Animated Solution for Physics - Properties of Solids and Liquids: A long cylindrical vessel is half-filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is $5\text{ cm}$ and its rotational speed is $2$ rotations per second, then the difference in the heights between the centre and the sides (in cm) will be

Have you ever stirred a cup of coffee really fast and noticed how the liquid climbs up the walls of the cup, leaving a dip in the center? This everyday phenomenon is a beautiful demonstration of fluid dynamics in a rotating frame of reference. When a cylindrical vessel containing a liquid is rotated about its vertical axis, the liquid doesn't just stay flat. It redistributes itself, forming a curved surface known as a paraboloid.

But why does this happen? Imagine you are a tiny observer sitting on a fluid particle inside this rotating vessel. From your perspective, you would feel a mysterious force pushing you radially outwards, away from the center of rotation. This is the centrifugal pseudo-force. To maintain equilibrium and prevent the fluid from spilling out indefinitely, the liquid surface must tilt. By rising at the edges, the fluid creates an inward pressure gradient (due to gravity) that perfectly balances this outward centrifugal push.

To translate this physical intuition into mathematics, we set up a coordinate system. Let's place our origin $(0,0)$ exactly at the lowest point of the curved liquid surface. The y-axis points vertically upwards, and the x-axis points radially outwards. Now, consider any arbitrary point $P(x,y)$ on this free surface. Because this point is in direct contact with the atmosphere, the pressure there must simply be the atmospheric pressure, $p_0$.

Now, let's calculate the pressure at point $P$ by navigating through the fluid from our origin. As we move horizontally from the center to a distance $x$, the rotation increases the pressure by $\frac{1}{2}\rho {\omega }^2{x}^2$. Then, as we move vertically upwards by a height $y$, the pressure drops by $\rho gy$ due to the loss of fluid column above us. Therefore, the total pressure at $P$ is given by the equation:

Since we already established that $P$ is on the free surface, we can set $p_P$ equal to $p_0$.

The atmospheric pressure $p_0$ beautifully cancels out from both sides. Rearranging the remaining terms to solve for $y$, we get the master equation for the surface profile:

Notice that $y$ is proportional to $x^2$. This confirms that the cross-section of the surface is indeed a parabola!

The question asks for the difference in height between the center and the sides of the vessel. In our coordinate system, the center is at $x=0$ (where $y=0$), and the side is at $x=r$, where $r$ is the radius of the vessel. Substituting $x=r$ into our master equation gives us the maximum height difference, $h$:

Now, it's time to crunch the numbers. We are given the radius $r=5\text{ cm}=0.05\text{ m}$. The rotational speed is given as $2$ rotations per second. To use our formula, we must convert this to angular velocity $\omega$ in radians per second. Since one rotation is $2\pi$ radians, $\omega =2\times 2\pi =4\pi \text{ rad/s}$. We also take the acceleration due to gravity $g=10{\text{ m/s}}^2$.

Substituting these values into our height equation:

In physics problems, it is a standard approximation to take ${\pi }^2\approx 10$. Using this, the numerator simplifies to $16\times 10\times 0.0025=0.4$.

Finally, converting this back to centimeters, we get $h=2\text{ cm}$. The liquid rises exactly 2 centimeters higher at the walls compared to the center. This elegant result shows how rotational kinematics and fluid statics intertwine to create predictable, geometric patterns in nature.