Video Transcript
Water emerges vertically downward
from a faucet that has a 1.600-centimeter diameter, moving at a speed of 0.400
meters per second. Because of the construction of the
faucet, there is no variation in speed across the stream. What is the flow rate of water from
the faucet? What is the diameter of the water
stream at a point 0.200 meters vertically below the faucet? Neglect any effects due to surface
tension.
In this two-part exercise, we want
to solve first for the flow rate of the water coming from the faucet. And then we wanna solve for the
diameter of the water stream a certain distance below the faucet. We’ll call this flow rate 𝑓 sub
𝑟. And the diameter we’ll name 𝑑. Let’s start by drawing a diagram of
the situation. Here we have water coming out of a
faucet, falling vertically downward, where the output diameter of the faucet we’ve
called 𝑑 one. It’s 1.600 centimeters. The water comes out with an initial
speed we’ve called 𝑣 one, 0.400 meters per second. And the first thing we want to
solve for is the flow rate of the water as it leaves the faucet.
We can recall that the volume flow
rate of a liquid is equal to the cross-sectional area of the pipe or tube it moves
through multiplied by its speed. We recall further that the
cross-sectional area of a circle is equal to 𝜋 divided by four times the diameter
of that circle squared. So we can write that 𝑓 sub 𝑟 is
equal to 𝜋 divided by four times 𝑑 one squared times 𝑣 one. When we plug in for these two
values, we keep 𝑑 one in units of centimeters and convert 𝑣 one to units of
centimeters per second. Entering these values on our
calculator, we find that 𝑓 sub 𝑟, to three significant figures, is 80.4 cubic
centimeters per second. That’s the flow rate of the water
from the faucet.
Next, we want to solve for the
diameter of the water stream 𝑑, a distance of 0.200 meters below the faucet
mouth. At first, we might think that 𝑑 is
equal to 𝑑 one, that the diameter of the stream doesn’t change. But we realize that the water, once
it leaves the faucet, is falling under the acceleration of gravity and will speed
up. The continuity equation tells us
that the cross-sectional area along some point of a fluid’s flow times its speed is
equal to the cross-sectional area at another point times its speed there. We can write then that 𝜋 over four
times 𝑑 one squared times 𝑣 one is equal to 𝜋 over four times 𝑑 squared times
the speed of the falling water at that point, which we can call 𝑣 two. We see the factor of 𝜋 over four
cancels from this expression. And if we rearrange it to solve for
𝑑, we find that 𝑑 is equal to 𝑑 one, the given diameter of the faucet, multiplied
by the square root of 𝑣 one divided by 𝑣 two, the speed of the falling water at
𝑑.
Since we’re given 𝑑 one and 𝑣
one, the question now becomes how do we solve for 𝑣 two. Since the water as it falls is
under a constant acceleration, that is, the acceleration due to gravity, the
kinematic equations apply for describing its motion. In particular, we can make use of
the kinematic equation that says final speed squared equals initial speed squared
plus two times acceleration times displacement. In our case, we can write that 𝑣
two squared equals 𝑣 one squared plus two times 𝑔 times ℎ. Where 𝑔, the acceleration due to
gravity, we’ll treat as exactly 9.8 meters per second squared. Now that we have an expression for
𝑣 two, we can substitute this in for 𝑣 two in our equation for 𝑑. And being given 𝑑 one and 𝑣 one,
as well as ℎ, and knowing that 𝑔 is a constant, we’re ready to plug in and solve
for 𝑑. When we do and enter this
expression on our calculator, we find that 𝑑 is 0.712 centimeters. That’s the diameter of the stream
of water after it’s fallen from the faucet a distance of 0.200 meters.
