Discussion
Given \( C_{m_q} = -12 \), \( \bar q = 4000 \ \mathrm{Pa} \), \( S = 30 \ \mathrm{m^2} \), \( \bar c = 3 \ \mathrm{m} \), \( V = 150 \ \mathrm{m/s} \) and \( I_{yy} = 100\,000 \ \mathrm{kg\,m^2} \), the dimensional \( M_q = C_{m_q}\bar q S \bar c^{2}/(2VI_{yy}) \) is about
Answer: D
Working through the conversion:
\[ \begin{aligned} \bar q S \bar c^{2} &= 4000 \times 30 \times 9 = 1.08 \times 10^{6} \\ 2VI_{yy} &= 3.0 \times 10^{7} \\ M_q &= -12 \times 0.036 = -0.43 \ \mathrm{s^{-1}} \end{aligned} \]
so -0.43 per second. -0.86 omits the factor of two. -12 is the coefficient itself, which has no units of inverse time and cannot enter the characteristic equation. -0.14 divides by the chord once instead of squaring it, which is the commonest slip in this particular conversion.
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