---
title: "Upward force on rotating drum top"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:rotating-drum-pipe-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/rotating-drum-pipe-force
site: BCT Engineering
---

# Upward force on rotating drum top

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2075 Chaitra, 8 marks
- Video: [PART 1: Solved Engineering Problem Involving Rotating Cylindrical Vessel (FLUID MECHANICS/MECHANICS)](https://www.youtube.com/watch?v=LcXfYfyRU8g) — Gillesania Engineering Videos
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/rotating-drum-pipe-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

A pipe 25 mm in diameter is connected to the centre of the top of a drum 0.5 m in diameter, the cylindrical axis of the pipe and the drum being vertical. Water is poured into the drum through the pipe until the water level stands in the pipe 0.6 m above the top of the drum. If the drum and pipe are now rotated about their vertical axis at 600 rev/min what will be the upward force exerted on the top of the drum.

## Short answer

Integrating pressure p = ρgh + ½ρω²r² over the annulus from pipe radius to drum radius yields an upward force of 13 265 N ≈ 13.3 kN.

## Answer

**Data:** drum radius $R=0.25$ m, pipe diameter 25 mm so $r_0=0.0125$ m, water level in the pipe $h=0.6$ m above the drum top, $N=600$ rpm so $\omega=\dfrac{2\pi\times600}{60}=62.83$ rad/s, $\rho=1000\ \text{kg/m}^3$.

```text
        |  | <- water, 0.6 m above top
        |  |
    ____|  |____   drum top
   |             |
   |   water     |   0.5 m dia
   |_____________|
```

The drum is full and the free surface lies in the pipe on the axis, so on the underside of the drum top (gauge pressure) at radius $r$:

$$
p=\rho g h+\tfrac{1}{2}\rho\omega^{2}r^{2}
$$

The upward force on the top (annulus from $r_0$ to $R$):

$$
F=\int_{r_0}^{R}p\,2\pi r\,dr=\rho g h\,\pi\left(R^{2}-r_0^{2}\right)+\frac{\pi\rho\omega^{2}}{4}\left(R^{4}-r_0^{4}\right)
$$

**Static part:**

$$
F_1=1000(9.81)(0.6)\,\pi\left(0.25^{2}-0.0125^{2}\right)=1152.8\ \text{N}
$$

**Rotational part:**

$$
F_2=\frac{\pi(1000)(62.83)^{2}}{4}\left(0.25^{4}-0.0125^{4}\right)=12\,111.8\ \text{N}
$$

$$
F=F_1+F_2=13\,264.6\ \text{N}
$$

**Answer: upward force on the drum top $\approx13.3$ kN** (13.27 kN if the small pipe area is ignored). Rotation contributes about 91 % of it.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the gauge pressure on the underside of the top at radius r?
   A. rho g h
   B. 1/2 rho omega^2 r^2
   C. rho g h + 1/2 rho omega^2 r^2
   D. rho g h - 1/2 rho omega^2 r^2
   Answer: C. Static head from the pipe level plus the rotational term add up. (Key point: Pressure on drum top)

2. How is the upward force on the top found?
   A. Pressure at the rim times drum area
   B. Pressure at the centre times area
   C. Weight of water in the drum
   D. Integrate p 2 pi r dr from the pipe radius to the drum radius
   Answer: D. The annulus from r0 to R carries the pressure that varies with r. (Key point: Force integration)

3. What is the static component of force?
   A. About 1153 N
   B. About 13 300 N
   C. About 12 100 N
   D. About 115 N
   Answer: A. F1 = 1000 x 9.81 x 0.6 x pi (0.25^2 - 0.0125^2) = 1152.8 N. (Key point: Static part)

4. Which part dominates the total force of about 13.3 kN?
   A. Static head, 91 %
   B. Rotation, about 91 %
   C. Both equal
   D. Pipe area
   Answer: B. The rotational part is 12.1 kN out of 13.3 kN. (Key point: Dominant part)

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