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Chapter 6: Problem 3

Length of horizontal arm of a uniform cross section \(U\)-tube is \(l=21 \mathrm{~cm}\) and ends of both of the vertical arms are open to surrounding of pressure \(10 \times 500 \mathrm{~N} / \mathrm{m}^{2}\). A liquid of density \(\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) is poured into the tube such that liquid just fillg is sealed and the tube is then rotated about a vertical axiy arm with angular val \(\omega_{0}=10 \mathrm{rad} / \mathrm{s} .\) If length of each vertical arm is \(a=6 \mathrm{~cm}\) calculate the length of air column in the sealed arm. (in cm).

Short Answer

Expert verified
The length of the air column in the sealed arm is approximately 11.53 cm.

Step by step solution

01

Understand the Given Problem

We have a U-tube with horizontal arm length of \(l = 21\ \mathrm{cm}\) and vertical arm lengths of \(a = 6\ \mathrm{cm}\). The pressure is initially \(10 \times 500\ \mathrm{N/m}^2\) with a liquid density \(\rho = 10^3\ \mathrm{kg/m}^3\). The wave number is \(\omega_0 = 10\ \mathrm{rad/s}\). We need to determine the air column length when the system is rotated.
02

Calculate the Rotation's Effect on Pressure

When the U-tube is rotated, centrifugal force affects the pressure distribution within the horizontal arm. The pressure difference due to rotation can be calculated using \(\Delta P = \rho \frac{\omega_0^2 l^2}{8}\). Substitute the given values: \(\rho = 1000\ \mathrm{kg/m}^3\), \(\omega_0 = 10\ \mathrm{rad/s}\), and \(l = 0.21\ \mathrm{m}\) to get \(\Delta P = \frac{1000 \times 10^2 \times 0.21^2}{8}\).
03

Solve for Pressure Difference

Calculate the pressure difference using the formula: \(\Delta P = \frac{1000 \times 100 \times 0.0441}{8} = \frac{4410}{8} = 551.25\ \mathrm{N/m}^2\). This is the extra pressure due to rotation.
04

Consider the Effect on the Air Column

The initial pressure of the air in the sealed arm is atmospheric, and the additional pressure due to rotation will compress this air column. The pressure balance gives: \(P_0 + \rho gh + \Delta P = P_0 + \rho gh'\), where \(h'\) is the new air column length, and \(h = 0.06\ \mathrm{m}\).
05

Calculate the New Air Column Length

Rearrange the equation for \(h'\): \(h' = h + \frac{\Delta P}{\rho g}\). Substitute \(\Delta P = 551.25\ \mathrm{N/m}^2\), \(\rho = 1000\ \mathrm{kg/m}^3\), and \(g = 9.8\ \mathrm{m/s}^2\) to find \(h' = 0.06 + \frac{551.25}{1000 \times 9.8}\).
06

