 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/7/b/solution

θ_{i}

from an unstressed configuration. The diagram has one torsional spring joining the frame to object 1, one joining objects 1 and 2, and one joining objects 2 and 3. The rotational kinetic energy and elastic potential energy are

T = \frac{1}{2} \sum_{i = 1}^{3} I_{i} \dot{θ}_{i}^{2}, U = \frac{1}{2} k_{1} θ_{1}^{2} + \frac{1}{2} k_{2} (θ_{2} - θ_{1})^{2} + \frac{1}{2} k_{3} (θ_{3} - θ_{2})^{2} .

Each spring contribution follows by integrating its restoring torque:

\int_{0}^{Δ θ} k_{j} s d s = k_{j} (Δ θ)^{2} /2

. Thus

L = T - U .

I_{1} \ddot{θ}_{1} + (k_{1} + k_{2}) θ_{1} - k_{2} θ_{2} I_{2} \ddot{θ}_{2} - k_{2} θ_{1} + (k_{2} + k_{3}) θ_{2} - k_{3} θ_{3} I_{3} \ddot{θ}_{3} - k_{3} θ_{2} + k_{3} θ_{3} = 0, = 0, = 0.

These hold for either interpretation of the square's moment of inertia; the geometry-based value from (a) is used first below.

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