A cylindrical vessel containing a liquid is rotated about its axis, so that the liquid rises at its

Q: Why do we use a rotating frame of reference instead of the ground frame?

Using the rotating frame allows us to treat the fluid particle as being at rest (in static equilibrium). This lets us use simple force balancing instead of dealing with centripetal acceleration and complex fluid dynamics in the ground frame.

Q: What exactly is the pseudo force $mx\omega^2$?

It's the centrifugal force! Whenever you are in a rotating (non-inertial) frame, every object experiences an outward push proportional to its mass, its distance from the axis of rotation ($x$), and the square of the angular velocity ($\omega^2$).

Q: Why is the slope $dy/dx$ equal to $\tan\theta$?

Geometrically, the derivative $dy/dx$ represents the slope of the tangent line to the curve at any point. By definition, the slope of a line is the tangent of the angle ($\theta$) it makes with the positive x-axis.

Q: Does the density of the liquid affect the height $h$?

Fascinatingly, no! If you look at our force balance equations, the mass $m$ (and thus the density) cancels out completely. A spinning cup of mercury will form the exact same paraboloid shape as a spinning cup of water at the same $\omega$.

Q: What happens if the vessel is spun extremely fast?

As $\omega$ increases, the height $h$ grows quadratically ($\omega^2$). Eventually, the liquid will reach the top rim of the vessel and start spilling over. The math perfectly predicts exactly when this spilling will occur!

Question

A cylindrical vessel containing a liquid is rotated about its axis, so that the liquid rises at its sides as shown in the figure. The radius of vessel is $5 	ext{ cm}$ and the angular speed of rotation is $\omega 	ext{ rad s}^{-1}$. The difference in the height $h$ (in cm) of liquid at the centre of vessel and at the side will be

Options:

(A) $\frac{25\omega^2}{2g}$

(B) $\frac{5\omega^2}{2g}$

(C) $\frac{2\omega^2}{25g}$

(D) $\frac{2\omega^2}{5g}$

Sigma Percentile · Accepted Answer

### The Spinning Vessel: A Thought Experiment

Imagine you are holding a cup of coffee and you start spinning it around its vertical axis. What happens? The coffee climbs up the walls of the cup, leaving a depression in the center. This isn't just a random splash; nature follows strict mathematical rules, and the surface forms a perfect geometric shape known as a **paraboloid**.

In this classic JEE problem, we are asked to find the exact height difference $h$ between the liquid at the center and the liquid at the edge of a cylindrical vessel of radius $R = 5 	ext{ cm}$, spinning with an angular velocity $\omega$. To solve this elegantly, we need to shift our perspective.

### Stepping into the Rotating Frame

Analyzing the fluid from the ground frame can be tricky because every particle is accelerating towards the center (centripetal acceleration). Instead, let's step *into* the rotating frame of the vessel.

In this rotating frame, the fluid appears completely at rest. This means every tiny fluid particle on the surface is in **dynamic equilibrium**. Let's pick a random fluid particle of mass $m$ at coordinates $(x, y)$ on the surface and draw its Free Body Diagram.

Three forces act on this particle:
1. The downward pull of gravity: $mg$.
2. The outward centrifugal pseudo force: $mx\omega^2$.
3. The normal reaction from the surrounding fluid: $N$, which acts strictly perpendicular to the liquid surface.

### The Geometry of the Surface

Because the particle is in equilibrium, the forces must balance perfectly. Let's say the normal force $N$ makes an angle $	heta$ with the vertical.

Balancing the vertical forces, the upward component of the normal force must support the weight:
$$N \cos	heta = mg$$

Balancing the horizontal forces, the inward component of the normal force must counteract the outward pseudo force:
$$N \sin	heta = mx\omega^2$$

Now, let's divide the horizontal equation by the vertical equation. The normal force $N$ and the mass $m$ beautifully cancel out, leaving us with:
$$	an	heta = \frac{\omega^2 x}{g}$$

Here is where the physics meets the calculus. Geometrically, the normal is perpendicular to the tangent of the surface. This means the angle $	heta$ the normal makes with the vertical is exactly equal to the angle the tangent makes with the horizontal. And we know that the tangent of this angle is simply the slope of the curve, $\frac{dy}{dx}$!

$$\frac{dy}{dx} = \frac{\omega^2 x}{g}$$

### The Final Calculation

We have a simple differential equation. Let's separate the variables and integrate to find the shape of the surface:
$$\int dy = \int \frac{\omega^2 x}{g} dx$$

Integrating both sides, we get the equation of our paraboloid:
$$y = \frac{\omega^2 x^2}{2g}$$

To find the maximum height difference $h$, we look at the edge of the vessel where $x = R = 5 	ext{ cm}$. At this point, the height $y$ is exactly $h$. Substituting these values into our master equation:
$$h = \frac{\omega^2 (5)^2}{2g} = \frac{25\omega^2}{2g}$$

And there we have it! The height difference is perfectly predicted by the interplay of rotation and gravity.

Select Answer:

Visualized Solution

$y = f (x)$

$Dynamic Equilibrium$

$Free Body Diagram$

$\sum F_{y} = 0$

$\sum F_{x} = 0$

$tan θ = \frac{ω ^{2} x}{g}$

$\frac{d y}{d x} = tan θ$

$\frac{d y}{d x} = \frac{ω ^{2} x}{g}$

$y = \frac{ω ^{2} x ^{2}}{2 g}$

$h = \frac{25 ω ^{2}}{2 g}$

$Conclusion$

The Sigma Insight: Fluid Pressure and Pascal's Law

The Spinning Vessel

Stepping into the Rotating Frame

The Geometry of the Surface

The Final Calculation

Similar Questions

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.

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

A vessel containing water is given a constant acceleration $a$ towards the right along a straight horizontal path. Which of the following diagrams represents the surface of the liquid?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=f(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dynamic Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Fy​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Fx​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ=gω2x​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​=tanθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​=gω2x​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2gω2x2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h=2g25ω2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion