Question

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. $3 M R^{2} / 32$
• B. $5 M R^{2} / 16$
• C. $11 M R^{2} / 64$
• D. None of these

Answer 1

Answer: (C)

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 $1$:

$I^{\prime}=\frac{1}{4} m_{1}\left(\frac{R}{2}\right)^{2}+m_{1}\left(\frac{R}{2}\right)^{2}$
or $I^{\prime}=\frac{5}{16} m_{1} R^{2}=\frac{5}{64} M R^{2}$
Required moment of inertia:
$I-I^{\prime}=\frac{1}{4} M R^{2}-\frac{5}{64} M R^{2}=\frac{11}{64} M R^{2}$