# Partially Filled Cylindrical Tank Rotated at 90 rpm

**Situation**

An open cylindrical vessel 1.3 m in diameter and 2.1 m high is 2/3 full of water. If rotated about the vertical axis at a constant angular speed of 90 rpm,

1. Determine how high is the paraboloid formed of the water surface.

A. 1.26 m | C. 2.46 m |

B. 1.91 m | D. 1.35 m |

2. Determine the amount of water that will be spilled out.

A. 140 L | C. 341 L |

B. 152 L | D. 146 L |

3. What should have been the least height of the vessel so that no water is spilled out?

A. 2.87 m | C. 3.15 m |

B. 2.55 m | D. 2.36 m |

**Answer Key**

Part 2: [ C ]

Part 3: [ D ]

**Solution**

$V_{tank} = \frac{1}{4}\pi D^2 H = \frac{1}{4}\pi (1.3^2)(2.1)$

$V_{tank} = 2.7874 ~ \text{m}^2$

$V_{w1} = \frac{1}{4}\pi D^2 d = \frac{1}{4}\pi (1.3^2)(1.4)$

$V_{w1} = 1.8582 ~ \text{m}^3$

$h = \dfrac{\omega^2 r^2}{2g} = \dfrac{(3\pi)^2 (0.65^2)}{2(9.81)}$

$h = 1.9128 ~ \text{m}$ ← [ B ] *answer for part 1*

$h/2 = 0.9564 ~ \text{m}$

Since *h*/2 > 0.7 m, water are spilled from the tank

$V_{air} = \frac{1}{2}\pi r^2 h = \frac{1}{2}\pi (0.65^2)(1.9128)$

$V_{air} = 1.2694 ~ \text{m}^3$

$V_{w2} = V_{tank} - V_{air} = 2.7874 - 1.2694$

$V_{w2} = 1.518 ~ \text{m}^3$

$V_{spilled} = V_{w1} - V_{w2} = 1.8582 - 1.518$

$V_{spilled} = 0.3402 ~ \text{m}^3 = 340.2 ~ \text{Liters}$ ← [ C ] *answer for part 2*

So that no water is spilled out

$H = 1.4 + h/2 = 1.4 + 0.9564$

$H = 2.3564 ~ \text{m}$ ← [ D ] *answer for part 3*

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