# Does a rigid body rotates in an irrotational flow?

Consider a given flow field such that vorticity $$\mathbf{\omega} = \text{Curl} ~\mathbf{u} = 0$$. In this case, we can consider an arbitrary shape fluid element and look at how it evolves in short time $$\Delta t$$ to check if the element deformed/rotated/translated in the given time $$\Delta t$$. I could verify this for several flow situations.

Now, let there be a finite sized solid body immersed in the flow field such $$\mathbf{\omega} = \text{Curl} ~\mathbf{u} = 0$$. Is it possible that the rigid body rotates? The differences in the two cases are 1) the rigid body is finite sized instead of being a small control volume 2) At the surface of solid and fluid there will be either free-slip (Potential type flow) or no-slip (viscous flow) boundary condition. For potential type of flow only the pressure forces are acting on the surface the rigid body. These forces are always normal to the surface. Since there are no tangential forces for the potential flow, the rotation of the rigid body seems unlikely. But then for the viscous flow, shear stress will generate tangential forces.

I feel that if the flow is not rotational, it should not give rise to any rotation if a solid body is gently placed inside the fluid (independent of the viscous/potential type). is it true? Any reference will be helpful.

• There exists an irrotational flow, called irrotational vortex, which (believe it or not) rotates around the origin. Friction at the surface of a body placed around the origin should lead to rotation. Commented Aug 29, 2023 at 7:52
• @KurtG., thanks for comment. In the irrotational vortex, rigid body does not change the direction of normal. for example, see youtube.com/watch?v=glfm3NMMxh0 Commented Aug 29, 2023 at 13:02
• Why should I watch youtube videos? You asked: "Is it possible that the rigid body rotates?" Why does a rotation require to change the direction of normal, whatever that means? Commented Aug 29, 2023 at 13:44

If a disk $$D$$ of radius one (wlog) is immersed in the irrotational flow $$\mathbf{u}$$ and there is friction between the fluid and the surface of the disk then at a point of the disk's boundary there is a tangential force whose magnitude is proportional to $$\mathbf{u}\cdot\mathbf{t}$$ where $$\mathbf{t}$$ is the unit tangent at the disk's boundary. That unit tangent is related to the unit normal at the disk's boundary by $$\mathbf{t}={-n_2\choose n_1}\,,\quad\mathbf{n}={n_1\choose n_2}\,.$$ The magnitude of the total torque acting on the disk is then proportional to $$\oint_{\partial D}\mathbf{u}\cdot\mathbf{t}\,ds=\oint_{\partial D}u_1\,dx+u_2\,dy\,.$$ By Green's theorem this is equal to $$\int_D(\partial_xu_2-\partial_yu_1)\,dx\,dy$$ which is zero when $${\rm curl}\,\mathbf{u}\equiv 0\,.$$

• The disk does not rotate.

When $$\mathbf{u}$$ is the irrotational vortex $$\mathbf{u}=\frac1{x^2+y^2}{-y\choose x}$$ then it has a singularity at the origin and Green's theorem is not applicable when the disk contains the origin.

This is the situation when the object placed into the center of the vortex rotates (regardless if the latter is irrotational or not) as shown in the video.

• thanks for the answer. although it does not include coupling between solid and fluid (in the sense that how given imposed fluid flow will change because of solid body presence), nice to see the connection between Green's identity and the curl of velocity field. Commented Aug 30, 2023 at 14:08

I wanted to place this as a comment, but I do not have the necessary reputation yet.

For an irrotational flow, the net local angular momentum should be zero. So, if a large size stationary object is placed in such a flow, the net angular momentum must still be conserved. For the object to rotate, the flow should rotate in the reverse direction for this. But since the flow is irrotational, that cannot happen.

So, I would say that the immersed stationary object (w.r.t. the flow) will not rotate in an irrotational flow