Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 11C / Solution

Let the heavier block move downward, so no slip gives linear acceleration $a\dot{\omega }$. Newton's equations and the pulley torque equation areEliminating the tensions gives the first-order equationWith $\omega (0)=0$, its solution is

For a thin slice of the solid of revolution, $dM=\rho \pi b(x{)}^{2}dx$ and its axial moment of inertia is $dI=\frac{1}{2}b(x{)}^{2}dM$. Consequently