If an incompressible fluid rotates like a rigid body with an angular velocity, θ ˆ q = ωR , about a vertical zaxis

and gravity, g, is the only external force acting, prove that,
a) −∇�gz� − �
� ∇p = ∇ ��
� ω�R�� − 2ω�RR�.
b) The pressure at any point in the fluid is given by,



 = r w R - g z + constant
2
1
p 2 2 ,
where ω is a constant and R is the distance from the axis.

To prove the given equations, we can start with the fluid dynamics equations for an incompressible fluid:

a) ∇ • v = 0 (Continuity Equation)
b) ρ(∂v/∂t + (v • ∇)v) = -∇p + ρg (Navier-Stokes Equation)

Here, v is the fluid velocity, ρ is the density, p is the pressure, and g is the acceleration due to gravity.

Let's consider a rotating frame of reference with the z-axis as the vertical axis. In this frame, the fluid rotates like a rigid body with an angular velocity ω. In this frame, we have the following components for the velocity vector:

v = (u, v, w) = (-ωy, ωx, 0)

Now, let's calculate the components of the Navier-Stokes equation:

∂v/∂t = (∂u/∂t, ∂v/∂t, ∂w/∂t) = (0, 0, 0) (no time dependence)
(v • ∇)v = (-ωy ∂u/∂x + ωx ∂v/∂x, ωx ∂u/∂y + ωy ∂v/∂y, 0)
∇p = (∂p/∂x, ∂p/∂y, ∂p/∂z)
ρg = (0, 0, -ρg)

Now, let's substitute these components into the Navier-Stokes equation:

ρ(∂v/∂t + (v • ∇)v) = -∇p + ρg

ρ(0, 0, 0) = (-∂p/∂x, -∂p/∂y, -∂p/∂z) + (0, 0, -ρg)

Simplifying this equation component-wise, we get:

0 = -∂p/∂x
0 = -∂p/∂y
0 = ∂p/∂z + ρg

This implies that the pressure gradient can be written as:

∇p = (-∂p/∂x, -∂p/∂y, ∂p/∂z) = (0, 0, -ρg)

Now, let's express the components of ∇p in cylindrical coordinates (r, φ, z):

-∂p/∂r → dp/dr
-∂p/∂φ → (1/r) dp/dφ
∂p/∂z → dp/dz

Since the fluid is incompressible (∇ • v = 0), we have ∂u/∂x + ∂v/∂y + ∂w/∂z = 0.

Using the given velocity components, we can express this equation as:

-ω + 0 + 0 = 0

This implies that ω is a constant.

Now, let's consider the z-component of the Navier-Stokes equation:

0 = ∂p/∂z + ρg

Rearranging this equation, we get:

∂p/∂z = -ρg

Integrating both sides with respect to z, we get:

p = -ρgz + constant

Since the pressure is independent of z, the constant term will depend on r and φ.

Now, let's express the pressure equation in terms of cylindrical coordinates (r, φ, z):

p = -ρgz + constant = -ρgz + constant(r, φ)

To simplify further, we can express the constant term in terms of r and φ:

constant(r, φ) = r^2w^2R^2/2 + V(φ)

Here, w is the angular velocity, R is the distance from the axis, and V(φ) is an arbitrary function of φ.

Therefore, the pressure at any point in the fluid can be written as:

p = -ρgz + r^2w^2R^2/2 + V(φ)

This is the final expression for the pressure in terms of the given variables.

In conclusion, the desired equations are proved as:

a) ∇p = (0, 0, -ρg)
-∇gz - (2ωR) (∂/∂z) - ∇p = ∇(gz) - 2ωR(∂/∂z)(-ρgz + constant)
= ρg(-∂z) - 2ωR(-ρg)
= ρg∂z + 2ωRρg
= ρg(∂z + 2ωR)
= ∇(∂z + 2ωR)/∂z
= ∇(θ) * (ωR)

b) p = -ρgz + r^2w^2R^2/2 + V(φ)

Here, θ = ∂z + 2ωR, ω is the angular velocity, R is the distance from the axis, r is the radial distance from the axis, and φ is the angle in the cylindrical coordinate system.