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. $3M{R}^2/32$
• B. $5M{R}^2/16$
• C. $11M{R}^2/64$
• 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:
$I-I^{\prime }=\frac{1}{4}M{R}^2-\frac{5}{64}M{R}^2=\frac{11}{64}M{R}^2$