Question

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?

• A. $13 MR^2/32$
• B. $11 MR^2/32$
• C. $9 MR^2/32$
• D. $15 MR^2/32$

Answer 1

Answer: (A)

$13 MR^2/32$

Answer 2

${I_{Total disc } = \frac{MR^2}{2}}$
${M_{Removed} = \frac{M}{4} (Mass \propto area) }$
${I_{removed} (about \; same \; Perpendicular \; axis)}$
${ = \frac{M}{4} \frac{(R /2)^2}{2} + \frac{M}{4} (\frac{R}{2})^2 = \frac{3MR^2}{32}}$
${I_{Remaing disc } = I_{Total} - I_{Removed}}$
${ = \frac{MR^2}{2} - \frac{3}{32} MR^2 = \frac{13}{32} MR^2}$