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A thin circular plate of mass M and radius R has its density varying $ρ (r) = ρ_{o} r$ with $ρ_{o}$ as constant and r is the distance from its center. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I = {aMR}^{2}$ . The value of the coefficient a is

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$\frac{3}{5}$

$\frac{8}{5}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

Mass of disc,

$\begin{array}{l} M = \int_{o}^{R} ρ_{o} r \times 2 πrdr = \frac{2 π ρ_{o} R^{3}}{3} \end{array}$

MOI about axis passing through center of disc,

$I_{c} = \int_{o}^{R} (ρ_{o} r \times 2 πrdr) r^{2} = \frac{2 π ρ_{o} R^{5}}{5}$

Using parallel axis theorem,

$I = I_{c} + {MR}^{2} = 2 π ρ_{o} R^{5} (\frac{1}{3} + \frac{1}{5}) = \frac{16 π ρ_{o} R^{5}}{15} = \frac{8}{5} {MR}^{2}$

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