 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/11c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let the heavier block move downward, so no slip gives linear acceleration

a \overset{ω}{˙}

T_{1} - M g = M a \overset{ω}{˙}, 2 M g - T_{2} = 2 M a \overset{ω}{˙},

(T_{2} - T_{1}) a - λ M ω = qM a^{2} \overset{ω}{˙} .

Eliminating the tensions gives the first-order equation

(q + 3) a^{2} \overset{ω}{˙} + λω = g a .

With

ω (0) = 0

, its solution is

ω (t) = \frac{g a}{λ} (1 - e^{- λ t / [(q + 3) a^{2}]}) .

For a thin slice of the solid of revolution,

d M = ρ πb (x)^{2} d x

and its axial moment of inertia is

d I = \frac{1}{2} b (x)^{2} d M

. Consequently

q = \frac{I}{M a ^{2}} = \frac{1}{2} \frac{\int _{- 1}^{1} ( 1 + ∣ x ∣ ) ^{4} d x}{\int _{- 1}^{1} ( 1 + ∣ x ∣ ) ^{2} d x} = \frac{1}{2} \frac{62/5}{14/3} = \frac{93}{70} .

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