OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 8B / a / Solution

At height

z

from the vertex, the cone has cross-sectional area

π R^{2} z^{2} / l^{2}

. Hence

z_{CM} = \frac{\int _{0}^{l} z ^{3} d z}{\int _{0}^{l} z ^{2} d z} = \frac{3 l}{4}, r_{CM} = a e_{3}, a = \frac{3 l}{4} .

The parallel-axis theorem is

I_{O} = I_{CM} + M {∣ ma t hb f a ∣^{2} 1 - a a^{T}} .

The displacement is along

e_{3}

, so it changes the two transverse moments but not the axial one. Thus, writing the vertex moments as

I_{1} = I_{2} = I

and

I_{3} = J

,

I = I_{1}^{CM} + M a^{2} = \frac{π R ^{2} lρ}{20} (R^{2} + 4 l^{2}) = \frac{3 M}{20} (R^{2} + 4 l^{2}),

and

J = I_{3}^{CM} = \frac{π R ^{4} lρ}{10} = \frac{3 M R ^{2}}{10} .

Solved by gpt-5.6-sol high.

 Ancestors (11)