/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 A tube is sealed at both ends an... [FREE SOLUTION] | 91Ó°ÊÓ

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A tube is sealed at both ends and contains a 0.0100-m-long portion of liquid. The length of the tube is large compared to \(0.0100 \mathrm{m}\). There is no air in the tube, and the vapor in the space above the liquid may be ignored. The tube is whirled around in a horizontal circle at a constant angular speed. The axis of rotation passes through one end of the tube, and during the motion, the liquid collects at the other end. The pressure experienced by the liquid is the same as it would experience at the bottom of the tube, if the tube were completely filled with liquid and allowed to hang vertically. Find the angular speed (in \(\mathrm{rad} / \mathrm{s}\) ) of the tube.

Short Answer

Expert verified
The angular speed is 31.3 rad/s.

Step by step solution

01

Understand the Problem

We have a closed tube with a 0.0100 m portion of liquid and no air. It is rotated in a circle with the rotation axis at one end of the tube. We need to find the angular speed at which the pressure at the liquid matches that at the bottom of the tube if it were vertically filled.
02

Define Variables and Known Values

Let the length of the liquid portion be \(L = 0.0100 \text{ m}\) and the density of the liquid be \(\rho\). The force involved here is the centrifugal force due to rotation.
03

Express Centrifugal Force

The centrifugal force experienced by the liquid when rotating is \(F = m r \omega^2\), where \(m\) is mass, \(r\) is the radius from rotation, and \(\omega\) is the angular speed. Here, \(r=L\) since the liquid collects at the end of the tube.
04

Relate Pressure to Centrifugal Force

The pressure from the centrifugal force is \(P_c = \rho g L\), equivalent to the pressure when upright, \(\Delta P = \rho g h\). Hence, \(\rho g h = \rho L \omega^2\).
05

Solve for Angular Speed

Cancel \(\rho\) and \(L\) from both sides of the equation: \(g = L \omega^2\), so \(\omega = \sqrt{\frac{g}{L}}\). Substitute \(g = 9.81 \text{ m/s}^2\) and \(L = 0.0100 \text{ m}\) to get \(\omega = \sqrt{\frac{9.81}{0.0100}} = 31.3 \mathrm{rad/s}\).

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Key Concepts

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

Centrifugal force in a rotating system
When an object is rotated in a circular path, it experiences a force that seems to push it away from the center of the circle. This is known as centrifugal force. Though not a true force in the Newtonian sense, it is a useful concept in understanding the effects of rotation from the perspective inside the rotating frame.

This force arises from the inertia of the object wanting to move in a straight line but being forced into a curved path. When dealing with a rotating tube, as in the exercise, the liquid is pushed outward to the end of the tube. This is because of the centrifugal force acting on it due to the rotation of the tube around its axis.

To calculate centrifugal force, we use the formula:
  • \(F = m \cdot r \cdot \omega^2\)
Here, \(m\) is the mass of the liquid, \(r\) is the radius (or length of the tube where the liquid is collected), and \(\omega\) is the angular speed. This formula helps us understand how fast the tube needs to be rotated for the liquid's pressure to equate to its pressure if it were vertically oriented.
Understanding angular speed
Angular speed refers to how fast something is rotating within a specific angle over time. In simple terms, it tells us how quickly an object is spinning or turning around its axis.

Angular speed is measured in radians per second (rad/s), which represents the angle traversed in a unit of time. In the context of the exercise, it is crucial to calculate the right angular speed to equate the centrifugal pressure with the pressure that would exist in a vertically suspended liquid column.The equation used is:
  • \(\omega = \sqrt{\frac{g}{L}}\)
Where \(g\) is the acceleration due to gravity (9.81 m/s\(^2\)), and \(L\) is the length of the liquid column (0.0100 m). By solving this, you arrive at the angular speed required for experiment conditions. It's a key part of finding out how forces distribute in a rotating system to maintain equilibrium.
Role of pressure in rotating systems
Pressure is a measure of force distribution across an area. In physics, it is often considered within the context of fluids, including liquids and gases.

In a rotating system, understanding pressure becomes important when determining how the forces exerted impact fluid distribution. In the given exercise, the pressure experienced by the liquid should equal that when the liquid is in a vertical column. This allows us to use concepts of pressure equilibrium to solve for the necessary angular speed.The centrifugal pressure is given by:
  • \(P_c = \rho \cdot g \cdot L\)
This formula ties the mass density of the fluid (\(\rho\)), gravitational acceleration (\(g\)), and the length of the liquid column (\(L\)). Maintaining this pressure ensures that rotation gives the same effects as gravity on a vertical column, balancing forces just right so that both situations are equivalent in terms of pressure.

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Most popular questions from this chapter

A small crack occurs at the base of a 15.0 -m-high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is \(15.2\) \(\mathrm{N}\). When completely submerged in water, its apparent weight is \(13.7\) \(\mathrm{N}\). What is the volume of the object?

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When an object moves through a fluid, the fluid exerts a viscous force \(\overrightarrow{\mathbf{F}}\) on the object that tends to slow it down. For a small sphere of radius \(R,\) moving slowly with a speed \(v,\) the magnitude of the viscous force is given by Stokes' law, \(F=6 \pi \eta R v,\) where \(\eta\) is the viscosity of the fluid. (a) What is the viscous force on a sphere of radius \(R=5.0 \times 10^{-4} \mathrm{m}\) that is falling through water \(\left(\eta=1.00 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) when the sphere has a speed of \(3.0 \mathrm{m} / \mathrm{s} ?\) (b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of \(1.0 \times 10^{-5} \mathrm{kg},\) what is its terminal speed?

A cylindrical storage tank has a radius of \(1.22\) \(\mathrm{m}\). When filled to a height of \(3.71\) \(\mathrm{m},\) it holds \(14300\) \(\mathrm{kg}\) of a liquid industrial solvent. What is the density of the solvent?

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