Questão 8. No desenvolvimento do problema do camber, precisamos obter a distribuição da intensidade da lâmina de vórtice \gamma_{c}(\theta) tal que
\frac{1}{\pi} \int_{0}^{c} \frac{\gamma_{c}\left(\theta_{0}\right)}{2 U_{\infty}} \frac{\operatorname{sen} \theta}{\cos \theta_{0}-\cos \theta} d \theta_{0}=\frac{d \eta_{c}}{d x}(\theta)=\frac{B_{0}}{2}+\sum_{n=1}^{\infty} B_{n} \cos n \theta
Para isso, decompomos \gamma_{c}=\gamma_{c, 1}+\gamma_{c, 2}+\gamma_{c, 3}. Mostre que, se
\begin{gathered} \frac{1}{\pi} \int_{0}^{c} \frac{\gamma_{c, 1}\left(\theta_{0}\right)}{2 U_{\infty}} \frac{\operatorname{sen} \theta_{0}}{\cos \theta_{0}-\cos \theta} d \theta_{0}=\sum_{n=1}^{\infty} B_{n} \cos n \theta \\\ \frac{1}{\pi} \int_{0}^{c} \frac{\gamma_{c, 2}\left(\theta_{0}\right)}{2 U_{\infty}} \frac{\operatorname{sen} \theta_{0}}{\cos \theta_{0}-\cos \theta} d \theta_{0}=\frac{B_{0}}{2} \\\ \frac{1}{\pi} \int_{0}^{c} \frac{\gamma_{c, 3}\left(\theta_{0}\right)}{2 U_{\infty}} \frac{\operatorname{sen} \theta_{0}}{\cos \theta_{0}-\cos \theta} d \theta_{0}=0 \end{gathered}
(a) \gamma_{c, 1}(\theta)=-2 U_{\infty} \sum_{n=1}^{\infty} B_{n} \operatorname{senn} \theta.
Dica: Reescreva \cos n \theta do lado direito em termos de uma das duas integrais mostradas na sala e compare os integrandos dos dois lados.
(b) \gamma_{c, 2}(\theta)=U_{\infty} B_{0}(\cos \theta / \operatorname{sen} \theta).
Dica: Reescreva o lado direito observando que 0,5 B_{0}=0,5 B_{0} \cdot 1 e escreva 1 em termos de uma das duas integrais mostradas na sala e compare os integrandos dos dois lados.
(c) \gamma_{c, 3}(\theta)=2 U_{\infty} K / \operatorname{sen} \theta, sendo K uma constante arbitrária.
Dica: Reescreva o lado direito observando que 0=0 \cdot K e escreva 0 em termos de uma das duas integrais mostradas na sala e compare os integrandos dos dois lados.
(d) Para que a \gamma_{c}(\theta=0)=\gamma_{c, 1}(\theta=0)+\gamma_{c, 2}(\theta=0)+\gamma_{c, 3}(\theta=0)=0, K=-0,5 B_{0}.

Passo 1

O enunciado pede para decompor
\gamma_{c}(\theta)=\gamma_{c,1}(\theta)+\gamma_{c,2}(\theta)+\gamma_{c,3}(\theta)
de modo que cada parcela satisfaça, via o núcleo \displaystyle \frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,\bigl[\cdot\bigr]\,d\theta_{0},
as igualdades
\frac{1}{\pi}\int_{0}^{c}\frac{\gamma_{c,1}(\theta_{0})}{2U_{\infty}} \frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,d\theta_{0} =\sum_{n=1}^{\infty}B_{n}\cos n\theta,
\frac{1}{\pi}\int_{0}^{c}\frac{\gamma_{c,2}(\theta_{0})}{2U_{\infty}} \frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,d\theta_{0} =\frac{B_{0}}{2},
\frac{1}{\pi}\int_{0}^{c}\frac{\gamma_{c,3}(\theta_{0})}{2U_{\infty}} \frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,d\theta_{0} =0.
Para isso usamos as identidades (vistase em sala):
\text{(I)}\quad \frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta} \sin n\theta_{0}\,d\theta_{0}=-\cos n\theta, \quad n\ge1,
\text{(II)}\quad \frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta} \,d\theta_{0}=0,
\text{(III)}\quad \frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}\,\cos\theta_{0}} {\cos\theta_{0}-\cos\theta}\,d\theta_{0}=1.
A partir daí comparamos integrandos para determinar cada \gamma_{c,i}.

