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Since the flow is irrotational, velocity vector may be written as $\mathbf{u}=\nabla\phi$, where $\phi$ is a scalar potential. The cylinder moves with constant velocity $\mathbf{U}$ (vectors are denoted by bold-face font). WhichAt the cylinder's surface, normal velocity of both cylinder and fluid must be the same. This means that if $\mathbf{n}$ is the normal vector at the cylinder's surface then $\mathbf{u}=\nabla\phi=\mathbf{U}$$\mathbf{n}\cdot\mathbf{u}=\mathbf{n}\cdot\nabla\phi=\mathbf{n}\cdot\mathbf{U}$.

enter image description here

Consider a cylindrical coordinate system whose origin is instantaneously coincident with the center of the cylinder. Now the normal to the surface of the cylinder is the unit radial vector $\mathbf{e}_r$. Forming the scalar product ofTherefore, at the previous expression with $\mathbf{e}_r$cylinder's surface, we gethave: $$\mathbf{e}_r\cdot\nabla\phi=\mathbf{e}_r\cdot\mathbf{U}\\\Rightarrow\quad\frac{\partial\phi}{\partial r}=U\cos\theta$$ in which $U$ is the magnitude of cylinder's velocity.

Since the flow is irrotational, velocity vector may be written as $\mathbf{u}=\nabla\phi$, where $\phi$ is a scalar potential. The cylinder moves with constant velocity $\mathbf{U}$ (vectors are denoted by bold-face font). Which means that at the cylinder's surface $\mathbf{u}=\nabla\phi=\mathbf{U}$.

enter image description here

Consider a cylindrical coordinate system whose origin is instantaneously coincident with the center of the cylinder. Now the normal to the surface of the cylinder is the unit radial vector $\mathbf{e}_r$. Forming the scalar product of the previous expression with $\mathbf{e}_r$ we get: $$\mathbf{e}_r\cdot\nabla\phi=\mathbf{e}_r\cdot\mathbf{U}\\\Rightarrow\quad\frac{\partial\phi}{\partial r}=U\cos\theta$$ in which $U$ is the magnitude of cylinder's velocity.

Since the flow is irrotational, velocity vector may be written as $\mathbf{u}=\nabla\phi$, where $\phi$ is a scalar potential. The cylinder moves with constant velocity $\mathbf{U}$ (vectors are denoted by bold-face font). At the cylinder's surface, normal velocity of both cylinder and fluid must be the same. This means that if $\mathbf{n}$ is the normal vector at the cylinder's surface then $\mathbf{n}\cdot\mathbf{u}=\mathbf{n}\cdot\nabla\phi=\mathbf{n}\cdot\mathbf{U}$.

enter image description here

Consider a cylindrical coordinate system whose origin is instantaneously coincident with the center of the cylinder. Now the normal to the surface of the cylinder is the unit radial vector $\mathbf{e}_r$. Therefore, at the cylinder's surface, we have: $$\mathbf{e}_r\cdot\nabla\phi=\mathbf{e}_r\cdot\mathbf{U}\\\Rightarrow\quad\frac{\partial\phi}{\partial r}=U\cos\theta$$ in which $U$ is the magnitude of cylinder's velocity.

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Deep
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  • 25

Since the flow is irrotational, velocity vector may be written as $\mathbf{u}=\nabla\phi$, where $\phi$ is a scalar potential. The cylinder moves with constant velocity $\mathbf{U}$ (vectors are denoted by bold-face font). Which means that at the cylinder's surface $\mathbf{u}=\nabla\phi=\mathbf{U}$.

enter image description here

Consider a cylindrical coordinate system whose origin is instantaneously coincident with the center of the cylinder. Now the normal to the surface of the cylinder is the unit radial vector $\mathbf{e}_r$. Forming the scalar product of the previous expression with $\mathbf{e}_r$ we get: $$\mathbf{e}_r\cdot\nabla\phi=\mathbf{e}_r\cdot\mathbf{U}\\\Rightarrow\quad\frac{\partial\phi}{\partial r}=U\cos\theta$$ in which $U$ is the magnitude of cylinder's velocity.