Question
Physics Question on System of Particles & Rotational Motion
A thin circular plate of mass M and radius R has its density varying as ρ(r)=ρ0r with ρ0 as constant and r 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=aMR2. The value of the coefficient a is :
A
23
B
21
C
53
D
58
Answer
58
Explanation
Solution
M=∫0Rρ0r(2πrdr)=3ρ0×2π×R3
{\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+MR2
=5ρ0×2πR5+3ρ0×2πR3×R2=ρ02πR5×158
=MR2×58