Calculate and Convert to Centimeters

Perform the calculation: \(h' = 0.06 + \frac{551.25}{9800} \approx 0.1153\ \mathrm{m}\). Convert this to centimeters: \(h' \approx 11.53\ \mathrm{cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Difference
Pressure difference is a core concept in fluid mechanics. It refers to the variation in pressure across different points within a fluid. This difference arises due to various factors including height, velocity changes, or applied forces.
In our scenario, the rotation of the U-tube creates a pressure variance along its horizontal arm. Let's break it down:
  • Initial State: The air and liquid in the U-tube are initially at atmospheric pressure.
  • During Rotation: As the U-tube rotates, centrifugal force acts on the liquid, causing a pressure gradient along the length of the tube.
  • Pressure Calculation: Here, the formula for pressure difference due to rotation is ewline ewline ewline \[ \Delta P = \rho \frac{\omega_0^2 l^2}{8} \]ewline which considers the density (\(\rho\)), angular velocity (\(\omega_0\)), and length of the horizontal arm (\(l\)).
  • Impact on Fluid: This pressure change causes a shift in the fluid's position, affecting the air column in one of the arms.
Understanding this concept helps in solving problems related to dynamic systems like rotating U-tubes and other applications in fluid dynamics.
Centrifugal Force
Centrifugal force is an apparent force experienced by objects moving in a circular path, directed away from the center of rotation. It's the force that makes you feel like you're pushed outward when you take a sharp turn in a car.
In the context of U-tube mechanics:
  • Origin: When the U-tube rotates about a vertical axis, each part of the liquid spins in a circular trajectory.
  • Effect on Fluid: The centrifugal force acts on the liquid, forcing it outward - towards the ends of the U-tube.
  • Pressure Result: This movement intensifies the pressure at the outer ends of the horizontal arm, as seen in the large value calculated in the original solution.
  • Calculation Formula: Directly influences pressure through the relation:ewline ewline \[ F_c = m \cdot r \cdot \omega^2 \]ewline where \(m\) is mass, \(r\) is radius, and \(\omega\) is angular velocity.
Centrifugal force is key to understanding how fluids and gases respond to rotation, crucial for U-tube applications and other rotational dynamics scenarios.
U-tube Mechanics
The U-tube is a fascinating device used to demonstrate principles of fluid mechanics. It has a U-shape, with one side filled, allowing observation of the pressure and displacement effects.
Let's see how the U-tube works:
  • Basic Structure: A typical U-tube has two vertical arms connected by a horizontal section. Liquids fill it to specific levels dictated by external pressures and forces.
  • In this Exercise: The U-tube is filled with liquid, and upon rotation about a vertical axis, liquid displaces due to changes in pressure. The displacement of the liquid also affects the air column within the tube.
  • Pressure Impacts: The rotation causes differing pressures on both ends of the horizontal arm, impacting fluid height and air column length.
  • Applications: U-tubes study fluid flow, pressure, and reaction to external forces under controlled environments.
Studying U-tube mechanics aids in understanding how liquids behave under different dynamic conditions, showcasing vital principles of pressure and displacement.
Rotational Dynamics
Rotational dynamics involves analyzing rotations and the forces that influence them. This aspect of physics is crucial in mastering fluid movement under rotational conditions.
Specifics related to our U-tube exercise include:
  • Rotation Definition: When a body rotates, each part of the object follows a circular path around an axis.
  • Angular Velocity: Denoted by \(\omega\), it's a measure of how quickly an object spins and is crucial in calculating centrifugal effects as seen in \ \[ \Delta P = \rho \frac{\omega^2 l^2}{8} \]\.
  • Effects on U-tube: When the tube rotates, centrifugal force changes the fluid dynamics, impacting pressure, liquid distribution, and the air column length.
  • Application: Understanding these principles is vital in fields like mechanical engineering, where rotational systems are customary, and their fluid interactions must be modeled accurately.
Rotational dynamics sheds light on the behavior of systems in motion, elucidating key principles of balanced and unbalanced forces within rotating bodies, like our U-tube.

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Most popular questions from this chapter

A hemispherical bowl just floats without sinking in a of density \(1.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). If the outer diameter a density of the bowl are \(1 \mathrm{~m}\) and \(2 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\), respectively, then the inner diameter of the bowl will be (1) \(0.94 \mathrm{~m}\) (2) \(0.97 \mathrm{~m}\) (3) \(0.98 \mathrm{~m}\) (4) \(0.99 \mathrm{~m}\)

A uniform \(\operatorname{rod} A B, 12 \mathrm{~m}\) long weighing \(24 \mathrm{~kg}\), is supported at end \(B\) by a flexible light string and a lead weight (of very small size) of \(12 \mathrm{~kg}\) attached at end \(A\). The rod floats in water with one-half of its length submerged. Find the volume of the rod (in \(\times 10^{-3} \mathrm{~m}^{3}\) ) [Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), density of water \(\left.=1000 \mathrm{~kg} / \mathrm{m}^{3}\right]\)

An empty balloon weighs \(W_{1}\). If air equal in weight to \(W\) is pumped into the balloon, the weight of the balloon becomes \(W_{2}\). Suppose that the density of air inside and outside the balloon is the same. Then (1) \(W_{2}=W_{1}+W\) (2) \(W_{2}=\sqrt{W_{1} W}\) (3) \(W_{2}=W_{1}\) (4) \(W_{2}=W_{1}-W\)

Equal volumes of a liquid are poured in the three vessels \(A\) \(B\) and \(C\left(h_{1}

Two unequal blocks placed over each other of different densities \(\sigma_{1}\) and \(\sigma_{2}\) are immersed in a fluid of density of \(\sigma\). The block of density \(\sigma_{1}\) is fully submerged and the block of density \(\sigma_{2}\) is partly submerged so that ratio of their masses is \(1 / 2\) and \(\sigma / \sigma_{1}=2\) and \(\sigma / \sigma_{2}=0.5\). Find the degree of submergence of the upper block of density \(\sigma_{2}\). (1) \(50 \%\) submerged (2) \(25 \%\) submerged (3) \(75 \%\) submerged (4) Fully submerged
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