 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/11c/c/ii/solution

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

t

L = \frac{1}{2} m R^{2} ω .

During

d t

, the ejected mass is

- d m

, and its signed tangential speed in the inertial frame is

R ω + u

. Conservation of angular momentum gives, to first order,

0 = d (\frac{1}{2} m R^{2} ω) + (- d m) R (R ω + u) .

After simplification, the angular rocket equation is

m \overset{ω}{˙} = \overset{m}{˙} (ω + \frac{2 u}{R}) .

Here

u

is signed relative to the direction of rotation; reversing the spray changes its sign.

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