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Physics Question on System of Particles & Rotational Motion

A thin disc of mass MM and radius RR has mass per unit area σ(r)=kr2\sigma (r) = kr^2 where rr 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

MR26\frac{MR^2}{6}

B

MR23\frac{MR^2}{3}

C

2MR23\frac{2MR^2}{3}

D

MR22\frac{MR^2}{2}

Answer

2MR23\frac{2MR^2}{3}

Explanation

Solution

IDisc=∫0R(dm)r2⇒IDisc=∫0R(σ2πrdr)r2I_{Disc } =\int^{R}_{0} \left(dm\right)r^{2} \Rightarrow I_{Disc } =\int^{R}_{0} \left(\sigma2\pi rdr\right)r^{2}
IDisc=∫0R(kr22πrdr)r2I_{Disc } = \int^{R}_{0} \left(kr^{2} 2\pi rdr\right)r^{2}
IDisc=2πk∫0Rr5drM=∫0R2πrdrkr2I_{Disc } =2\pi k \int^{R}_{0} r^{5} dr M = \int^{R}_{0} 2\pi rdr kr^{2}
IDics=2πk(r66)0RM=2πk∫0Rr3drI_{Dics } = 2\pi k\left(\frac{r^{6}}{6}\right)^{R}_{0} M = 2\pi k \int^{R}_{0} r^{3}dr
IDisc=2πkR66M=2πkr44∣0RI_{Disc } = 2\pi k \frac{R^{6}}{6} M = 2\pi k \frac{r^{4}}{4} \bigg|^{R}_{0}
IDisc=πkR63=(πkR42)R223M=2πkR44I_{Disc } =\frac{\pi kR^{6}}{3} = \left(\frac{\pi kR^{4}}{2}\right) \frac{R^{2}2}{3} M = 2\pi k \frac{R^{4}}{4}
IDisc=M2R23I_{Disc } = \frac{M2R^{2}}{3}
IDisc=23MR2I_{Disc} = \frac{2}{3} MR^{2}