Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 3 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii
- iii c
  - Solution iii

i

Solution

Local force balance and moment balance on a slender rod are

\partial_{s} F + f = 0, \partial_{s} M + t \times F + m = 0 .

In a planar rod,

M

and

m

point perpendicular to the plane.

ii

Solution

Substitute

f = - γ \cdot r_{t} + f^{A}

into force balance. With

μ = γ^{- 1}

,

r_{t} = μ \cdot (\partial_{s} F + f^{A}) .

Projection onto the local frame gives

W = t \cdot μ (\partial_{s} F + f^{A}), U = n \cdot μ (\partial_{s} F + f^{A}) .

iii

Solution

Resolve

F = F_{∥} t + F_{⊥} n .

The Frenet-Serret formulas give

\partial_{s} F = (\partial_{s} F_{∥} - k F_{⊥}) t + (\partial_{s} F_{⊥} + k F_{∥}) n .

Using

μ = μ_{∥} tt + μ_{⊥} nn, f^{A} = f_{∥}^{A} t + f_{⊥}^{A} n,

therefore yields

W = μ_{∥} (\partial_{s} F_{∥} - k F_{⊥} + f_{∥}^{A}),

U = μ_{⊥} (\partial_{s} F_{⊥} + k F_{∥} + f_{⊥}^{A}) .

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