Problem 22 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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 1: 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: \(Q = \rho * \pi {r_0}^2 * v_0\)

Step 2: 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 \(y\).

Step 3: 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 \(y\): \(y = v_0\frac{v(y) - v_0}{g} + \frac{1}{2}g\left(\frac{v(y) - v_0}{g}\right)^2\).

Step 4: 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 \(y\): \(v(y) = v_0 + \sqrt{2gy}\).

Step 5: 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: \(\rho * \pi {r_0}^2 * v_0 = \rho * \pi {r(y)}^2 * \left(v_0 + \sqrt{2gy}\right)\).

Step 6: 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, \(r(y)\).

Key Concepts

Mass Flow Rate

Mass flow rate is an important concept in fluid mechanics. It's essentially the quantity of mass passing through a given cross-sectional area per unit time. In simpler terms, it tells you how much of a fluid, like water, moves through a space each second. The mass flow rate is calculated by the formula:
\( Q = \rho \times A \times v \)
where:

• \(Q\) is the mass flow rate,
• \(\rho\) is the fluid's density,
• \(A\) is the cross-sectional area the fluid is flowing through, and
• \(v\) is the velocity of the fluid.

For a circular opening, the area \(A\) is calculated as \(\pi r^2\), making it easy to see why the formula includes circles in problems involving faucets or pipes. This formula helps us determine how the speed and size of the opening impact how much water flows out.

Conservation of Mass

In fluid mechanics, the conservation of mass principle is crucial. It states that mass in a closed system remains constant over time. For fluids, this means that if you know how much fluid enters a system, you can determine how much will leave, assuming nothing is added or lost along the way.
In the context of our exercise, the mass flow rate of water at the faucet's opening must remain the same as it falls through the air, even as the stream narrows. Mathematically, it can be expressed as:
\( \rho \times \pi {r_0}^2 \times v_0 = \rho \times \pi {r(y)}^2 \times v(y) \)
This equation reflects how any changes in the stream's velocity \(v(y)\) due to gravity must be balanced by corresponding changes in its cross-sectional area \(\pi r(y)^2\) to ensure mass continuity. By keeping the mass flow rate constant, we ensure that the fluid's behavior is predictable without loss or gain of material.

Equations of Motion

Equations of motion are a set of formulas used in physics to predict how objects move under various forces, particularly due to gravity. When a fluid such as water falls under the influence of gravity, its speed will increase. This acceleration can be modeled using basic kinematic equations.
For our falling water stream, the speed \(v(y)\) at a certain distance \(y\) can be calculated using:

• \( y = v_0 t + \frac{1}{2}gt^2 \), which relates distance to time and acceleration, and
• \( v(y) = v_0 + gt \), which shows how velocity changes with time.

These equations account for gravity’s constant pull, represented by \(g\), facilitating the acceleration from the initial speed \(v_0\). By substituting equations and solving the system, we can find \(v(y) = v_0 + \sqrt{2gy}\), capturing how increased distance due to the effect of gravity results in increased speed, and in turn, a change in the water stream’s radius.