Coin race

Conservation of mechanical energy for the rolling coin suggests that $mgy = \tfrac12mv^2 + \tfrac12I\omega^2 = \tfrac12m(1+\gamma)v^2,$ where $I = \gamma m r^2$ is the expression for the coin's moments of inertia. Therefore the time needed to go down the incline is proportional to $t \propto \frac 1 v \propto \sqrt{1+\gamma}.$ For the solid coin we have $\gamma_0 = \tfrac12$ .

Let $R$ be the radius of the second coin, $d$ its thickness, and define $m_0 = \pi d R^2 \rho_B$ . Then for the masses of the brass and gold parts we have $m_B = \pi d (0.9R)^2 \rho_B = 0.81\:m_0;$ $m_G = \pi d (R^2 - (0.9R)^2) \rho_G = (1 - 0.81)\cdot 2.28\:m_0 = 0.4332\:m_0;$ $m = m_B + m_G = 1.2432\:m_0.$ For their moments of inertia, use $I = \tfrac12m(R_{\text{inner}}^2 + R_{\text{outer}}^2)$ for a thick ring: $I_B = \tfrac12m_B(0.9R)^2 = \tfrac12\cdot 0.81\cdot 0.81\:m_0R^2 = 0.3281\:m_0R^2;$ $I_G = \tfrac12m_G((0.9R)^2+R^2) = \tfrac12\cdot 0.4332\cdot 1.81\:m_0R^2 = 0.3920\:m_0R^2;$ $I = I_B + I_G = 0.7201\:m_0 R^2.$ We see that $\gamma = \frac{I}{mR^2} = \frac{0.7201}{1.2432} = 0.5792,$ so that $\frac{t}{t_0} = \sqrt{\frac{1+\gamma}{1+\gamma_0}} = \sqrt{\frac{1.5792}{1.5000}} = 1.0261$ and $t = 1.0261\cdot\SI{2.00}{s} = \SI{2.05}{s}$ .