Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 22

Water flows from a circular faucet opening of radius \(r_{0}\) directed vertically downward, at speed \(v_{0}\). As the stream of water falls, it narrows. Find an expression for the radius of the stream as a function of distance fallen, \(r(y),\) where \(y\) is measured downward from the opening. Neglect the eventual breakup of the stream into droplets, and any resistance due to drag or viscosity.

Short Answer

Expert verified
Answer: The expression for the radius of the water stream as a function of distance fallen is \(r(y) = r_0 * \sqrt{\frac{v_0}{v_0 + \sqrt{2gy}}}\).

Step by step solution

01

Determine the mass flow rate at the faucet opening

Using the given initial conditions, we can calculate the mass flow rate at the faucet opening, which is given by the product of the cross-sectional area, the velocity, and the density of the water: \(Q = \rho * A_0 * v_0\) Here, \(Q\) is the mass flow rate, \(\rho\) is the density of water (assumed constant), \(A_0\) is the initial cross-sectional area, and \(v_0\) is the initial speed. Since the faucet opening is circular, its area can be expressed as \(A_0 = \pi {r_0}^2\). Therefore, the initial mass flow rate is: \(Q = \rho * \pi {r_0}^2 * v_0\)
02

Conservation of mass flow rate

Due to the conservation of mass and neglecting the eventual breakup of the stream and any resistance due to drag or viscosity, the mass flow rate must remain constant along the falling water stream. Hence, at any distance \(y\), we can write: \(\rho * \pi {r_0}^2 * v_0 = \rho * \pi {r(y)}^2 * v(y)\) Here, \(r(y)\) is the radius of the water stream as a function of the distance fallen, and \(v(y)\) is the speed of the water at distance \(y\).
03

Find the relationship between velocities \(v_0\) and \(v(y)\)

The water accelerates due to gravity as it falls, which can be described by: \(y = v_0t + \frac{1}{2}gt^2\), where \(t\) is the time it takes for the water to fall this distance and \(g\) is the acceleration due to gravity. We can also write the speed \(v(y)\) in terms of the initial velocity and gravity: \(v(y) = v_0 + gt\). Now, we need to eliminate the variable \(t\). To do this, we can solve the first equation for \(t\): \(t = \frac{v(y) - v_0}{g}\), and substitute this back into the equation for \(y\): \(y = v_0\frac{v(y) - v_0}{g} + \frac{1}{2}g\left(\frac{v(y) - v_0}{g}\right)^2\).
04

Solve for \(v(y)\) in terms of \(y\)

From the final equation in step 3 above, we can solve for \(v(y)\) as a function of \(y\): \(v(y) = v_0 + \sqrt{2gy}\).
05

Substitute \(v(y)\) back into the conservation of mass flow rate equation

We can now substitute the expression for \(v(y)\) back into the conservation of mass flow rate equation from step 2: \(\rho * \pi {r_0}^2 * v_0 = \rho * \pi {r(y)}^2 * \left(v_0 + \sqrt{2gy}\right)\).
06

Solve for \(r(y)\)

Finally, we can solve for the radius of the water stream as a function of the distance fallen: \({r(y)}^2 = {r_0}^2 * \frac{v_0}{v_0 + \sqrt{2gy}}\) \(r(y) = r_0 * \sqrt{\frac{v_0}{v_0 + \sqrt{2gy}}}\). This is the expression for the radius of the water stream as a function of distance fallen, \(r(y)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block of cherry wood that is \(20.0 \mathrm{~cm}\) long, \(10.0 \mathrm{~cm}\) wide, and \(2.00 \mathrm{~cm}\) thick has a density of \(800 . \mathrm{kg} / \mathrm{m}^{3}\). What is the volume of a piece of iron that, if glued to the bottom of the block makes the block float in water with its top just at the surface of the water? The density of iron is \(7860 \mathrm{~kg} / \mathrm{m}^{3},\) and the density of water is \(1000 . \mathrm{kg} / \mathrm{m}^{3}\).

Donald Duck and his nephews manage to sink Uncle Scrooge's yacht \((m=4500 \mathrm{~kg}),\) which is made of steel \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\right)\). In typical comic-book fashion, they decide to raise the yacht by filling it with ping-pong balls. A pingpong ball has a mass of \(2.7 \mathrm{~g}\) and a volume of \(3.35 \cdot 10^{-5} \mathrm{~m}^{3}\) a) What is the buoyant force on one ping-pong ball in water? b) How many balls are required to float the ship?

Blood pressure is usually reported in millimeters of mercury (mmHg) or the height of a column of mercury producing the same pressure value. Typical values for an adult human are \(130 / 80\); the first value is the systolic pressure, during the contraction of the ventricles of the heart, and the second is the diastolic pressure, during the contraction of the auricles of the heart. The head of an adult male giraffe is \(6.0 \mathrm{~m}\) above the ground; the giraffe's heart is \(2.0 \mathrm{~m}\) above the ground. What is the minimum systolic pressure (in \(\mathrm{mmHg}\) ) required at the heart to drive blood to the head (neglect the additional pressure required to overcome the effects of viscosity)? The density of giraffe blood is \(1.00 \mathrm{~g} / \mathrm{cm}^{3},\) and that of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\)

The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(945 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan ) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

In many problems involving application of Newton's Second Law to the motion of solid objects, friction is neglected for the sake of making the solution easier. The counterpart of friction between solids is viscosity of liquids. Do problems involving fluid flow become simpler if viscosity is neglected? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Energy Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Wave Optics

Read Explanation

Kinematics Physics

Read Explanation

Waves Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free