A thin disc of mass M and radius R has mass per unit area σ(r)=kr2, where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is
A
MR22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
MR26
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
MR23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2MR23
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D2MR23 Given, surface mass density σ=kr2
So, mass of the disc can be calculated by considering small element of area 2rπdr on it and then integrating it for complete disc, i.e,
Moment of inertia about the axis of the disc, I=∫dI=∫dmr2=∫σdAr2 =∫R0kr2(2πrdr)r2 ⇒I=2πk∫R0r5dr=2πkR66=πkR63....(ii)
From Eqs. (i) and (ii), we get I=23MR2