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Question

Physics Question on System of Particles & Rotational Motion

A thin circular plate of mass MM and radius RR has its density varying as ρ(r)=ρ0r\rho (r) = \rho_0 r with ρ0\rho_0 as constant and rr is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I=aMR2I = aMR^2. The value of the coefficient aa is :

A

32\frac{3}{2}

B

12\frac{1}{2}

C

35\frac{3}{5}

D

85\frac{8}{5}

Answer

85\frac{8}{5}

Explanation

Solution

M=0Rρ0r(2πrdr)=ρ0×2π×R33M= \int^{R}_{0} \rho_{0} r \left(2\pi rdr\right) = \frac{\rho_{0} \times2 \pi \times R^{3}}{3}
{\underset{\text{ (MOI about COM) }}{{ I_0 }}= \int^{R}_{0} \rho_{0} r (2 \pi rdr) \times r^{2} = \frac{\rho_{0} \times2\pi R^{5}}{5} }
by parallel axis theorem
I=I0+MR2I = I_{0} + MR^{2}
=ρ0×2πR55+ρ0×2πR33×R2=ρ02πR5×815= \frac{\rho_{0} \times2 \pi R^{5}}{5} + \frac{\rho_{0} \times2\pi R^{3}}{3} \times R^{2} = \rho_{0} 2\pi R^{5} \times\frac{8}{15}
=MR2×85= MR^{2} \times\frac{8}{5}