Step 1: Lift Calculation
The total lift \( L \) on a lifting surface is related to the circulation distribution by the Kutta-Joukowski theorem: \[ L = \rho V_\infty \int_{-b/2}^{b/2} \Gamma(y) \, dy \] However, in this problem, the circulation distribution \( \Gamma(\theta) = A \sin 3\theta \) is given with odd symmetry (because of the \( \sin 3\theta \) term), and when integrated over the span, the total circulation results in zero: \[ \int_{-b/2}^{b/2} \Gamma(y) \, dy = 0 \] Therefore, the total lift \( L \) is zero.
Step 2: Induced Drag Calculation
The induced drag \( D_i \) is related to the downwash distribution, which is given by: \[ w(\theta) = V_\infty \left( \frac{3A \sin 3\theta}{\sin \theta} \right) \] Since the downwash is nonzero, the interaction between the circulation and the downwash will produce a nonzero induced drag. The induced drag \( D_i \) is given by: \[ D_i = \int_{-b/2}^{b/2} \frac{\Gamma(y) w(\theta)}{V_\infty} \, dy \] This results in a nonzero induced drag because the downwash \( w(\theta) \) is nonzero and varies along the span. Thus, \( L = 0 \) and \( D_i \neq 0 \).
The lift per unit span for a spinning circular cylinder in a potential flow is 6 N/m. The free-stream velocity is 30 m/s, and the density of air is 1.225 kg/m\(^3\). The circulation around the cylinder is __________ m\(^2\)/s (rounded off to two decimal places).