Question

For a rotational flow, the ratio of normal velocity vector to the angular velocity vector is

• A. 1.0
• B. 2.0
• C. 0.667
• D. 0.5

Hint:

In fluid dynamics, the vorticity \( \vec{\zeta} \) is defined as the curl of the velocity vector \( \nabla \times \mathbf{V} \). The angular velocity \( \vec{\omega} \) of a fluid particle is exactly half of its vorticity: \( \vec{\omega} = \frac{1}{2} \vec{\zeta} \). This means \( |\vec{\zeta}| = 2 |\vec{\omega}| \). This fundamental relationship is key to understanding rotational flows.

Answer 1

The Correct Option is B

Step 1: Understand the velocity distribution in rotational flow.
In a rotational flow (solid body rotation), the tangential velocity \( v \) at a radius \( r \) is given by: \[ v = \omega r \] where \( \omega \) is the angular velocity.
Step 2: Define vorticity.
Vorticity \( \vec{\zeta} \) in 2D flow is: \[ \zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \] But in cylindrical coordinates for rotational flow (where velocity is purely tangential): \[ \zeta = 2\omega \]
Step 3: Relate normal velocity gradient to vorticity.
The angular velocity is: \[ \omega = \frac{1}{2} \zeta \] So, the ratio of normal velocity vector (or tangential speed per unit radius \( \frac{v}{r} \)) to angular velocity is: \[ \frac{v/r}{\omega} = \frac{\omega r / r}{\omega} = \frac{\omega}{\omega} = 1 \] But the question likely refers to the ratio of the vorticity to angular velocity, which is: \[ \frac{\zeta}{\omega} = \frac{2\omega}{\omega} = 2 \] Hence, the required ratio is: \[ \boxed{2.0} \]