An open circular cylinder of 1m diameter and 2m depth is completely filled with water and rotated about its axis about 45 rpm. Determine the depth at the axis and amount of water spilled. Also find the speed of rotation at which the central axial depth is zero.

Rotating Open Cylinder – Free Surface & Spillage Analysis

Problem Statement

An open circular cylinder (tank) with a 1 m diameter and 2 m depth is completely filled with water and rotated about its vertical axis at 45 rpm.

(a) Determine the depth at the axis and the amount of water spilled.

(b) Find the speed of rotation (in rpm) at which the central (axial) depth becomes zero.

Solution

Given:

Diameter = 1 m ⇒ Radius, $ r = 0.5\,\text{m} $
Depth, $ H = 2\,\text{m} $
Rotation speed, $ N = 45\,\text{rpm} $

The angular velocity is calculated as:

$ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 45}{60} \approx 4.712\,\text{rad/s} $
Part (a): Depth at the Axis & Water Spilled

For a rotating liquid, the free surface becomes a paraboloid. The depression at the axis is given by:

$ z = \frac{r^2\,\omega^2}{2g} $

Substituting $ r = 0.5\,\text{m} $, $ \omega \approx 4.712\,\text{rad/s} $, and $ g = 9.81\,\text{m/s}^2 $:

$ z = \frac{(0.5)^2 \times (4.712)^2}{2 \times 9.81} \approx 0.283\,\text{m} $

Thus, the water depth at the axis is:

Depth at axis $ = H – z \approx 2 – 0.283 \approx 1.717\,\text{m} $

The volume of water spilled is the volume of the parabolic segment that overflows:

$ V = \frac{1}{2}\,\pi\,r^2\,z = \frac{1}{2}\,\pi\,(0.5)^2 \times 0.283 \approx 0.111\,\text{m}^3 $
Part (b): Rotation Speed for Zero Central Depth

For the central depth to become zero, the free surface must be depressed by 2 m (i.e. $ z = 2\,\text{m} $). Setting:

$ 2 = \frac{r^2\,\omega^2}{2g} $

Solve for $ \omega $:

$ \omega^2 = \frac{2g \times 2}{r^2} = \frac{4 \times 9.81}{0.25} = 156.96 \quad\Longrightarrow\quad \omega \approx 12.53\,\text{rad/s} $

Converting $ \omega $ to rpm:

$ N = \frac{60\,\omega}{2\pi} = \frac{60 \times 12.53}{2\pi} \approx 120\,\text{rpm} $

Figure: Free Surface in a Rotating Open Cylinder

The figure shows a vertical cross‑section of the open cylindrical tank (scale: 1 m = 100 px). The container, with a 1 m diameter and 2 m depth, is drawn with its top rim at y = 50 and bottom at y = 250. Under rotation at 45 rpm (i.e. $ \omega \approx 4.712\,\text{rad/s} $), the free surface becomes parabolic—touching the rim at the edges while being depressed by $ 0.283\,\text{m} $ at the center. This yields a central water depth of approximately $ 1.717\,\text{m} $ and results in exactly 0.111 m³ of water being spilled.

Explanation

When the cylinder is rotated, centrifugal forces drive the water outward, causing the free surface to form a paraboloid. The depression at the center is given by $$ z = \frac{r^2\,\omega^2}{2g} $$ where $ z $ is the depression relative to the container rim. At 45 rpm, $ z \approx 0.283\,\text{m} $ so the water level at the center is $ 2 – 0.283 = 1.717\,\text{m} $ deep. The spilled volume, calculated as $$ V = \frac{1}{2}\,\pi\,r^2\,z, $$ comes out to be exactly 0.111 m³.

Physical Meaning

This problem demonstrates the impact of rotation on the free surface of a liquid in an open container. As the container rotates, centrifugal force pushes the liquid outward, causing the free surface to form a paraboloid. When the rotation rate is high enough, the depression at the center becomes significant, and excess liquid spills over the rim. The analysis shows that at 45 rpm, the center is depressed by 0.283 m, yielding a central depth of 1.717 m and spilling exactly 0.111 m³ of water. The calculation in part (b) shows that increasing the rotation speed to about 120 rpm would completely depress the central surface to the bottom.