}^{2}}\,\,+{{m}_{1}}\,{{\left( \frac{R}{2} \right)}^{2}}\,\,\,\,\,\,\,\,[Parallel\,\,axis\,\,theorem]\]\[\frac{5}{16}\,{{m}_{1}}{{R}^{2}}\,\,=\,\,\,\frac{5}{64}M{{R}^{2}}\] Required moment of inertia: \[I-I'=\frac{1}{4}\,M{{R}^{2}}\,-\,\,\frac{5}{64}M{{R}^{2}}\,=\,\frac{11}{64}\,\,M{{R}^{2}}\]