A circular hole of radius (R/4) is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

Question

A circular hole of radius (R/4) is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is : (A) (219 MR2/256) (B) (237 MR2/512) (C) (19 MR2/512) (D) (197 MR2/256)

Tardigrade Admin · Accepted Answer

ID=(m r2/2) I text removed =(1/2) (m/16) (r2/16)+(m/16) (9 r2/16)(I m+m d) =(m r2+18 m r2/512) =(19 m r2/512) I text remaining =(m r2/2)-(19/512) m r2 =(237/512) m r2

Q. A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$ , as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is :

$\frac{219 M R ^{2}}{256}$

$\frac{237 M R ^{2}}{512}$

$\frac{19 M R ^{2}}{512}$

$\frac{197 M R ^{2}}{256}$

$I_{D} = \frac{m r ^{2}}{2}$
$I_{removed} = \frac{1}{2} \frac{m}{16} \frac{r ^{2}}{16} + \frac{m}{16} \frac{9 r ^{2}}{16} (I m + m d)$
$= \frac{m r ^{2} + 18 m r ^{2}}{512}$
$= \frac{19 m r ^{2}}{512}$
$I_{remaining} = \frac{m r ^{2}}{2} - \frac{19}{512} m r^{2}$
$= \frac{237}{512} m r^{2}$

Q. A circular hole of radius 4R​ is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

256219MR2​

512237MR2​

51219MR2​

256197MR2​