# Question #89cc5

Mar 4, 2017

We know that the moment of inertia $I$ of a disk is related with its mass $M$ and radius $R$ as follows

$\textcolor{b l u e}{I = \frac{1}{2} M {R}^{2.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left[1\right]}$

Now $M = \frac{4}{3} \pi {R}^{3} \mathrm{dr} h o$

where $m \to \text{mass" , d->"thickness" and rho ->"density}$

$\implies R = {\left(\frac{3 M}{4 \pi \mathrm{dr} h o}\right)}^{\frac{1}{3}}$

$\implies {R}^{2} = {\left(\frac{3 M}{4 \pi \mathrm{dr} h o}\right)}^{\frac{2}{3}}$

Hence equation [1] becomes

$\textcolor{g r e e n}{I = \frac{1}{2} M \times {\left(\frac{3 M}{4 \pi \mathrm{dr} h o}\right)}^{\frac{2}{3}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left[2\right]}$

Now if $M \mathmr{and} d$ are remaining constant as per given condition then

$\textcolor{red}{I \propto \frac{1}{\rho} ^ \left(\frac{2}{3}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left[3\right]}$

If the moment of inertia of first disk of density ${\rho}_{1} = 7.2 \text{g/} c {m}^{3}$ be ${I}_{1}$ and the moment of inertia of 2nd disk of density ${\rho}_{2} = 8.9 \text{g/} c {m}^{3}$ be ${I}_{2}$ then the ratio of moment of inertia of two disks is given by

$\textcolor{b l u e}{{I}_{1} / {I}_{2} = {\left({\rho}_{2} / {\rho}_{1}\right)}^{\frac{2}{3}} = {\left(\frac{8.9}{7.2}\right)}^{\frac{2}{3}} \approx 1.15}$