Rotation 1 Question 3

3. 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

(2019 Main, 10 April I)

(a) MR22

(b) MR26

(c) MR23

(d) 2MR23

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Answer:

Correct Answer: 3. (d)

Solution:

  1. Given, Surface mass density, σ=kr2

So, mass of the disc can be calculated by considering small element of area 2πrdr on it and then integrating it for complete disc, i.e.

dm=σdA=σ×2πrdrdm=M=0R(kr2)2πrdrM=2πkR44=12πkR4

Moment of inertia about the axis of the disc,

I=dI=dmr2=σdAr2=0Rkr2(2πrdr)r2I=2πk0Rr5dr=2πkR66=πkR63

From Eqs. (i) and (ii), we get

I=23MR2



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