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,


  • 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...

AAPT members receive access to The Physics Teacher and the American Journal of Physics as a member benefit. To learn more about this member benefit and becoming an AAPT member, visit the Joining AAPT page.