A U-shaped tube contains a liquid of density rho and it is rotated about the vertical axis passing t

Q: Why does the cross-sectional area of the tube not affect the height difference?

The cross-sectional area $A$ cancels out because both the mass of the liquid element (which determines the required centripetal force) and the pressure force acting on it are directly proportional to $A$. Thus, the geometry of the tube's cross-section has no effect on the equilibrium height difference.

Q: Does the density of the liquid affect the height difference?

No, the density $\rho$ cancels out when we equate the centrifugal pressure difference to the hydrostatic pressure difference. Therefore, the height difference $H$ is the same whether you use water, mercury, or oil.

Q: What would happen if the axis of rotation was in the middle of the U-tube?

If the axis of rotation was exactly in the middle, both limbs would be at an equal distance $L/2$ from the axis. By symmetry, the liquid levels in both limbs would remain equal ($H = 0$).

Q: Is the shape of the liquid surface in the limbs flat?

In reality, the liquid surface in each limb is slightly curved (parabolic) due to rotation. However, if the limbs are narrow ($r \ll L$), we can assume the surface in each limb is approximately flat, and $H$ represents the difference between their average heights.

Q: How does this relate to the equation of a paraboloid?

In a rotating frame, the effective gravity is the vector sum of real gravity downwards and centrifugal acceleration outwards. The free surface of a rotating liquid is always perpendicular to this effective gravity, which mathematically results in a paraboloid of revolution described by $y = \frac{\omega^2 x^2}{2g}$.

Question

A U-shaped tube contains a liquid of density $\rho$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing next to a U-shaped tube filled with a liquid of density $\rho$.

When the tube is at rest, the liquid levels in both vertical limbs are perfectly equal due to hydrostatic equilibrium.

But what happens when we spin the tube about the vertical axis of its left limb with a constant angular velocity $\omega$?

As the tube rotates, the liquid in the horizontal connecting tube is thrown outward due to the centrifugal effect.

This causes the liquid level in the right limb to rise and the level in the left limb to fall, creating a height difference $H$.

Our goal is to find this height difference $H$ at equilibrium.

### The Master Equation: Horizontal Pressure Variation

To understand how the pressure varies horizontally, let us analyze a small horizontal element of liquid of length $dx$ at a distance $x$ from the axis of rotation.

Let $A$ be the cross-sectional area of the tube. The mass $dm$ of this tiny element is:

$$dm = \rho A dx$$

Now, let's look at the forces acting on this element in the horizontal direction.

The pressure on the left side of the element is $p$, and the pressure on the right side is slightly higher, $p + dp$.

This pressure difference creates a net force pointing towards the left (towards the axis of rotation):

$$dF = (p + dp)A - pA = dp \cdot A$$

This net force must provide the necessary centripetal force for the circular motion of the element:

$$dp \cdot A = (dm) x \omega^2$$

Substituting the mass $dm = \rho A dx$ into this equation:

$$dp \cdot A = (\rho A dx) x \omega^2$$

Notice how the cross-sectional area $A$ beautifully cancels out from both sides! We are left with:

$$dp = \rho \omega^2 x \, dx$$

### Integrating the Pressure Difference

To find the total pressure difference between the left limb ($x = 0$) and the right limb ($x = L$), we integrate this differential equation:

$$\int_{p_1}^{p_2} dp = \rho \omega^2 \int_{0}^{L} x \, dx$$

$$p_2 - p_1 = \rho \omega^2 \left[ \frac{x^2}{2} \right]_{0}^{L}$$

$$p_2 - p_1 = \frac{\rho \omega^2 L^2}{2}$$

This equation tells us that the pressure at the bottom of the right limb is higher than the pressure at the bottom of the left limb by $\frac{\rho \omega^2 L^2}{2}$ due to the rotation.

### Vertical Hydrostatic Equilibrium

Now, let's relate this pressure difference to the heights of the liquid columns in the vertical limbs.

The pressure at the bottom of the left limb ($p_1$) and the right limb ($p_2$) are given by hydrostatic pressure:

$$p_1 = p_0 + \rho g h_1$$

$$p_2 = p_0 + \rho g h_2$$

Subtracting these two equations gives:

$$p_2 - p_1 = \rho g (h_2 - h_1) = \rho g H$$

where $H = h_2 - h_1$ is the height difference between the two liquid columns.

### Final Calculation

Now, we equate our two independent expressions for the pressure difference $p_2 - p_1$:

$$\rho g H = \frac{\rho \omega^2 L^2}{2}$$

Notice that the density $\rho$ cancels out completely! This means the height difference is independent of the liquid used.

Solving for $H$, we get the final elegant result:

$$H = \frac{\omega^2 L^2}{2g}$$

This is a classic result in fluid dynamics, showing how rotational motion and gravity cooperate to shape the liquid surface.

Animated Solution for Physics - Properties of Solids and Liquids: A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

Visualized Solution

Understanding the Rotating U-Tube System

Analyzing a Small Liquid Element

Mass of the Element

Centripetal Force Equation

Substituting Mass into the Force Equation

Integrating to Find Horizontal Pressure Difference

Relating Pressure to Height Difference

Equating the Pressure Differences

Conceptual Takeaway

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

The Master Equation

Integrating the Pressure Difference

Vertical Hydrostatic Equilibrium

Final Calculation

Similar Questions

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d _{1}}{d _{2}}$ is

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $d_{1} / d_{2}$ is