From the geometry of the problem,
ω(θ)=2tan1[ W2y2+h2 ]=2tan1[ 12μsinθ ]2tan1μθ2.
Retaining the first term only in the power series for /, we find that, as θ → 0,
A2tan1W2h.

Therefore,

  • A2tan11.43.60.7rad40. Note that the approximation tan−1 xx for small values of x works quite well in this case.

  • A2tan11.42.61rad57.

  • As h → 0, Aπ = 180°! (It’s a very flat mouse!)

The estimate for Question 1(a) is somewhat...

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