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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, \(r(y)\).

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