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
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B
MR26
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C
MR23
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D
2MR23
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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, dm=σdA=σ×2πrdr ∫dm=M=∫R0(kr2)2πrdr ⇒M=2πkR44=12πkR4
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