Passo 2

Queremos
\sum_{n=1}^{\infty}B_{n}\cos n\theta = \frac{1}{\pi}\int_{0}^{c}\frac{\gamma_{c,1}(\theta_{0})}{2U_{\infty}} \frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,d\theta_{0}.
Pela identidade (I) temos
-\cos n\theta =\frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta} \sin n\theta_{0}\,d\theta_{0}.
Logo
\sum_{n=1}^{\infty}B_{n}\cos n\theta =\frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta} \Bigl[-\sum_{n=1}^{\infty}B_{n}\sin n\theta_{0}\Bigr]\,d\theta_{0}.
Comparando com o lado direito, identificamos
\frac{\gamma_{c,1}(\theta_{0})}{2U_{\infty}} =-\sum_{n=1}^{\infty}B_{n}\sin n\theta_{0} \;\;\Longrightarrow\;\; \gamma_{c,1}(\theta) =-2U_{\infty}\sum_{n=1}^{\infty}B_{n}\sin n\theta.

Passo 3

Queremos
\frac{B_{0}}{2} =\frac{1}{\pi}\int_{0}^{c}\frac{\gamma_{c,2}(\theta_{0})}{2U_{\infty}} \frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,d\theta_{0}.
Escrevemos B_{0}/2=(B_{0}/2)\cdot1 e usamos (III):
1 =\frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}\cos\theta_{0}} {\cos\theta_{0}-\cos\theta}\,d\theta_{0}.
Assim
\frac{B_{0}}{2} =\frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta} \Bigl[\frac{B_{0}}{2}\,\frac{\cos\theta_{0}}{\sin\theta_{0}}\Bigr]\,d\theta_{0}
e, comparando,
\frac{\gamma_{c,2}(\theta_{0})}{2U_{\infty}} =\frac{B_{0}}{2}\,\frac{\cos\theta_{0}}{\sin\theta_{0}} \;\;\Longrightarrow\;\; \gamma_{c,2}(\theta) =U_{\infty}B_{0}\,\frac{\cos\theta}{\sin\theta}.

Passo 4

Queremos
0 =\frac{1}{\pi}\int_{0}^{c}\frac{\gamma_{c,3}(\theta_{0})}{2U_{\infty}} \frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}\,d\theta_{0}.
Mas pela identidade (II) \frac{1}{\pi}\int_{0}^{c}\frac{\sin\theta_{0}}{\cos\theta_{0}-\cos\theta}d\theta_{0}=0.
Basta ter \gamma_{c,3}(\theta_{0})/(2U_{\infty})=K constante, o que dá
\gamma_{c,3}(\theta)=2U_{\infty}\,\frac{K}{\sin\theta}.

Passo 5

Para impor \gamma_{c}(0)=0 usamos \gamma_{c}(0)=\gamma_{c,1}(0)+\gamma_{c,2}(0)+\gamma_{c,3}(0).
Como \gamma_{c,1}(0)=0 e \gamma_{c,2},\gamma_{c,3}\sim1/\theta no limite \theta\to0, somam-se os coeficientes de 1/\theta:
U_{\infty}B_{0}+2U_{\infty}K=0 \;\;\Longrightarrow\;\; K=-\tfrac{1}{2}B_{0}.

Resposta

a) \displaystyle \gamma_{c,1}(\theta)=-2U_{\infty}\sum_{n=1}^{\infty}B_{n}\sin n\theta.
b) \displaystyle \gamma_{c,2}(\theta)=U_{\infty}B_{0}\,\frac{\cos\theta}{\sin\theta}.
c) \displaystyle \gamma_{c,3}(\theta)=2U_{\infty}\,\frac{K}{\sin\theta}.
d) \displaystyle K=-\tfrac{1}{2}B_{0}.

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