 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-4/11c/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The inertial acceleration obtained by differentiating the position used in part (i) is

\overset{r}{¨} = R (\ddot{θ} cos θ - \dot{θ}^{2} sin θ - ω^{2} sin θ) e_{r} + 2 R ω \dot{θ} cos θ e_{ϕ} + R (\ddot{θ} sin θ + \dot{θ}^{2} cos θ) e_{z} .

Since

N = m \overset{r}{¨} + m g e_{z}

, substitution of the equation of motion gives the wire's force on the bead as

N = m (R \dot{θ}^{2} + R ω^{2} sin^{2} θ + g cos θ) (- sin θ e_{r} + cos θ e_{z}) + 2 m R ω \dot{θ} cos θ e_{ϕ} .

The first component is normal to the circle within its plane; the second is perpendicular to the rotating plane.

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