= Solution
Let the heavier block move downward, so no slip gives linear acceleration $a\dot\omega$. Newton's equations and the pulley <torque> equation are
$$
T_1-Mg=Ma\dot\omega,\qquad
2Mg-T_2=2Ma\dot\omega,
$$
$$
(T_2-T_1)a-\lambda M\omega=qMa^2\dot\omega.
$$
Eliminating the tensions gives the first-order equation
$$
\boxed{(q+3)a^2\dot\omega+\lambda\omega=ga}.
$$
With $\omega(0)=0$, its solution is
$$
\boxed{\omega(t)=\frac{ga}{\lambda}
\left(1-e^{-\lambda t/[(q+3)a^2]}\right)}.
$$
For a thin slice of the solid of revolution, $dM=\rho\pi b(x)^2dx$ and its axial <moment of inertia> is $dI=\tfrac12b(x)^2dM$. Consequently
$$
q=\frac{I}{Ma^2}
=\frac12\frac{\int_{-1}^1(1+|x|)^4dx}
{\int_{-1}^1(1+|x|)^2dx}
=\frac12\frac{62/5}{14/3}
=\boxed{\frac{93}{70}}.
$$
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