= Solution
For rolling without slipping, $\dot s=a\Omega$. <Conservation of energy> gives
$$
Mg\,s\sin\alpha=\frac12M\dot s^2+\frac12I\frac{\dot s^2}{a^2},
$$
and differentiation yields
$$
\boxed{\ddot s=\frac{g\sin\alpha}{1+I/(Ma^2)}.}
$$
For a sphere with density proportional to $r^c$,
$$
M=4\pi\rho_0\frac{a^{c+3}}{c+3},\qquad
I=\frac{8\pi\rho_0}{3}\frac{a^{c+5}}{c+5},
$$
so $I/(Ma^2)=2(c+3)/(3(c+5))$. A uniform disc has $I/(Ma^2)=1/2$. Equal launch speeds give equal return times exactly when the accelerations, and hence these ratios, agree. Solving gives \b[$c=3$].
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