Question

# A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in figure. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is:

A
1532MR2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
716MR2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
1332MR2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
38MR2
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C 1332MR2Let I be the Moment of inertia of the disc about O without removal of the circular portion.Moment of inertia of remaining portion Iremainder=I−Icut−offI=12MR2 (using the formula for MI of a solid disc about the centre and perpendicular to the plane of the disc).Radius of the cut-off disc is Rcut−off=R2By using parallel axis theorem, Icut−off=I′O+(Mcut−off)d2where I′O is the Moment of inertia of cut-off disc about the axis passing through O' (which is cm) and d is the distance between the centres O and O'.Mass of cut-off portion Mcut−off=πρ(R2−(R2)2)=π×ρ×34R2∴ Moment of inertia of cut-off disc Icut−off=12Mcut−off×(R2)2=12πρ×34R2×(R2)2=3ρπ32R4∴Iremainder=12πρR2×R2−3πρ32R4=1332πρR2×R2=1332×(πρR2)×R2=1332MR2

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Integrating Solids into the Picture
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program