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? [2016]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 [2016]

1 $11 M R^{2} / 32$

2 $9 M R^{2} / 32$

3 $15 M R^{2} / 32$

4 $13 M R^{2} / 32$

Ans.
(4)

Mass per unit area of disc $= \frac{M}{π R^{2}}$

Mass of removed portion of disc,

$M' = \frac{M}{π R^{2}} \times π {(\frac{R}{2})}^{2} = \frac{M}{4}$

Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,

$I'_{O} = I_{o'} + M' d^{2}$

$= \frac{1}{2} \times \frac{M}{4} \times {(\frac{R}{2})}^{2} + \frac{M}{4} \times {(\frac{R}{2})}^{2}$

$= \frac{M R^{2}}{32} + \frac{M R^{2}}{16} = \frac{3 M R^{2}}{32}$

The moment of inertia of complete disc about centre O before removing the portion of the disc

$I_{O} = \frac{1}{2} M R^{2}$

So, moment of inertia of the disc with removed portion is

$I = I_{O} - I'_{O} = \frac{1}{2} M R^{2} - \frac{3 M R^{2}}{32} = \frac{13 M R^{2}}{32}$

Want Us to Email you About Special Offers & Updates?